Chapter 9 Differential Equations

Part (i)

The given differential equation is $\frac{dy}{dx} - \cos x = 0$.

The highest order derivative present in the equation is $\frac{dy}{dx}$.

The order of this derivative is 1.

The equation can be written as a polynomial in the derivatives. The power of the highest order derivative ($\frac{dy}{dx}$) is 1.

Therefore, the degree of the differential equation is 1.

Order: 1

Degree: 1

Part (ii)

The given differential equation is $xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0$.

The derivatives present in the equation are $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

The highest order derivative present is $\frac{d^2y}{dx^2}$.

The order of this derivative is 2.

The equation is a polynomial equation in the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

The highest order derivative is $\frac{d^2y}{dx^2}$, and its power in the equation is 1 (in the term $xy \frac{d^2y}{dx^2}$).

Therefore, the degree of the differential equation is 1.

Order: 2

Degree: 1

Part (iii)

The given differential equation is $y’’’ + y^2 + e^{y’} = 0$.

This can be written using Leibniz notation as $\frac{d^3y}{dx^3} + y^2 + e^{\frac{dy}{dx}} = 0$.

The derivatives present are $y'''$ (or $\frac{d^3y}{dx^3}$) and $y'$ (or $\frac{dy}{dx}$).

The highest order derivative is $y'''$ (or $\frac{d^3y}{dx^3}$).

The order of this derivative is 3.

The equation contains the term $e^{y'}$, which involves the derivative $y'$ as the exponent of the exponential function. This means the equation is not a polynomial equation in the derivatives ($y'$ or $y'''$).

Therefore, the degree of the differential equation is not defined.

Order: 3

Degree: Not defined

Given:

The differential equation: $\frac{d^4y}{dx^4} + \sin (y’’’) = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $\frac{d^4y}{dx^4} + \sin \left( \frac{d^3y}{dx^3} \right) = 0$.

The derivatives present in the equation are $\frac{d^4y}{dx^4}$ and $\frac{d^3y}{dx^3}$.

The highest order derivative in the equation is $\frac{d^4y}{dx^4}$.

The order of the differential equation is the order of the highest order derivative present.

Thus, the Order of the differential equation is 4.

To determine the degree, we need to check if the equation is a polynomial in the derivatives. The equation contains the term $\sin \left( \frac{d^3y}{dx^3} \right)$. This term is a function of a derivative ($\frac{d^3y}{dx^3}$) and is not a polynomial in the derivative.

Since the equation is not a polynomial equation in the derivatives, the degree is not defined.

Thus, the Degree of the differential equation is not defined.

Conclusion:

Order = 4

Degree = Not defined

Given:

The differential equation: $y' + 5y = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $y' + 5y = 0$.

Using Leibniz notation, this can be written as $\frac{dy}{dx} + 5y = 0$.

The only derivative present in the equation is $y'$ (or $\frac{dy}{dx}$).

The highest order derivative is $y'$, which has an order of 1.

Thus, the Order of the differential equation is 1.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation involves the derivative $y'$ raised to the power of 1, and no transcendental functions of derivatives are present. Thus, the equation is a polynomial in the derivative $y'$.

The power of the highest order derivative ($y'$) in this polynomial form is 1.

Therefore, the Degree of the differential equation is 1.

Conclusion:

Order = 1

Degree = 1

Given:

The differential equation: $\left( \frac{ds}{dt} \right)^4 + 3s \frac{d^2s}{dt^2} = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $\left( \frac{ds}{dt} \right)^4 + 3s \frac{d^2s}{dt^2} = 0$.

The derivatives present in the equation are $\frac{ds}{dt}$ and $\frac{d^2s}{dt^2}$.

The highest order derivative present is $\frac{d^2s}{dt^2}$.

The order of the differential equation is the order of the highest order derivative.

The order of $\frac{d^2s}{dt^2}$ is 2.

Thus, the Order of the differential equation is 2.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation is a sum of terms involving derivatives raised to integer powers. The term $\left( \frac{ds}{dt} \right)^4$ involves the first derivative raised to the power 4. The term $3s \frac{d^2s}{dt^2}$ involves the second derivative raised to the power 1. The equation can be considered a polynomial in $\frac{ds}{dt}$ and $\frac{d^2s}{dt^2}$.

The degree is the highest power of the highest order derivative when the equation is written as a polynomial in derivatives.

The highest order derivative is $\frac{d^2s}{dt^2}$, and its power in the equation is 1.

Therefore, the Degree of the differential equation is 1.

Conclusion:

Order = 2

Degree = 1

Given:

The differential equation: $\left( \frac{d^2y}{dx^2} \right)^2 + \cos \left( \frac{dx}{dy} \right) = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $\left( \frac{d^2y}{dx^2} \right)^2 + \cos \left( \frac{dx}{dy} \right) = 0$.

The derivatives present in the equation are $\frac{d^2y}{dx^2}$ and $\frac{dx}{dy}$.

The term $\frac{dx}{dy}$ is the reciprocal of $\frac{dy}{dx}$. However, standard differential equations are expressed in terms of derivatives of the dependent variable (usually $y$) with respect to the independent variable (usually $x$).

Regardless of how $\frac{dx}{dy}$ is related to $\frac{dy}{dx}$, it is a derivative. The orders of the derivatives present are 2 (for $\frac{d^2y}{dx^2}$) and 1 (for $\frac{dy}{dx}$, if we consider $\frac{dx}{dy}$ as related to the first derivative). The highest order derivative explicitly written is $\frac{d^2y}{dx^2}$.

The order of the differential equation is the order of the highest order derivative present.

Thus, the Order of the differential equation is 2.

To determine the degree, we need to check if the equation is a polynomial equation in the derivatives. The equation contains the term $\cos \left( \frac{dx}{dy} \right)$. This term is a transcendental function (cosine) of a derivative.

For the degree of a differential equation to be defined, the equation must be expressible as a polynomial in all the derivatives. Since the term $\cos \left( \frac{dx}{dy} \right)$ is not a polynomial in the derivative $\frac{dx}{dy}$ (or $\frac{dy}{dx}$), the equation is not a polynomial in derivatives.

Therefore, the degree of the differential equation is not defined.

Thus, the Degree of the differential equation is not defined.

Conclusion:

Order = 2

Degree = Not defined

Given:

The differential equation: $\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$.

This can be rearranged as $\frac{d^2y}{dx^2} - \cos 3x - \sin 3x = 0$.

The only derivative present in the equation is $\frac{d^2y}{dx^2}$.

The highest order derivative in the equation is $\frac{d^2y}{dx^2}$.

The order of the differential equation is the order of the highest order derivative present.

The order of $\frac{d^2y}{dx^2}$ is 2.

Thus, the Order of the differential equation is 2.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation involves the derivative $\frac{d^2y}{dx^2}$ raised to the power 1. The terms $\cos 3x$ and $\sin 3x$ are functions of the independent variable $x$, not derivatives. The equation is a polynomial in the derivative $\frac{d^2y}{dx^2}$.

The degree is the highest power of the highest order derivative when the equation is written as a polynomial in derivatives.

The highest order derivative is $\frac{d^2y}{dx^2}$, and its power in the equation is 1.

Therefore, the Degree of the differential equation is 1.

Conclusion:

Order = 2

Degree = 1

Given:

The differential equation: $(y’’’)^2 + (y’’)^3 + (y’)^4 + y^5 = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $(y’’’)^2 + (y’’)^3 + (y’)^4 + y^5 = 0$.

Using Leibniz notation, this can be written as $\left( \frac{d^3y}{dx^3} \right)^2 + \left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^4 + y^5 = 0$.

The derivatives present in the equation are $y'$ (or $\frac{dy}{dx}$), $y''$ (or $\frac{d^2y}{dx^2}$), and $y'''$ (or $\frac{d^3y}{dx^3}$).

The orders of these derivatives are 1, 2, and 3, respectively.

The highest order derivative present in the equation is $y'''$ (or $\frac{d^3y}{dx^3}$).

The order of the differential equation is the order of the highest order derivative present.

Thus, the Order of the differential equation is 3.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation involves the derivatives $y'$, $y''$, and $y'''$ raised to integer powers, and no non-polynomial functions of derivatives are present. Therefore, the equation is a polynomial in the derivatives.

The degree is the highest power of the highest order derivative when the equation is written as a polynomial in derivatives.

The highest order derivative is $y'''$. The term containing the highest power of $y'''$ is $(y''')^2$.

The power of $y'''$ in this term is 2.

Therefore, the Degree of the differential equation is 2.

Conclusion:

Order = 3

Degree = 2

Given:

The differential equation: $y’’’ + 2y’’ + y’ = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $y’’’ + 2y’’ + y’ = 0$.

Using Leibniz notation, this can be written as $\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$.

The derivatives present in the equation are $y'$ (or $\frac{dy}{dx}$), $y''$ (or $\frac{d^2y}{dx^2}$), and $y'''$ (or $\frac{d^3y}{dx^3}$).

The orders of these derivatives are 1, 2, and 3, respectively.

The highest order derivative present in the equation is $y'''$ (or $\frac{d^3y}{dx^3}$).

The order of the differential equation is the order of the highest order derivative present.

Thus, the Order of the differential equation is 3.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation is a sum of terms where each term is a constant times a derivative raised to an integer power (in this case, $y'''$ to power 1, $y''$ to power 1, and $y'$ to power 1). No non-polynomial functions of derivatives are present. Therefore, the equation is a polynomial in the derivatives.

The degree is the highest power of the highest order derivative when the equation is written as a polynomial in derivatives.

The highest order derivative is $y'''$. The term containing $y'''$ is $y'''$, and its power is 1.

Therefore, the Degree of the differential equation is 1.

Conclusion:

Order = 3

Degree = 1

Given:

The differential equation: $y’ + y = e^x$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $y’ + y = e^x$.

Using Leibniz notation, this can be written as $\frac{dy}{dx} + y = e^x$.

The only derivative present in the equation is $y'$ (or $\frac{dy}{dx}$).

The highest order derivative is $y'$, which has an order of 1.

Thus, the Order of the differential equation is 1.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation can be written as $y' + y - e^x = 0$. It involves the derivative $y'$ raised to the power 1. The terms $y$ (which is the dependent variable) and $e^x$ (which is a function of the independent variable $x$) do not involve any derivatives in a way that would make the equation non-polynomial in derivatives. The equation is a polynomial in the derivative $y'$.

The degree is the highest power of the highest order derivative when the equation is written as a polynomial in derivatives.

The highest order derivative is $y'$, and its power in the equation is 1.

Therefore, the Degree of the differential equation is 1.

Conclusion:

Order = 1

Degree = 1

Given:

The differential equation: $y’’ + (y’)^2 + 2y = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $y’’ + (y’)^2 + 2y = 0$.

Using Leibniz notation, this can be written as $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 + 2y = 0$.

The derivatives present in the equation are $y'$ (or $\frac{dy}{dx}$) and $y''$ (or $\frac{d^2y}{dx^2}$).

The orders of these derivatives are 1 and 2, respectively.

The highest order derivative present in the equation is $y''$ (or $\frac{d^2y}{dx^2}$).

The order of the differential equation is the order of the highest order derivative present.

Thus, the Order of the differential equation is 2.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation is a sum of terms where each term is a constant times a derivative (or the dependent variable) raised to an integer power. No non-polynomial functions of derivatives are present. Therefore, the equation is a polynomial in the derivatives $y'$ and $y''$.

The degree is the highest power of the highest order derivative when the equation is written as a polynomial in derivatives.

The highest order derivative is $y''$. The term containing $y''$ is $y''$, and its power is 1.

Therefore, the Degree of the differential equation is 1.

Conclusion:

Order = 2

Degree = 1

Given:

The differential equation: $y’’ + 2y’ + \sin y = 0$.

To Find:

The order and degree (if defined) of the given differential equation.

Solution:

The given differential equation is $y’’ + 2y’ + \sin y = 0$.

Using Leibniz notation, this can be written as $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \sin y = 0$.

The derivatives present in the equation are $y'$ (or $\frac{dy}{dx}$) and $y''$ (or $\frac{d^2y}{dx^2}$).

The orders of these derivatives are 1 and 2, respectively.

The highest order derivative present in the equation is $y''$ (or $\frac{d^2y}{dx^2}$).

The order of the differential equation is the order of the highest order derivative present.

Thus, the Order of the differential equation is 2.

To determine the degree, we check if the equation is a polynomial equation in the derivatives. The given equation involves the derivatives $y''$ (power 1) and $y'$ (power 1). The term $\sin y$ involves the dependent variable $y$, but not a derivative. The equation is expressible as a polynomial in the derivatives $y'$ and $y''$.

The degree is the highest power of the highest order derivative when the equation is written as a polynomial in derivatives.

The highest order derivative is $y''$. The term containing $y''$ is $y''$, and its power is 1.

Therefore, the Degree of the differential equation is 1.

Conclusion:

Order = 2

Degree = 1

Given:

The differential equation: $\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin \left( \frac{dy}{dx} \right) + 1 = 0$.

To Find:

The degree of the given differential equation.

Solution:

The given differential equation is $\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin \left( \frac{dy}{dx} \right) + 1 = 0$.

To determine the degree of a differential equation, it must be a polynomial equation in the derivatives. The degree is then the highest power of the highest order derivative in that polynomial form.

In the given equation, we have the derivatives $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$.

The highest order derivative is $\frac{d^2y}{dx^2}$, and its order is 2.

The term $\left( \frac{d^2y}{dx^2} \right)^3$ involves the highest order derivative raised to the power 3.

However, the equation contains the term $\sin \left( \frac{dy}{dx} \right)$. This term is a transcendental function (sine) of a derivative $\frac{dy}{dx}$.

Because of the presence of $\sin \left( \frac{dy}{dx} \right)$, the equation cannot be expressed as a polynomial in the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

Therefore, the degree of the differential equation is not defined.

Conclusion:

The degree of the differential equation $\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^2 + \sin \left( \frac{dy}{dx} \right) + 1 = 0$ is not defined.

Comparing with the given options:

(A) 3

(B) 2

(C) 1

(D) not defined

The correct answer is (D).

Given:

The differential equation: $2x^2 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0$.

To Find:

The order of the given differential equation.

Solution:

The given differential equation is $2x^2 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0$.

The order of a differential equation is the order of the highest order derivative appearing in the equation.

The derivatives present in the equation are:

1. $\frac{dy}{dx}$: This is the first derivative of $y$ with respect to $x$. Its order is 1.

2. $\frac{d^2y}{dx^2}$: This is the second derivative of $y$ with respect to $x$. Its order is 2.

Comparing the orders of the derivatives, the highest order derivative is $\frac{d^2y}{dx^2}$.

The order of $\frac{d^2y}{dx^2}$ is 2.

Therefore, the order of the differential equation is 2.

Conclusion:

The order of the differential equation $2x^2 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + y = 0$ is 2.

Comparing with the given options:

(A) 2

(B) 1

(C) 0

(D) not defined

The correct answer is (A).

Given:

Function: $y = e^{-3x}$

Differential Equation: $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$

To Verify:

Whether the given function is a solution of the given differential equation.

Solution:

We are given the function $y = e^{-3x}$.

To verify if it is a solution to the differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$, we need to find the first and second derivatives of $y$ with respect to $x$ and substitute them into the differential equation.

Find the first derivative $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{d}{dx}(e^{-3x})$

Using the chain rule, $\frac{d}{dx}(e^{au}) = ae^{au} \frac{du}{dx}$. Here $u=-3x$, so $\frac{du}{dx} = -3$.

$\frac{dy}{dx} = e^{-3x} \cdot (-3)$

Find the second derivative $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-3e^{-3x})$

$\frac{d^2y}{dx^2} = -3 \frac{d}{dx}(e^{-3x})$

Using the result from the first derivative calculation:

$\frac{d^2y}{dx^2} = -3 (-3e^{-3x})$

Now, substitute $y$, $\frac{dy}{dx}$ (from (i)), and $\frac{d^2y}{dx^2}$ (from (ii)) into the left-hand side (LHS) of the given differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.

LHS $= \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y$

LHS $= (9e^{-3x}) + (-3e^{-3x}) - 6(e^{-3x})$

LHS $= 9e^{-3x} - 3e^{-3x} - 6e^{-3x}$

Combine the terms:

LHS $= (9 - 3 - 6)e^{-3x}$

LHS $= (6 - 6)e^{-3x}$

LHS $= 0 \cdot e^{-3x}$

LHS $= 0$

The right-hand side (RHS) of the differential equation is 0.

Since LHS = RHS, the given function $y = e^{-3x}$ satisfies the differential equation.

Conclusion:

The function $y = e^{-3x}$ is a solution of the differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.

Given:

Function: $y = a \cos x + b \sin x$, where $a, b \in \mathbb{R}$.

Differential Equation: $\frac{d^2y}{dx^2} + y = 0$.

To Verify:

Whether the given function is a solution of the given differential equation.

Solution:

We are given the function $y = a \cos x + b \sin x$.

To verify if it is a solution to the differential equation $\frac{d^2y}{dx^2} + y = 0$, we need to find the first and second derivatives of $y$ with respect to $x$ and substitute them into the differential equation.

Find the first derivative $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{d}{dx}(a \cos x + b \sin x)$

Using the differentiation rules $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\sin x) = \cos x$:

$\frac{dy}{dx} = a(-\sin x) + b(\cos x)$

Find the second derivative $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-a \sin x + b \cos x)$

Using the differentiation rules again:

$\frac{d^2y}{dx^2} = -a(\cos x) + b(-\sin x)$

Now, substitute $y = a \cos x + b \sin x$ and $\frac{d^2y}{dx^2} = -a \cos x - b \sin x$ (from (ii)) into the left-hand side (LHS) of the given differential equation $\frac{d^2y}{dx^2} + y = 0$.

LHS $= \frac{d^2y}{dx^2} + y$

LHS $= (-a \cos x - b \sin x) + (a \cos x + b \sin x)$

Rearrange the terms:

LHS $= (-a \cos x + a \cos x) + (-b \sin x + b \sin x)$

LHS $= 0 + 0$

LHS $= 0$

The right-hand side (RHS) of the differential equation is 0.

Since LHS = RHS, the given function $y = a \cos x + b \sin x$ satisfies the differential equation.

Conclusion:

The function $y = a \cos x + b \sin x$, where $a, b \in \mathbb{R}$, is a solution of the differential equation $\frac{d^2y}{dx^2} + y = 0$.

The given function is $y = e^x + 1$.

To verify if this function is a solution to the differential equation $y'' - y' = 0$, we need to find the first and second derivatives of $y$ with respect to $x$.

The first derivative of $y$ with respect to $x$ is $y'$.

$y' = \frac{dy}{dx} = \frac{d}{dx}(e^x + 1)$

$y' = e^x + 0$

$y' = e^x$

The second derivative of $y$ with respect to $x$ is $y''$.

$y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}(y')$

$y'' = \frac{d}{dx}(e^x)$

$y'' = e^x$

Now, substitute the values of $y'$ and $y''$ into the given differential equation $y'' - y' = 0$.

Left Hand Side (LHS) = $y'' - y'$

LHS = $e^x - e^x$

LHS = $0$

The Right Hand Side (RHS) of the differential equation is $0$.

Since LHS = RHS ($0 = 0$), the given function satisfies the differential equation.

Therefore, the given function $y = e^x + 1$ is a solution of the corresponding differential equation $y'' - y' = 0$.

Given Function:

$y = x^2 + 2x + C$

Differential Equation:

$y' - 2x - 2 = 0$

To verify if the given function is a solution to the differential equation, we need to find the first derivative $y'$ of the function and substitute it into the differential equation.

Find $y'$:

$y' = \frac{dy}{dx} = \frac{d}{dx}(x^2 + 2x + C)$

$y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(C)$

$y' = 2x + 2(1) + 0$

$y' = 2x + 2$

Substitute $y' = 2x + 2$ into the differential equation $y' - 2x - 2 = 0$:

LHS = $y' - 2x - 2$

LHS = $(2x + 2) - 2x - 2$

LHS = $2x + 2 - 2x - 2$

LHS = $(2x - 2x) + (2 - 2)$

LHS = $0 + 0$

LHS = $0$

RHS = $0$

Since LHS = RHS ($0 = 0$), the given function satisfies the differential equation.

Therefore, the given function $y = x^2 + 2x + C$ is a solution of the corresponding differential equation $y' - 2x - 2 = 0$.

Given Function:

$y = \cos x + C$

Differential Equation:

$y' + \sin x = 0$

To verify if the given function is a solution to the differential equation, we need to find the first derivative $y'$ of the function and substitute it into the differential equation.

Find $y'$:

$y' = \frac{dy}{dx} = \frac{d}{dx}(\cos x + C)$

$y' = \frac{d}{dx}(\cos x) + \frac{d}{dx}(C)$

$y' = -\sin x + 0$

$y' = -\sin x$

Substitute $y' = -\sin x$ into the differential equation $y' + \sin x = 0$:

LHS = $y' + \sin x$

LHS = $(-\sin x) + \sin x$

LHS = $0$

RHS = $0$

Since LHS = RHS ($0 = 0$), the given function satisfies the differential equation.

Therefore, the given function $y = \cos x + C$ is a solution of the corresponding differential equation $y' + \sin x = 0$.

Given Function:

$y = \sqrt{1+x^2}$

Differential Equation:

$y' = \frac{xy}{1+x^2}$

To verify if the given function is a solution to the differential equation, we need to find the first derivative $y'$ of the function and check if it satisfies the given differential equation.

Find $y'$:

$y = \sqrt{1+x^2} = (1+x^2)^{1/2}$

Using the chain rule, we differentiate $y$ with respect to $x$:

$y' = \frac{dy}{dx} = \frac{d}{dx}((1+x^2)^{1/2})$

$y' = \frac{1}{2}(1+x^2)^{1/2 - 1} \cdot \frac{d}{dx}(1+x^2)$

$y' = \frac{1}{2}(1+x^2)^{-1/2} \cdot (2x)$

$y' = \frac{1}{2\sqrt{1+x^2}} \cdot (2x)$

$y' = \frac{2x}{2\sqrt{1+x^2}}$

$y' = \frac{x}{\sqrt{1+x^2}}$

Now, let's evaluate the Right Hand Side (RHS) of the differential equation using the given function $y = \sqrt{1+x^2}$.

RHS = $\frac{xy}{1+x^2}$

Substitute $y = \sqrt{1+x^2}$ into the RHS expression:

RHS = $\frac{x(\sqrt{1+x^2})}{1+x^2}$

We know that $1+x^2 = (\sqrt{1+x^2})^2$. Substitute this into the denominator:

RHS = $\frac{x\sqrt{1+x^2}}{(\sqrt{1+x^2})^2}$

Cancel out one factor of $\sqrt{1+x^2}$ from the numerator and denominator:

RHS = $\frac{x}{\sqrt{1+x^2}}$

Compare the calculated $y'$ (LHS) with the evaluated RHS:

LHS = $y' = \frac{x}{\sqrt{1+x^2}}$

RHS = $\frac{x}{\sqrt{1+x^2}}$

Since LHS = RHS, the given function satisfies the differential equation.

Therefore, the given function $y = \sqrt{1+x^2}$ is a solution of the corresponding differential equation $y' = \frac{xy}{1+x^2}$.

Given Function:

$y = Ax$

Differential Equation:

$xy' = y \quad (x \neq 0)$

To verify if the given function is a solution to the differential equation, we need to find the first derivative $y'$ of the function and substitute it along with the function $y$ into the differential equation.

Find $y'$:

$y = Ax$

Differentiate $y$ with respect to $x$:

$y' = \frac{dy}{dx} = \frac{d}{dx}(Ax)$

$y' = A \frac{d}{dx}(x)$

$y' = A \cdot 1$

$y' = A$

Substitute $y' = A$ and $y = Ax$ into the differential equation $xy' = y$:

LHS = $xy'$

LHS = $x(A)$

LHS = $Ax$

RHS = $y$

RHS = $Ax$

Since LHS = RHS ($Ax = Ax$), the given function satisfies the differential equation for $x \neq 0$.

Therefore, the given function $y = Ax$ is a solution of the corresponding differential equation $xy' = y \quad (x \neq 0)$.

Given Function:

$y = x \sin x$

Differential Equation:

$xy' = y + x \sqrt{x^2 - y^2}$

Domain Condition: $x \neq 0$ and $(x > y \text{ or } x < -y)$.

The condition $x > y$ or $x < -y$ implies $x^2 > y^2$.

Substitute $y = x \sin x$ into $x^2 > y^2$:

$x^2 > (x \sin x)^2$

$x^2 > x^2 \sin^2 x$

Since $x \neq 0$, $x^2 > 0$. We can divide by $x^2$:

$1 > \sin^2 x$

$\sin^2 x < 1$

This means $\sin x \neq 1$ and $\sin x \neq -1$, which implies $\cos^2 x = 1 - \sin^2 x > 0$. Thus, $\cos x \neq 0$.

So the domain is $x \neq 0$ and $\cos x \neq 0$.

To Verify: The given function is a solution to the differential equation.

Solution:

We need to find the first derivative $y'$ of the function $y = x \sin x$.

Using the product rule, $\frac{d}{dx}(uv) = u'v + uv'$, with $u=x$ and $v=\sin x$:

$y' = \frac{d}{dx}(x \sin x)$

$y' = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x)$

$y' = 1 \cdot \sin x + x \cdot \cos x$

$y' = \sin x + x \cos x$

Now, substitute $y$ and $y'$ into the differential equation $xy' = y + x \sqrt{x^2 - y^2}$.

Consider the Left Hand Side (LHS):

LHS = $xy'$

Substitute the expression for $y'$:

LHS = $x(\sin x + x \cos x)$

LHS = $x \sin x + x^2 \cos x$

Consider the Right Hand Side (RHS):

RHS = $y + x \sqrt{x^2 - y^2}$

Substitute the expression for $y$:

RHS = $(x \sin x) + x \sqrt{x^2 - (x \sin x)^2}$

RHS = $x \sin x + x \sqrt{x^2 - x^2 \sin^2 x}$

Factor out $x^2$ under the square root:

RHS = $x \sin x + x \sqrt{x^2 (1 - \sin^2 x)}$

Use the identity $1 - \sin^2 x = \cos^2 x$:

RHS = $x \sin x + x \sqrt{x^2 \cos^2 x}$

Since $\sqrt{a^2} = |a|$, we have $\sqrt{x^2 \cos^2 x} = |x \cos x|$.

RHS = $x \sin x + x |x \cos x|$

For the given function to be a solution, we must have LHS = RHS:

$x \sin x + x^2 \cos x = x \sin x + x |x \cos x|$

Subtract $x \sin x$ from both sides:

$x^2 \cos x = x |x \cos x|$

This equality holds if and only if $x \cos x \geq 0$, because if $x \cos x \geq 0$, then $|x \cos x| = x \cos x$, and $x^2 \cos x = x(x \cos x) = x^2 \cos x$. If $x \cos x < 0$, then $|x \cos x| = -(x \cos x)$, and $x^2 \cos x = x(-(x \cos x)) = -x^2 \cos x$, which implies $2x^2 \cos x = 0$. Since $x \neq 0$ in the domain, this requires $\cos x = 0$, but this contradicts $x \cos x < 0$ and the domain condition $\cos x \neq 0$.

Thus, the function $y = x \sin x$ satisfies the differential equation $xy' = y + x \sqrt{x^2 - y^2}$ if and only if $x \cos x \geq 0$. Within the given domain ($x \neq 0$ and $\cos x \neq 0$), this means the verification holds on the subset where $x$ and $\cos x$ have the same sign (i.e., $x \cos x > 0$). Assuming the context of the problem intends verification within regions where the square root simplifies conveniently to $x \cos x$ while satisfying the domain constraints:

If $x \cos x > 0$, then $\sqrt{x^2 \cos^2 x} = x \cos x$. The RHS becomes:

RHS = $x \sin x + x (x \cos x)$

RHS = $x \sin x + x^2 \cos x$

This is equal to the LHS.

Therefore, the given function $y = x \sin x$ is a solution of the corresponding differential equation $xy' = y + x \sqrt{x^2 - y^2}$ in the domain where $x \neq 0$, $\cos x \neq 0$, and $x \cos x > 0$.

Given Implicit Function:

$xy = \log y + C$

Differential Equation:

$y' = \frac{y^2}{1 - xy} \quad (xy \neq 1)$

To verify if the given implicit function is a solution to the differential equation, we need to differentiate the implicit function with respect to $x$ to find $y'$ and check if it matches the given differential equation.

Differentiate both sides of the equation $xy = \log y + C$ with respect to $x$. Remember that $y$ is a function of $x$, so we will use the chain rule for terms involving $y$ and the product rule for the $xy$ term.

$\frac{d}{dx}(xy) = \frac{d}{dx}(\log y + C)$

Applying the product rule on the left side and the sum rule on the right side:

$\frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y) = \frac{d}{dx}(\log y) + \frac{d}{dx}(C)$

Calculate the derivatives:

$1 \cdot y + x \cdot y' = \frac{1}{y} \cdot y' + 0$

$y + xy' = \frac{y'}{y}$

Now, we need to rearrange this equation to solve for $y'$. Multiply both sides by $y$ to eliminate the fraction (note that this step requires $y \neq 0$. If $y=0$, $\log y$ is undefined anyway, so we assume $y \neq 0$).

$y(y + xy') = y'$

$y^2 + xy y' = y'$

Move all terms containing $y'$ to one side and other terms to the other side:

$y^2 = y' - xy y'$

Factor out $y'$ from the terms on the right side:

$y^2 = y'(1 - xy)$

Solve for $y'$ by dividing both sides by $(1 - xy)$. This step is valid because the problem statement specifies the condition $xy \neq 1$, which means $1 - xy \neq 0$.

$y' = \frac{y^2}{1 - xy}$

This derived expression for $y'$ matches the given differential equation.

Therefore, the given implicit function $xy = \log y + C$ is a solution of the corresponding differential equation $y' = \frac{y^2}{1 - xy} \quad (xy \neq 1)$.

Given Implicit Function:

$y - \cos y = x$

Differential Equation:

$(y \sin y + \cos y + x) y' = y$

To verify if the given implicit function is a solution to the differential equation, we need to differentiate the implicit function with respect to $x$ to find $y'$ and then check if it satisfies the given differential equation.

Differentiate both sides of the equation $y - \cos y = x$ with respect to $x$. Remember that $y$ is a function of $x$, so we will use the chain rule for terms involving $y$.

$\frac{d}{dx}(y - \cos y) = \frac{d}{dx}(x)$

Applying the difference rule on the left side:

$\frac{d}{dx}(y) - \frac{d}{dx}(\cos y) = \frac{d}{dx}(x)$

Calculate the derivatives. The derivative of $y$ with respect to $x$ is $y'$. The derivative of $\cos y$ with respect to $x$, using the chain rule, is $\frac{d}{dy}(\cos y) \cdot \frac{dy}{dx} = (-\sin y) \cdot y'$. The derivative of $x$ with respect to $x$ is $1$.

$y' - (-\sin y) \cdot y' = 1$

$y' + y' \sin y = 1$

Factor out $y'$ from the terms on the left side:

$y'(1 + \sin y) = 1$

Solve for $y'$:

$y' = \frac{1}{1 + \sin y}$

Now, substitute this expression for $y'$ into the given differential equation $(y \sin y + \cos y + x) y' = y$.

LHS = $(y \sin y + \cos y + x) y'$

Substitute the expression for $x$ from the given function ($x = y - \cos y$) and the expression for $y'$:

LHS = $(y \sin y + \cos y + (y - \cos y)) \left(\frac{1}{1 + \sin y}\right)$

Simplify the term in the parenthesis:

LHS = $(y \sin y + \cos y + y - \cos y) \left(\frac{1}{1 + \sin y}\right)$

The $\cos y$ terms cancel out:

LHS = $(y \sin y + y) \left(\frac{1}{1 + \sin y}\right)$

Factor out $y$ from the terms in the parenthesis:

LHS = $y(\sin y + 1) \left(\frac{1}{1 + \sin y}\right)$

Since $\sin y + 1 = 1 + \sin y$, and assuming $1 + \sin y \neq 0$ (which means $\sin y \neq -1$), we can cancel the $(1 + \sin y)$ terms:

LHS = $y \cdot 1$

LHS = $y$

RHS = $y$

Since LHS = RHS ($y = y$), the given implicit function satisfies the differential equation, provided $1 + \sin y \neq 0$.

Therefore, the given implicit function $y - \cos y = x$ is a solution of the corresponding differential equation $(y \sin y + \cos y + x) y' = y$.

Given Implicit Function:

$x + y = \tan^{-1} y$

Differential Equation:

$y^2 y' + y^2 + 1 = 0$

To verify if the given implicit function is a solution to the differential equation, we need to differentiate the implicit function with respect to $x$ to find $y'$ and then check if it satisfies the given differential equation.

Differentiate both sides of the equation $x + y = \tan^{-1} y$ with respect to $x$. Remember that $y$ is a function of $x$, so we will use the chain rule for terms involving $y$.

$\frac{d}{dx}(x + y) = \frac{d}{dx}(\tan^{-1} y)$

Applying the sum rule on the left side:

$\frac{d}{dx}(x) + \frac{d}{dx}(y) = \frac{d}{dx}(\tan^{-1} y)$

Calculate the derivatives. The derivative of $x$ with respect to $x$ is $1$. The derivative of $y$ with respect to $x$ is $y'$. The derivative of $\tan^{-1} y$ with respect to $x$, using the chain rule, is $\frac{d}{dy}(\tan^{-1} y) \cdot \frac{dy}{dx} = \frac{1}{1+y^2} \cdot y'$.

$1 + y' = \frac{1}{1+y^2} y'$

Now, we need to rearrange this equation to solve for $y'$. Gather the $y'$ terms on one side:

$1 = \frac{y'}{1+y^2} - y'$

Factor out $y'$ on the right side:

$1 = y' \left(\frac{1}{1+y^2} - 1\right)$

Combine the terms inside the parenthesis:

$1 = y' \left(\frac{1 - (1+y^2)}{1+y^2}\right)$

$1 = y' \left(\frac{1 - 1 - y^2}{1+y^2}\right)$

$1 = y' \left(\frac{-y^2}{1+y^2}\right)$

Solve for $y'$ by multiplying both sides by $\frac{1+y^2}{-y^2}$ (assuming $y \neq 0$):

$y' = \frac{1 \cdot (1+y^2)}{-y^2}$

$y' = -\frac{1+y^2}{y^2}$

$y' = -\left(\frac{1}{y^2} + 1\right)$

Now, substitute this expression for $y'$ into the given differential equation $y^2 y' + y^2 + 1 = 0$.

LHS = $y^2 y' + y^2 + 1$

Substitute the expression for $y'$:

LHS = $y^2 \left(-\frac{1+y^2}{y^2}\right) + y^2 + 1$

Assuming $y \neq 0$, we can cancel out the $y^2$ terms:

LHS = $-(1+y^2) + y^2 + 1$

LHS = $-1 - y^2 + y^2 + 1$

LHS = $(-1 + 1) + (-y^2 + y^2)$

LHS = $0 + 0$

LHS = $0$

RHS = $0$

Since LHS = RHS ($0 = 0$), the given implicit function satisfies the differential equation (for $y \neq 0$). If $y=0$, the original equation becomes $x+0 = \tan^{-1}(0)$, which means $x=0$. The differential equation becomes $0^2 y' + 0^2 + 1 = 0$, i.e., $1=0$, which is false. So the point $(0,0)$ is not on the solution curve. Thus $y \neq 0$ is implied.

Therefore, the given implicit function $x + y = \tan^{-1} y$ is a solution of the corresponding differential equation $y^2 y' + y^2 + 1 = 0$.

Given Function:

$y = \sqrt{a^2 - x^2}$

Given Differential Equation:

$x + y \frac{dy}{dx} = 0 \quad (y \neq 0)$

To verify if the given function is a solution to the differential equation, we need to find the first derivative $\frac{dy}{dx}$ of the function and substitute it along with the function $y$ into the differential equation.

Find $\frac{dy}{dx}$:

The given function is $y = \sqrt{a^2 - x^2}$. We can rewrite this as $y = (a^2 - x^2)^{1/2}$.

Differentiate $y$ with respect to $x$ using the chain rule:

$\frac{dy}{dx} = \frac{d}{dx}((a^2 - x^2)^{1/2})$

$\frac{dy}{dx} = \frac{1}{2}(a^2 - x^2)^{1/2 - 1} \cdot \frac{d}{dx}(a^2 - x^2)$

$\frac{dy}{dx} = \frac{1}{2}(a^2 - x^2)^{-1/2} \cdot (0 - 2x)$

$\frac{dy}{dx} = \frac{1}{2\sqrt{a^2 - x^2}} \cdot (-2x)$

$\frac{dy}{dx} = \frac{-2x}{2\sqrt{a^2 - x^2}}$

$\frac{dy}{dx} = \frac{-x}{\sqrt{a^2 - x^2}}$

Substitute $y = \sqrt{a^2 - x^2}$ and $\frac{dy}{dx} = \frac{-x}{\sqrt{a^2 - x^2}}$ into the Left Hand Side (LHS) of the differential equation $x + y \frac{dy}{dx} = 0$:

LHS = $x + y \frac{dy}{dx}$

LHS = $x + (\sqrt{a^2 - x^2}) \left(\frac{-x}{\sqrt{a^2 - x^2}}\right)$

Since $x \in (-a, a)$, $a^2 - x^2 > 0$, which means $\sqrt{a^2 - x^2} > 0$. Also, the condition $y \neq 0$ is given, and $y = \sqrt{a^2 - x^2}$, so $\sqrt{a^2 - x^2} \neq 0$. Thus, we can cancel the $\sqrt{a^2 - x^2}$ terms in the numerator and denominator.

LHS = $x + (-x)$

LHS = $x - x$

LHS = $0$

The Right Hand Side (RHS) of the differential equation is $0$.

RHS = $0$

Since LHS = RHS ($0 = 0$), the given function satisfies the differential equation for $x \in (-a, a)$ and $y \neq 0$.

Therefore, the given function $y = \sqrt{a^2 - x^2}$ is a solution of the corresponding differential equation $x + y \frac{dy}{dx} = 0 \quad (y \neq 0)$.

Concept:

The general solution of a differential equation of order $n$ contains $n$ arbitrary constants.

Given:

We are given a differential equation of fourth order.

To Find:

The number of arbitrary constants in the general solution.

Solution:

According to the concept, the number of arbitrary constants in the general solution of a differential equation is equal to its order.

The order of the given differential equation is 4.

Therefore, the number of arbitrary constants in the general solution is 4.

Conclusion:

The number of arbitrary constants in the general solution of a differential equation of fourth order is 4.

The correct option is (D).

Concept:

The general solution of a differential equation of order $n$ contains $n$ arbitrary constants.

A particular solution is obtained from the general solution by assigning specific values to the arbitrary constants (usually based on initial or boundary conditions).

Given:

We are given a differential equation of third order.

We are asked about the number of arbitrary constants in the particular solution.

To Find:

The number of arbitrary constants in the particular solution.

Solution:

The order of the differential equation is 3.

The general solution of a third-order differential equation contains 3 arbitrary constants.

A particular solution is derived from the general solution by fixing the values of these arbitrary constants.

By definition, a particular solution does not contain any arbitrary constants.

Conclusion:

The number of arbitrary constants in the particular solution of a differential equation of third order is 0.

The correct option is (D).

Given Differential Equation:

$\frac{dy}{dx} = \frac{x + 1}{2 − y} \quad (y \neq 2)$

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is a first-order separable differential equation.

We can separate the variables $x$ and $y$ by multiplying both sides by $(2-y)$ and $dx$:

$(2 - y) dy = (x + 1) dx$

Integrate both sides of the separated equation:

$\int (2 - y) dy = \int (x + 1) dx$

Perform the integration on both sides:

$\int 2 \, dy - \int y \, dy = \int x \, dx + \int 1 \, dx$

$2y - \frac{y^2}{2} + C_1 = \frac{x^2}{2} + x + C_2$

where $C_1$ and $C_2$ are integration constants.

Combine the constants of integration into a single constant $C$, where $C = C_2 - C_1$:

$2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C$

To eliminate the fractions, multiply the entire equation by 2:

$2 \left(2y - \frac{y^2}{2}\right) = 2 \left(\frac{x^2}{2} + x + C\right)$

$4y - y^2 = x^2 + 2x + 2C$

Let $K = 2C$, which is also an arbitrary constant:

$4y - y^2 = x^2 + 2x + K$

Rearranging the terms to express the general solution in a common form:

$x^2 + 2x + y^2 - 4y + K = 0$

This is the general solution of the given differential equation. The domain $y \neq 2$ ensures that the denominator in the original differential equation is not zero and the separation step is valid.

General Solution:

$x^2 + 2x + y^2 - 4y + K = 0$

where $K$ is an arbitrary constant.

Given Differential Equation:

$\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is a first-order separable differential equation.

We can separate the variables $x$ and $y$. Divide both sides by $(1+y^2)$ and multiply by $dx$:

$\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$

Integrate both sides of the separated equation:

$\int \frac{dy}{1 + y^2} = \int \frac{dx}{1 + x^2}$

Perform the integration on both sides. The integral of $\frac{1}{1+u^2}$ with respect to $u$ is $\tan^{-1} u + C$.

$\tan^{-1} y = \tan^{-1} x + C$

where $C$ is the arbitrary constant of integration.

This is the general solution of the given differential equation. We can leave the solution in this form or express $\tan^{-1} y$ in terms of $\tan^{-1} x$ and the constant.

$\tan^{-1} y - \tan^{-1} x = C$

Using the tangent subtraction formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, we can write:

$\tan(\tan^{-1} y - \tan^{-1} x) = \tan(C)$

$\frac{\tan(\tan^{-1} y) - \tan(\tan^{-1} x)}{1 + \tan(\tan^{-1} y) \tan(\tan^{-1} x)} = \tan(C)$

$\frac{y - x}{1 + xy} = \tan(C)$

Let $A = \tan(C)$, where $A$ is another arbitrary constant:

$\frac{y - x}{1 + xy} = A$

$y - x = A(1 + xy)$

$y - x = A + Axy$

$y - Axy = A + x$

$y(1 - Ax) = A + x$

$y = \frac{A + x}{1 - Ax}$ (assuming $1 - Ax \neq 0$)

Both $\tan^{-1} y = \tan^{-1} x + C$ and $y = \frac{A + x}{1 - Ax}$ are valid forms of the general solution, representing the same family of curves. The first form is simpler and directly obtained from the integration.

General Solution:

$\tan^{-1} y = \tan^{-1} x + C$

where $C$ is an arbitrary constant.

Given Differential Equation:

$\frac{dy}{dx} = −4xy^2$

Initial Condition:

$y = 1$ when $x = 0$.

To Find:

The particular solution of the differential equation satisfying the given initial condition.

Solution:

The given differential equation is a first-order separable differential equation. We can separate the variables $x$ and $y$ by dividing by $y^2$ (assuming $y \neq 0$) and multiplying by $dx$.

$\frac{dy}{y^2} = -4x \, dx$

$y^{-2} \, dy = -4x \, dx$

Integrate both sides of the separated equation:

$\int y^{-2} \, dy = \int -4x \, dx$

Perform the integration on both sides:

$\frac{y^{-2+1}}{-2+1} = -4 \int x \, dx$

$\frac{y^{-1}}{-1} = -4 \frac{x^{1+1}}{1+1} + C$

$-y^{-1} = -4 \frac{x^2}{2} + C$

$-\frac{1}{y} = -2x^2 + C$

This is the general solution of the differential equation, where $C$ is the arbitrary constant of integration.

$2x^2 - \frac{1}{y} = C$

Now, we use the initial condition $y = 1$ when $x = 0$ to find the value of $C$.

Substitute $x=0$ and $y=1$ into the general solution:

$2(0)^2 - \frac{1}{1} = C$

$0 - 1 = C$

$-1 = C$

Substitute the value of $C = -1$ back into the general solution to obtain the particular solution:

$2x^2 - \frac{1}{y} = -1$

We can rearrange the particular solution to express $y$ explicitly if desired (assuming $y \neq 0$):

$2x^2 + 1 = \frac{1}{y}$

$y = \frac{1}{2x^2 + 1}$

Note that $2x^2 + 1$ is always greater than 0 for any real $x$, so $y$ is never zero. Also, when $x=0$, $y = \frac{1}{2(0)^2 + 1} = \frac{1}{1} = 1$, which matches the initial condition.

Particular Solution:

$2x^2 - \frac{1}{y} = -1$

or equivalently,

$y = \frac{1}{2x^2 + 1}$

Given Differential Equation:

$x \, dy = (2x^2 + 1) \, dx \quad (x \neq 0)$

Initial Condition:

The curve passes through the point $(1, 1)$. This means $y = 1$ when $x = 1$.

To Find:

The particular solution of the differential equation satisfying the given initial condition (the equation of the curve).

Solution:

The given differential equation can be written in the form $\frac{dy}{dx} = \frac{2x^2 + 1}{x}$.

This is a first-order separable differential equation. We can separate the variables $x$ and $y$ by dividing the equation by $x$:

$dy = \frac{2x^2 + 1}{x} \, dx$

$dy = \left(\frac{2x^2}{x} + \frac{1}{x}\right) \, dx$

$dy = \left(2x + \frac{1}{x}\right) \, dx$

Integrate both sides of the separated equation to find the general solution:

$\int dy = \int \left(2x + \frac{1}{x}\right) \, dx$

$\int dy = \int 2x \, dx + \int \frac{1}{x} \, dx$

Perform the integration:

$y = 2 \left(\frac{x^{1+1}}{1+1}\right) + \log |x| + C$

$y = 2 \left(\frac{x^2}{2}\right) + \log |x| + C$

$y = x^2 + \log |x| + C$

This is the general solution, where $C$ is the arbitrary constant of integration.

Now, we use the initial condition $(1, 1)$ to find the value of $C$. Substitute $x=1$ and $y=1$ into the general solution:

$1 = (1)^2 + \log |1| + C$

$1 = 1 + 0 + C$

$1 = 1 + C$

$C = 1 - 1$

$C = 0$

Substitute the value of $C = 0$ back into the general solution to obtain the particular solution:

$y = x^2 + \log |x| + 0$

$y = x^2 + \log |x|$

Since the point $(1, 1)$ is given and $x=1$ is positive, we can write $\log |x|$ as $\log x$ in the particular solution as the curve passes through a region where $x > 0$.

Equation of the curve (Particular Solution):

$y = x^2 + \log x$ (for $x > 0$)

Given:

The slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.

The curve passes through the point $(-2, 3)$.

Differential Equation:

The slope of the tangent is given by $\frac{dy}{dx}$. So, the differential equation is:

$\frac{dy}{dx} = \frac{2x}{y^2}$

To Find:

The equation of the curve (the particular solution) passing through the point $(-2, 3)$.

Solution:

The given differential equation is a first-order separable differential equation. We can separate the variables $x$ and $y$ by multiplying both sides by $y^2$ and $dx$:

$y^2 \, dy = 2x \, dx$

Integrate both sides of the separated equation to find the general solution:

$\int y^2 \, dy = \int 2x \, dx$

Perform the integration:

$\frac{y^{2+1}}{2+1} = 2 \frac{x^{1+1}}{1+1} + C$

$\frac{y^3}{3} = 2 \frac{x^2}{2} + C$

$\frac{y^3}{3} = x^2 + C$

This is the general solution, where $C$ is the arbitrary constant of integration.

Now, we use the initial condition that the curve passes through the point $(-2, 3)$. Substitute $x=-2$ and $y=3$ into the general solution to find the value of $C$.

$\frac{(3)^3}{3} = (-2)^2 + C$

$\frac{27}{3} = 4 + C$

$9 = 4 + C$

$C = 9 - 4$

$C = 5$

Substitute the value of $C=5$ back into the general solution $\frac{y^3}{3} = x^2 + C$ to obtain the particular solution.

$\frac{y^3}{3} = x^2 + 5$

We can multiply by 3 to clear the fraction:

$y^3 = 3(x^2 + 5)$

$y^3 = 3x^2 + 15$

This is the equation of the curve passing through the point $(-2, 3)$.

Equation of the curve (Particular Solution):

$y^3 = 3x^2 + 15$

Given:

Initial Principal $P_0 = \textsf{₹} 1000$.

Rate of increase $r = 5\%$ per year, compounded continuously.

The phrase "principal increases continuously at the rate of 5% per year" implies that the rate of change of principal with respect to time is proportional to the current principal amount.

Differential Equation:

Let $P$ be the principal amount at time $t$ (in years).

The rate of increase of the principal is $\frac{dP}{dt}$.

According to the problem, $\frac{dP}{dt}$ is $5\%$ of $P$.

$\frac{dP}{dt} = 5\% \text{ of } P = \frac{5}{100} P = 0.05 P$

Initial Condition:

At time $t = 0$, the initial principal is $\textsf{₹} 1000$.

$P(0) = 1000$

To Find:

The time $t$ in years when the principal doubles, i.e., when $P(t) = 2 \times P_0 = 2 \times 1000 = \textsf{₹} 2000$.

Solution:

The given differential equation is $\frac{dP}{dt} = 0.05 P$.

This is a first-order separable differential equation.

Separate the variables by dividing by $P$ (assuming $P \neq 0$, which is true for principal in a bank) and multiplying by $dt$:

$\frac{dP}{P} = 0.05 \, dt$

Integrate both sides:

$\int \frac{dP}{P} = \int 0.05 \, dt$

$\log |P| = 0.05 t + C_1$

where $C_1$ is the integration constant.

Since the principal $P$ is always positive, $|P| = P$. Exponentiate both sides:

$P = e^{0.05 t + C_1} = e^{C_1} e^{0.05 t}$

Let $C = e^{C_1}$. Since $e^{C_1}$ is always positive, $C$ is a positive constant.

$P(t) = C e^{0.05 t}$

Use the initial condition $P(0) = 1000$ to find the value of $C$:

$1000 = C e^{0.05 \times 0}$

$1000 = C e^0$

$1000 = C \times 1$

$C = 1000$

Substitute the value of $C$ into the general solution to get the particular solution for this problem:

$P(t) = 1000 e^{0.05 t}$

We need to find the time $t$ when the principal doubles, i.e., $P(t) = 2000$.

Set the expression for $P(t)$ equal to 2000:

$1000 e^{0.05 t} = 2000$

Divide both sides by 1000:

$e^{0.05 t} = \frac{2000}{1000}$

$e^{0.05 t} = 2$

Take the natural logarithm ($\log_e$) of both sides to solve for $t$:

$\log_e(e^{0.05 t}) = \log_e(2)$

$0.05 t = \log_e(2)$

Isolate $t$:

$t = \frac{\log_e(2)}{0.05}$

$t = \frac{\log_e(2)}{5/100}$

$t = \frac{100 \log_e(2)}{5}$

$t = 20 \log_e(2)$

The time taken for the principal to double is $20 \log_e(2)$ years.

If a numerical value is required, using $\log_e(2) \approx 0.6931$:

$t \approx 20 \times 0.6931 \approx 13.862$ years.

Answer:

The principal will double itself in $20 \log_e(2)$ years.

The given differential equation is:

$\frac{dy}{dx} = \frac{1 − \cos x}{1 + \cos x}$

This is a variable separable differential equation. We can rewrite it as:

$dy = \frac{1 − \cos x}{1 + \cos x} dx$

Integrating both sides, we get:

$\int dy = \int \frac{1 − \cos x}{1 + \cos x} dx$

Let's simplify the integrand using the half-angle trigonometric identities:

$1 - \cos x = 2 \sin^2 \frac{x}{2}$

$1 + \cos x = 2 \cos^2 \frac{x}{2}$

Substituting these into the integrand:

$\frac{1 − \cos x}{1 + \cos x} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \tan^2 \frac{x}{2}$

Using the identity $\tan^2 \theta = \sec^2 \theta - 1$, we have:

$\tan^2 \frac{x}{2} = \sec^2 \frac{x}{2} - 1$

Now the integral on the right side becomes:

$\int \frac{1 − \cos x}{1 + \cos x} dx = \int \left( \sec^2 \frac{x}{2} - 1 \right) dx$

We can split this into two separate integrals:

$\int \sec^2 \frac{x}{2} dx - \int 1 dx$

Integrating the left side of the original equation:

$\int dy = y$

Integrating the terms on the right side:

For the first term, use the substitution $u = \frac{x}{2}$, so $du = \frac{1}{2} dx$, which means $dx = 2 du$.

$\int \sec^2 \frac{x}{2} dx = \int \sec^2 u (2 du) = 2 \int \sec^2 u du = 2 \tan u + C_1 = 2 \tan \frac{x}{2} + C_1$

For the second term:

$\int 1 dx = x + C_2$

Combining the integrals on the right side:

$2 \tan \frac{x}{2} - x + C$, where $C = C_1 - C_2$ is the constant of integration.

Equating the results from both sides of the integrated equation, the general solution is:

$y = 2 \tan \frac{x}{2} - x + C$

The given differential equation is:

$\frac{dy}{dx} = \sqrt{4 − y^2}$

The restriction $(-2 < y < 2)$ implies that $\sqrt{4 - y^2}$ is real and non-zero, allowing us to divide by it.

This is a variable separable differential equation. We can separate the variables by rearranging the terms:

$\frac{dy}{\sqrt{4 − y^2}} = dx$

Integrate both sides of the equation:

$\int \frac{dy}{\sqrt{4 − y^2}} = \int dx$

The integral on the left side is of the form $\int \frac{dx}{\sqrt{a^2 - x^2}}$, which is equal to $\sin^{-1} \left( \frac{x}{a} \right) + C$.

In this case, $y$ is the variable and $a^2 = 4$, so $a = 2$.

$\int \frac{dy}{\sqrt{4 − y^2}} = \sin^{-1} \left( \frac{y}{2} \right)$

The integral on the right side is straightforward:

$\int dx = x$

Combining the results of the integration and adding the constant of integration, C, to the right side, we get the general solution:

$\sin^{-1} \left( \frac{y}{2} \right) = x + C$

We can express $y$ explicitly in terms of $x$ and C by taking the sine of both sides:

$\frac{y}{2} = \sin(x + C)$

Multiply by 2:

$y = 2 \sin(x + C)$

This is the general solution to the given differential equation.

General Solution:

$y = 2 \sin(x + C)$

or

$\sin^{-1} \left( \frac{y}{2} \right) = x + C$

The given differential equation is:

$\frac{dy}{dx} + y = 1$

The restriction $(y \neq 1)$ is given to ensure that we do not divide by zero during the separation of variables.

Rearrange the equation to separate the variables:

$\frac{dy}{dx} = 1 - y$

Assuming $y \neq 1$, we can divide by $(1-y)$ and multiply by $dx$:

$\frac{dy}{1 - y} = dx$

Integrate both sides of the separated equation:

$\int \frac{dy}{1 - y} = \int dx$

Let's evaluate the integrals.

For the left side, let $u = 1 - y$. Then $du = -dy$, so $dy = -du$.

$\int \frac{-du}{u} = - \int \frac{du}{u} = - \log_e|u| + C_1 = - \log_e|1 - y| + C_1$

For the right side:

$\int dx = x + C_2$

Equating the results of the integration:

$- \log_e|1 - y| = x + C$

where $C = C_2 - C_1$ is the constant of integration.

To find the general solution, we solve for $y$.

Multiply by -1:

$\log_e|1 - y| = -(x + C)$

Take the exponential of both sides:

$|1 - y| = e^{-(x + C)}$

$|1 - y| = e^{-x} e^{-C}$

Let $A = e^{-C}$. Since $e^{-C}$ is always positive, $A > 0$.

$|1 - y| = A e^{-x}$

$1 - y = \pm A e^{-x}$

Let $B = \pm A$. Since $A > 0$, $B$ can be any non-zero real number.

$1 - y = B e^{-x}$

$y = 1 - B e^{-x}$

Note that the case $y=1$ (a constant function) is also a solution to the original differential equation, as $\frac{d(1)}{dx} + 1 = 0 + 1 = 1$. This corresponds to the case where $B=0$. Thus, we can include the case $B=0$ and state that $B$ is any real constant.

General Solution:

$y = 1 - C e^{-x}$

(where $C$ is an arbitrary constant, which we have replaced $B$ with)

The given differential equation is:

$\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$

This is a variable separable differential equation. We can rearrange the terms to separate the variables:

$\sec^2 y \tan x \, dy = - \sec^2 x \tan y \, dx$

Assuming $\tan x \neq 0$ and $\tan y \neq 0$, we can divide both sides by $(\tan x \tan y)$:

$\frac{\sec^2 y \tan x}{\tan x \tan y} dy = - \frac{\sec^2 x \tan y}{\tan x \tan y} dx$

Simplifying, we get:

$\frac{\sec^2 y}{\tan y} dy = - \frac{\sec^2 x}{\tan x} dx$

Now, integrate both sides of the separated equation:

$\int \frac{\sec^2 y}{\tan y} dy = \int - \frac{\sec^2 x}{\tan x} dx$

To evaluate the integrals, we can use the substitution method. The integral is of the form $\int \frac{f'(u)}{f(u)} du = \log_e|f(u)| + C$.

For the left side:

Let $u = \tan y$. Then $du = \sec^2 y \, dy$.

$\int \frac{1}{u} du = \log_e|u| + C_1 = \log_e|\tan y| + C_1$

For the right side:

Let $v = \tan x$. Then $dv = \sec^2 x \, dx$.

$\int - \frac{1}{v} dv = - \int \frac{1}{v} dv = - \log_e|v| + C_2 = - \log_e|\tan x| + C_2$

Equating the results from both sides of the integration:

$\log_e|\tan y| = - \log_e|\tan x| + C$

where $C = C_2 - C_1$ is the constant of integration.

Rearrange the terms to express the general solution:

$\log_e|\tan y| + \log_e|\tan x| = C$

Using the logarithm property $\log_e a + \log_e b = \log_e(ab)$:

$\log_e|\tan x \tan y| = C$

To eliminate the logarithm, take the exponential of both sides:

$|\tan x \tan y| = e^C$

Let $K = e^C$. Since $e^C$ is always positive, $K > 0$.

$\tan x \tan y = \pm K$

Let $C'$ be an arbitrary constant that can be positive or negative ($\pm K$). We can write the solution as:

$\tan x \tan y = C'$

where $C'$ is a non-zero arbitrary constant derived from the integration.

Note that if we allow $C'$ to be zero, the equation $\tan x \tan y = 0$ covers the cases where $\tan x = 0$ or $\tan y = 0$, which correspond to the trivial solutions $x = m\pi$ or $y = n\pi$ (for integers $m, n$). These are also solutions to the original differential equation.

General Solution:

$\tan x \tan y = C$

where $C$ is an arbitrary constant.

The given differential equation is:

$(e^x + e^{-x}) dy – (e^x – e^{-x}) dx = 0$

This is a variable separable differential equation. We can rearrange the terms to separate the variables:

$(e^x + e^{-x}) dy = (e^x – e^{-x}) dx$

Divide both sides by $(e^x + e^{-x})$ to isolate the $dy$ term (note that $e^x + e^{-x}$ is always positive for real $x$, so we don't divide by zero):

$dy = \frac{e^x – e^{-x}}{e^x + e^{-x}} dx$

Integrate both sides of the separated equation:

$\int dy = \int \frac{e^x – e^{-x}}{e^x + e^{-x}} dx$

Evaluate the integrals.

The left side integral is simply:

$\int dy = y$

For the right side integral, we can use a substitution. Let $u = e^x + e^{-x}$.

Differentiating $u$ with respect to $x$ gives:

$\frac{du}{dx} = \frac{d}{dx}(e^x + e^{-x}) = e^x - e^{-x}$

So, $du = (e^x - e^{-x}) dx$.

Substituting $u$ and $du$ into the right side integral:

$\int \frac{1}{u} du$

This standard integral evaluates to $\log_e|u| + C'$, where $C'$ is the constant of integration.

Substituting back $u = e^x + e^{-x}$:

$\int \frac{e^x – e^{-x}}{e^x + e^{-x}} dx = \log_e|e^x + e^{-x}| + C'$

Since $e^x > 0$ and $e^{-x} > 0$ for all real $x$, $e^x + e^{-x}$ is always positive. Therefore, $|e^x + e^{-x}| = e^x + e^{-x}$.

The right side integral is $\log_e(e^x + e^{-x}) + C'$.

Equating the results from both sides of the integration, we get the general solution:

$y = \log_e(e^x + e^{-x}) + C$

where $C$ is an arbitrary constant of integration.

General Solution:

$y = \log_e(e^x + e^{-x}) + C$

The given differential equation is:

$\frac{dy}{dx} = (1 + x^2) (1 + y^2)$

This is a variable separable differential equation. We can separate the variables by rearranging the terms. Since $(1+y^2)$ is always positive for real $y$, we can safely divide by it:

$\frac{dy}{1 + y^2} = (1 + x^2) dx$

Integrate both sides of the separated equation:

$\int \frac{dy}{1 + y^2} = \int (1 + x^2) dx$

Evaluate the integrals on both sides.

The left side integral is a standard form:

$\int \frac{dy}{1 + y^2} = \tan^{-1} y + C_1$

The right side integral can be split into two parts:

$\int (1 + x^2) dx = \int 1 \, dx + \int x^2 \, dx$

$\int 1 \, dx = x + C_2$

$\int x^2 \, dx = \frac{x^{2+1}}{2+1} + C_3 = \frac{x^3}{3} + C_3$

So, $\int (1 + x^2) dx = x + \frac{x^3}{3} + C_2 + C_3$

Equating the results from both sides of the integration and combining the constants $C_1, C_2, C_3$ into a single arbitrary constant $C = C_2 + C_3 - C_1$:

$\tan^{-1} y = x + \frac{x^3}{3} + C$

This is the general solution in implicit form. We can also express $y$ explicitly:

$y = \tan \left( x + \frac{x^3}{3} + C \right)$

General Solution:

$\tan^{-1} y = x + \frac{x^3}{3} + C$

or

$y = \tan \left( x + \frac{x^3}{3} + C \right)$

where $C$ is an arbitrary constant.

Given:

The differential equation is $y \log y \, dx – x \, dy = 0$.

(We assume $\log y$ denotes the natural logarithm, $\log_e y$).

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$y \log_e y \, dx – x \, dy = 0$

Rearrange the terms to separate the variables:

$y \log_e y \, dx = x \, dy$

This is a variable separable differential equation. Assuming $x \neq 0$, $y > 0$ (for $\log_e y$ to be defined) and $\log_e y \neq 0$ (i.e., $y \neq 1$), we can divide both sides by $x y \log_e y$ to separate the variables:

$\frac{y \log_e y \, dx}{x y \log_e y} = \frac{x \, dy}{x y \log_e y}$

$\frac{dx}{x} = \frac{dy}{y \log_e y}$

Integrate both sides of the separated equation:

$\int \frac{dx}{x} = \int \frac{dy}{y \log_e y}$

Evaluate the integrals.

For the left side:

$\int \frac{dx}{x} = \log_e|x| + C_1$

For the right side, we use the substitution method. Let $u = \log_e y$. Then, the differential $du = \frac{1}{y} dy$. The integral becomes:

$\int \frac{1}{u} du = \log_e|u| + C_2 = \log_e|\log_e y| + C_2$

Equating the results from both sides of the integration:

$\log_e|x| + C_1 = \log_e|\log_e y| + C_2$

Combine the constants and rearrange the equation to solve for $y$.

$\log_e|\log_e y| = \log_e|x| + (C_1 - C_2)$

Let $C = C_1 - C_2$ be an arbitrary constant.

$\log_e|\log_e y| = \log_e|x| + C$

Using logarithm properties, $\log_e a - \log_e b = \log_e(a/b)$:

$\log_e|\log_e y| - \log_e|x| = C$

$\log_e\left|\frac{\log_e y}{x}\right| = C$

Take the exponential of both sides:

$\left|\frac{\log_e y}{x}\right| = e^C$

Let $A = e^C$. Since $e^C > 0$, $A$ is a positive constant.

$\left|\frac{\log_e y}{x}\right| = A$

This implies $\frac{\log_e y}{x} = \pm A$. Let $K = \pm A$. $K$ is a non-zero constant.

$\frac{\log_e y}{x} = K$

$\log_e y = Kx$

Taking the exponential again to solve explicitly for $y$:

$y = e^{Kx}$

This solution was derived under the assumptions $x \neq 0$, $y > 0$, and $y \neq 1$.

Let's consider the cases where the denominators in the separation step might be zero.

If $y=1$, then $dy=0$ and $\log_e y = \log_e 1 = 0$. Substituting into the original equation $y \log_e y \, dx – x \, dy = 0$:

$1 \cdot 0 \, dx – x \cdot 0 = 0$

$0 = 0$

Thus, $y=1$ is a constant solution to the differential equation. This solution corresponds to the line $y=1$.

The general solution obtained by separation is $y = e^{Kx}$ for $K \neq 0$. If we allow the constant $K$ to be 0, the solution becomes $y = e^{0 \cdot x} = e^0 = 1$, which is the constant solution $y=1$. Therefore, the form $y = e^{Cx}$ where $C$ is any real constant encompasses all solutions obtained, including the singular solution $y=1$ (for $y>0$).

General Solution:

The general solution is $y = e^{Cx}$, where $C$ is an arbitrary real constant.

(Equivalently, for $y>0$, the solution can be written as $\log_e y = Cx$).

The given differential equation is:

$x^5 \frac{dy}{dx} = -y^5$

This is a variable separable differential equation. We can rearrange the terms to separate the variables. Assuming $x \neq 0$ and $y \neq 0$, divide both sides by $x^5 y^5$:

$\frac{1}{y^5} dy = -\frac{1}{x^5} dx$

$y^{-5} dy = -x^{-5} dx$

Integrate both sides of the separated equation:

$\int y^{-5} dy = \int -x^{-5} dx$

Evaluate the integrals using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1}$ (for $n \neq -1$).

For the left side ($\int y^{-5} dy$):

$\frac{y^{-5+1}}{-5+1} = \frac{y^{-4}}{-4} = -\frac{1}{4y^4}$

For the right side ($\int -x^{-5} dx$):

$- \int x^{-5} dx = - \left( \frac{x^{-5+1}}{-5+1} \right) = - \left( \frac{x^{-4}}{-4} \right) = - \left( -\frac{1}{4x^4} \right) = \frac{1}{4x^4}$

Adding the constant of integration $C'$ to one side, we get:

$-\frac{1}{4y^4} = \frac{1}{4x^4} + C'$

Rearrange the terms to get the general solution in implicit form:

$-\frac{1}{4y^4} - \frac{1}{4x^4} = C'$

Multiply the entire equation by -4 to simplify the form of the constant:

$(-4) \left( -\frac{1}{4y^4} - \frac{1}{4x^4} \right) = (-4) C'$

$\frac{1}{y^4} + \frac{1}{x^4} = -4C'$

Let $C = -4C'$. Since $C'$ is an arbitrary constant, $C$ is also an arbitrary constant.

$\frac{1}{x^4} + \frac{1}{y^4} = C$

This equation represents the general solution for $x \neq 0$ and $y \neq 0$.

We should also check if $y=0$ is a solution. If $y=0$, then $\frac{dy}{dx} = 0$. Substituting into the original equation:

$x^5 (0) = -(0)^5$

$0 = 0$

This is true for all $x$. So, $y=0$ is a solution to the differential equation. This is a singular solution not included in the implicit form obtained by separation of variables when $y \neq 0$.

The implicit form is typically presented as the general solution obtained by this method.

General Solution:

$\frac{1}{x^4} + \frac{1}{y^4} = C$

where $C$ is an arbitrary constant.

The solution $y=0$ is also a solution.

The given differential equation is:

$\frac{dy}{dx} = \sin^{-1} x$

This equation can be written in the separated variable form by multiplying both sides by $dx$:

$dy = \sin^{-1} x \, dx$

To find the general solution, we integrate both sides with respect to their respective variables:

$\int dy = \int \sin^{-1} x \, dx$

The left side integral is straightforward:

$\int dy = y$

For the right side integral, $\int \sin^{-1} x \, dx$, we use integration by parts. Recall the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

Let $u = \sin^{-1} x$ and $dv = dx$.

Then, $du = \frac{1}{\sqrt{1 - x^2}} dx$ and $v = \int dx = x$.

Applying the integration by parts formula:

$\int \sin^{-1} x \, dx = (\sin^{-1} x)(x) - \int x \left(\frac{1}{\sqrt{1 - x^2}}\right) dx$

$\int \sin^{-1} x \, dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} dx$

Now, we evaluate the remaining integral, $\int \frac{x}{\sqrt{1 - x^2}} dx$. We use a substitution.

Let $w = 1 - x^2$.

Then $dw = \frac{d}{dx}(1 - x^2) dx = -2x \, dx$.

So, $x \, dx = -\frac{1}{2} dw$.

Substituting these into the integral:

$\int \frac{x}{\sqrt{1 - x^2}} dx = \int \frac{-\frac{1}{2} dw}{\sqrt{w}} = -\frac{1}{2} \int w^{-1/2} dw$

Using the power rule for integration, $\int w^n dw = \frac{w^{n+1}}{n+1}$:

$-\frac{1}{2} \left( \frac{w^{-1/2 + 1}}{-1/2 + 1} \right) = -\frac{1}{2} \left( \frac{w^{1/2}}{1/2} \right) = -\frac{1}{2} (2 \sqrt{w}) = - \sqrt{w}$

Substitute back $w = 1 - x^2$:

$- \sqrt{1 - x^2}$

Now substitute this result back into the integration by parts equation:

$\int \sin^{-1} x \, dx = x \sin^{-1} x - (-\sqrt{1 - x^2}) + C'$

$\int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C'$

where $C'$ is the constant of integration.

Equating the results from both sides of the original integrated equation:

$y = x \sin^{-1} x + \sqrt{1 - x^2} + C$

where $C$ is an arbitrary constant.

General Solution:

$y = x \sin^{-1} x + \sqrt{1 - x^2} + C$

Given:

The differential equation is $e^x \tan y \, dx + (1 – e^x) \sec^2 y \, dy = 0$.

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is:

$e^x \tan y \, dx + (1 – e^x) \sec^2 y \, dy = 0$

We can rearrange the terms to separate the variables ($x$ terms with $dx$ and $y$ terms with $dy$). Move the first term to the right side:

$ (1 – e^x) \sec^2 y \, dy = -e^x \tan y \, dx $

Now, divide both sides by $(1 – e^x) \tan y$. This separates the $y$ terms with $dy$ and the $x$ terms with $dx$:

$ \frac{(1 – e^x) \sec^2 y}{(1 – e^x) \tan y} dy = \frac{-e^x \tan y}{(1 – e^x) \tan y} dx $

Simplifying both sides, assuming $(1 – e^x) \tan y \neq 0$:

$ \frac{\sec^2 y}{\tan y} dy = \frac{-e^x}{1 – e^x} dx $

Now that the variables are separated, we integrate both sides of the equation:

$ \int \frac{\sec^2 y}{\tan y} dy = \int \frac{-e^x}{1 – e^x} dx $

Let's evaluate each integral.

For the integral on the left side, $\int \frac{\sec^2 y}{\tan y} dy$, we use a substitution.

Let $u = \tan y$.

Differentiating $u$ with respect to $y$, we get $du = \sec^2 y \, dy$.

The left integral becomes $\int \frac{1}{u} du$. The integral of $\frac{1}{u}$ is $\log|u|$.

$ \int \frac{\sec^2 y}{\tan y} dy = \int \frac{1}{u} du = \log|u| + C_1 = \log|\tan y| + C_1 $

For the integral on the right side, $\int \frac{-e^x}{1 – e^x} dx$, we use another substitution.

Let $v = 1 – e^x$.

Differentiating $v$ with respect to $x$, we get $dv = \frac{d}{dx}(1 - e^x) \, dx = (0 - e^x) \, dx = -e^x \, dx$.

The right integral becomes $\int \frac{1}{v} dv$. The integral of $\frac{1}{v}$ is $\log|v|$.

$ \int \frac{-e^x}{1 – e^x} dx = \int \frac{1}{v} dv = \log|v| + C_2 = \log|1 – e^x| + C_2 $

Equating the results from the integration of both sides:

$ \log|\tan y| + C_1 = \log|1 – e^x| + C_2 $

Move the constant terms to one side:

$ \log|\tan y| - \log|1 – e^x| = C_2 - C_1 $

Let the combined constant be $C' = C_2 - C_1$.

$ \log|\tan y| - \log|1 – e^x| = C' $

Using the property of logarithms, $\log a - \log b = \log(a/b)$:

$ \log\left|\frac{\tan y}{1 – e^x}\right| = C' $

Exponentiate both sides (using base $e$) to remove the logarithm:

$ \left|\frac{\tan y}{1 – e^x}\right| = e^{C'} $

Let $K = e^{C'}$. Since $e^{C'}$ is always positive, $K$ is a positive constant ($K > 0$).

$ \frac{\tan y}{1 – e^x} = \pm K $

Let $C = \pm K$. Here, $C$ is a non-zero arbitrary constant. To include the potential singular solutions where $\tan y = 0$ (which are $y = n\pi$, valid solutions as $dy=0$ and $\tan(n\pi)=0$), we allow $C$ to be zero as well. Thus, $C$ is an arbitrary constant.

$ \tan y = C(1 – e^x) $

General Solution:

The general solution of the given differential equation is:

$ \tan y = C(1 – e^x) $

where $C$ is an arbitrary constant.

Given:

The differential equation is $(x^3 + x^2 + x + 1) \frac{dy}{dx} = 2x^2 + x$.

The initial condition is $y = 1$ when $x = 0$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is:

$(x^3 + x^2 + x + 1) \frac{dy}{dx} = 2x^2 + x$

Factor the denominator term $(x^3 + x^2 + x + 1)$:

$x^3 + x^2 + x + 1 = x^2(x + 1) + 1(x + 1) = (x^2 + 1)(x + 1)$

So the equation becomes:

$(x^2 + 1)(x + 1) \frac{dy}{dx} = 2x^2 + x$

Separate the variables. Divide both sides by $(x^2 + 1)(x + 1)$ and multiply by $dx$:

$dy = \frac{2x^2 + x}{(x^2 + 1)(x + 1)} dx$

Integrate both sides:

$\int dy = \int \frac{2x^2 + x}{(x^2 + 1)(x + 1)} dx$

The left side integral is $\int dy = y$.

For the right side integral, we use partial fraction decomposition for the integrand $\frac{2x^2 + x}{(x^2 + 1)(x + 1)}$.

Let $\frac{2x^2 + x}{(x^2 + 1)(x + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}$.

Multiply both sides by $(x^2 + 1)(x + 1)$:

$2x^2 + x = A(x^2 + 1) + (Bx + C)(x + 1)$

$2x^2 + x = Ax^2 + A + Bx^2 + Bx + Cx + C$

$2x^2 + x = (A + B)x^2 + (B + C)x + (A + C)$

Comparing the coefficients of the powers of $x$:

Coefficient of $x^2$: $A + B = 2$ ...(1)

Coefficient of $x$: $B + C = 1$ ...(2)

Constant term: $A + C = 0$ ...(3)

From equation (3), $C = -A$. Substitute this into equation (2):

$B + (-A) = 1 \implies B - A = 1$ ...(4)

Now we have a system of two equations with A and B:

$A + B = 2$ ...(1)

$B - A = 1$ ...(4)

Add equation (1) and equation (4):

$(A + B) + (B - A) = 2 + 1$

$2B = 3 \implies B = \frac{3}{2}$

Substitute $B = \frac{3}{2}$ into equation (1):

$A + \frac{3}{2} = 2 \implies A = 2 - \frac{3}{2} = \frac{4 - 3}{2} = \frac{1}{2}$

Substitute $A = \frac{1}{2}$ into equation (3):

$\frac{1}{2} + C = 0 \implies C = -\frac{1}{2}$

So, the partial fraction decomposition is:

$\frac{2x^2 + x}{(x^2 + 1)(x + 1)} = \frac{1/2}{x + 1} + \frac{(3/2)x - 1/2}{x^2 + 1} = \frac{1}{2(x + 1)} + \frac{3x}{2(x^2 + 1)} - \frac{1}{2(x^2 + 1)}$

Now, integrate the terms:

$\int \frac{1}{2(x + 1)} dx = \frac{1}{2} \int \frac{1}{x + 1} dx = \frac{1}{2} \log|x + 1|$

$\int \frac{3x}{2(x^2 + 1)} dx = \frac{3}{2} \int \frac{x}{x^2 + 1} dx$

Let $u = x^2 + 1$, $du = 2x \, dx$, so $x \, dx = \frac{1}{2} du$.

$\frac{3}{2} \int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{3}{4} \int \frac{1}{u} du = \frac{3}{4} \log|u| = \frac{3}{4} \log|x^2 + 1|$

Since $x^2 + 1$ is always positive, $\log|x^2 + 1| = \log(x^2 + 1)$.

$\int -\frac{1}{2(x^2 + 1)} dx = -\frac{1}{2} \int \frac{1}{x^2 + 1} dx = -\frac{1}{2} \tan^{-1} x$

Combining the integrals for the right side:

$\int \frac{2x^2 + x}{(x^2 + 1)(x + 1)} dx = \frac{1}{2} \log|x + 1| + \frac{3}{4} \log(x^2 + 1) - \frac{1}{2} \tan^{-1} x + K$

where $K$ is the constant of integration for the general solution.

So, the general solution is:

$y = \frac{1}{2} \log|x + 1| + \frac{3}{4} \log(x^2 + 1) - \frac{1}{2} \tan^{-1} x + K$

Now, apply the initial condition $y = 1$ when $x = 0$ to find the value of $K$.

$1 = \frac{1}{2} \log|0 + 1| + \frac{3}{4} \log(0^2 + 1) - \frac{1}{2} \tan^{-1} 0 + K$

$1 = \frac{1}{2} \log|1| + \frac{3}{4} \log(1) - \frac{1}{2} (0) + K$

$1 = \frac{1}{2} (0) + \frac{3}{4} (0) - 0 + K$

$1 = 0 + 0 - 0 + K$

$K = 1$

Substitute the value of $K$ back into the general solution to get the particular solution:

$y = \frac{1}{2} \log|x + 1| + \frac{3}{4} \log(x^2 + 1) - \frac{1}{2} \tan^{-1} x + 1$

Given:

The differential equation is $x(x^2 - 1) \frac{dy}{dx} = 1$.

The initial condition is $y = 0$ when $x = 2$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is:

$x(x^2 - 1) \frac{dy}{dx} = 1$

This is a variable separable differential equation. Separate the variables by dividing by $x(x^2-1)$ and multiplying by $dx$, assuming $x \neq 0$, $x \neq 1$, and $x \neq -1$:

$dy = \frac{1}{x(x^2 - 1)} dx$

Integrate both sides:

$\int dy = \int \frac{1}{x(x^2 - 1)} dx$

The left side integral is $\int dy = y$.

For the right side integral, we use partial fraction decomposition for the integrand $\frac{1}{x(x^2 - 1)}$. First, factor the denominator:

$x(x^2 - 1) = x(x - 1)(x + 1)$

Let $\frac{1}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$.

Multiply both sides by $x(x - 1)(x + 1)$:

$1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)$

$1 = A(x^2 - 1) + B(x^2 + x) + C(x^2 - x)$

$1 = (A + B + C)x^2 + (B - C)x - A$

Comparing the coefficients of the powers of $x$:

Coefficient of $x^2$: $A + B + C = 0$

Coefficient of $x$: $B - C = 0 \implies B = C$

Constant term: $-A = 1 \implies A = -1$

Substitute $A = -1$ and $B = C$ into the first equation:

$-1 + C + C = 0 \implies 2C = 1 \implies C = \frac{1}{2}$

Since $B = C$, $B = \frac{1}{2}$.

So, the partial fraction decomposition is:

$\frac{1}{x(x^2 - 1)} = \frac{-1}{x} + \frac{1/2}{x - 1} + \frac{1/2}{x + 1}$

Now, integrate the terms:

$\int \left(\frac{-1}{x} + \frac{1/2}{x - 1} + \frac{1/2}{x + 1}\right) dx = \int -\frac{1}{x} dx + \int \frac{1}{2(x - 1)} dx + \int \frac{1}{2(x + 1)} dx$

$= -\log|x| + \frac{1}{2} \log|x - 1| + \frac{1}{2} \log|x + 1| + K'$

where $K'$ is the constant of integration.

Using logarithm properties, $\frac{1}{2} \log|x - 1| + \frac{1}{2} \log|x + 1| = \frac{1}{2} (\log|x - 1| + \log|x + 1|) = \frac{1}{2} \log|(x - 1)(x + 1)| = \frac{1}{2} \log|x^2 - 1|$.

So the integral is:

$-\log|x| + \frac{1}{2} \log|x^2 - 1| + K'$

The general solution is:

$y = -\log|x| + \frac{1}{2} \log|x^2 - 1| + K$

(where $K$ is the arbitrary constant of integration)

Now, apply the initial condition $y = 0$ when $x = 2$ to find the value of $K$.

$0 = -\log|2| + \frac{1}{2} \log|2^2 - 1| + K$

$0 = -\log 2 + \frac{1}{2} \log|4 - 1| + K$

$0 = -\log 2 + \frac{1}{2} \log 3 + K$

$K = \log 2 - \frac{1}{2} \log 3$

Using logarithm properties, $K = \log 2 - \log 3^{1/2} = \log 2 - \log \sqrt{3} = \log \left(\frac{2}{\sqrt{3}}\right)$.

Substitute the value of $K$ back into the general solution to get the particular solution:

$y = -\log|x| + \frac{1}{2} \log|x^2 - 1| + \log \left(\frac{2}{\sqrt{3}}\right)$

We can also write this as:

$y = \log\left|\frac{\sqrt{x^2-1}}{x}\right| + \log \left(\frac{2}{\sqrt{3}}\right)$

$y = \log\left|\frac{\sqrt{x^2-1}}{x} \cdot \frac{2}{\sqrt{3}}\right|$

For $x=2$, the terms are positive, so the absolute values can be removed in the final form specific to the domain around $x=2$ where the solution is sought.

$y = \log\left(\frac{2\sqrt{x^2-1}}{x\sqrt{3}}\right)$

Given:

The differential equation is $\cos \left( \frac{dy}{dx} \right) = a$.

The initial condition is $y = 1$ when $x = 0$.

$a$ is a real constant ($a \in \mathbb{R}$).

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is:

$\cos \left( \frac{dy}{dx} \right) = a$

For this equation to have real solutions for $\frac{dy}{dx}$, the value of $a$ must be within the range of the cosine function, which is $[-1, 1]$.

Thus, a real solution exists only if $-1 \leq a \leq 1$.

If $-1 \leq a \leq 1$, we can take the inverse cosine (arccosine) of both sides. The general solution for $\cos \theta = a$ is $\theta = 2n\pi \pm \cos^{-1}(a)$, where $n$ is an integer and $\cos^{-1}(a)$ is the principal value in $[0, \pi]$.

So, we have:

Let $C_1 = 2n\pi \pm \cos^{-1}(a)$. Since $a$ is a constant and $n$ is an integer, $C_1$ is a constant value for a fixed integer $n$ and choice of sign.

The differential equation simplifies to:

$\frac{dy}{dx} = C_1$

This is a variable separable equation. Separate the variables by multiplying by $dx$:

$dy = C_1 \, dx$

Integrate both sides:

$\int dy = \int C_1 \, dx$

$y = C_1 x + C_2$

where $C_2$ is the constant of integration.

Substituting $C_1$ back, the general solution is:

$y = (2n\pi \pm \cos^{-1}(a)) x + C_2$

where $n$ is an integer and $C_2$ is an arbitrary constant.

Now, apply the initial condition $y = 1$ when $x = 0$ to find the value of $C_2$.

Substitute the value of $C_2$ back into the general solution to get the particular solution:

$y = (2n\pi \pm \cos^{-1}(a)) x + 1$

This is a family of particular solutions, one for each integer value of $n$ and choice of sign.

This solution is valid only if $-1 \leq a \leq 1$.

Given:

The differential equation is $\frac{dy}{dx} = y \tan x$.

The initial condition is $y = 1$ when $x = 0$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is:

$\frac{dy}{dx} = y \tan x$

This is a variable separable differential equation. Assuming $y \neq 0$, separate the variables by dividing by $y$ and multiplying by $dx$:

$\frac{dy}{y} = \tan x \, dx$

Integrate both sides:

$\int \frac{dy}{y} = \int \tan x \, dx$

Evaluate the integrals.

The left side integral is:

$\int \frac{dy}{y} = \log|y| + C_1$

The right side integral is:

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} dx$

Let $u = \cos x$, then $du = -\sin x \, dx$. So $\sin x \, dx = -du$.

$\int \frac{-du}{u} = -\int \frac{1}{u} du = -\log|u| + C_2 = -\log|\cos x| + C_2$

Equating the results of the integration:

$\log|y| + C_1 = -\log|\cos x| + C_2$

Rearrange the terms and combine the constants:

$\log|y| + \log|\cos x| = C_2 - C_1$

$\log|y| + \log|\cos x| = C$

where $C = C_2 - C_1$ is the arbitrary constant of integration.

Using the logarithm property $\log a + \log b = \log(ab)$:

$\log|y \cos x| = C$

Take the exponential of both sides to solve for $y$:

$|y \cos x| = e^C$

Let $A = \pm e^C$. Since $e^C > 0$, $A$ is a non-zero arbitrary constant.

$y \cos x = A$

$y = \frac{A}{\cos x} = A \sec x$

The case $y=0$ is also a solution, which is included when $A=0$. So, the general solution is $y = C \sec x$, where $C$ is an arbitrary constant.

Now, apply the initial condition $y = 1$ when $x = 0$ to find the value of $C$.

Substitute the value of $C$ back into the general solution to obtain the particular solution:

$y = 1 \cdot \sec x$

$y = \sec x$

Given:

The differential equation is $y’ = e^x \sin x$, which can be written as $\frac{dy}{dx} = e^x \sin x$.

The curve passes through the point (0, 0), which means $y = 0$ when $x = 0$.

To Find:

The equation of the curve (the particular solution of the differential equation) passing through the point (0, 0).

Solution:

The given differential equation is a variable separable equation:

$\frac{dy}{dx} = e^x \sin x$

Separate the variables:

$dy = e^x \sin x \, dx$

Integrate both sides to find the general solution:

$\int dy = \int e^x \sin x \, dx$

The left side integral is $\int dy = y$.

For the right side integral, let $I = \int e^x \sin x \, dx$. We use integration by parts, $\int u \, dv = uv - \int v \, du$.

Let $u = \sin x$ and $dv = e^x \, dx$.

Then $du = \cos x \, dx$ and $v = \int e^x \, dx = e^x$.

$I = e^x \sin x - \int e^x \cos x \, dx$

Now, integrate $\int e^x \cos x \, dx$ using integration by parts again.

Let $u = \cos x$ and $dv = e^x \, dx$.

Then $du = -\sin x \, dx$ and $v = e^x$.

$\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$

Substitute this back into the equation for $I$:

$I = e^x \sin x - (e^x \cos x + I)$

$I = e^x \sin x - e^x \cos x - I$

Add $I$ to both sides:

$2I = e^x \sin x - e^x \cos x$

$2I = e^x (\sin x - \cos x)$

Divide by 2 and add the constant of integration $C$:

$I = \frac{1}{2} e^x (\sin x - \cos x) + C$

So, the general solution is:

$y = \frac{1}{2} e^x (\sin x - \cos x) + C$

Now, apply the initial condition that the curve passes through the point (0, 0), i.e., $y = 0$ when $x = 0$, to find the value of $C$.

Substitute the value of $C$ back into the general solution to get the particular solution (the equation of the curve):

$y = \frac{1}{2} e^x (\sin x - \cos x) + \frac{1}{2}$

This can also be written as:

$y = \frac{1}{2} (e^x (\sin x - \cos x) + 1)$

Given:

The differential equation is $xy \frac{dy}{dx} = (x + 2) (y + 2)$.

The curve passes through the point (1, –1), which means $y = -1$ when $x = 1$.

To Find:

The equation of the solution curve passing through the point (1, –1).

Solution:

The given differential equation is:

$xy \frac{dy}{dx} = (x + 2) (y + 2)$

This is a variable separable differential equation. Separate the variables by dividing both sides by $xy(y+2)$ and multiplying by $dx$. Since the initial condition is at $(1, -1)$, where $x=1 \neq 0$, $y=-1 \neq 0$, and $y+2 = 1 \neq 0$, we can perform the separation.

$\frac{y \, dy}{y + 2} = \frac{x + 2}{x} dx$

Integrate both sides of the separated equation:

$\int \frac{y}{y + 2} dy = \int \frac{x + 2}{x} dx$

Evaluate the integrals.

For the left side integral, we can rewrite the integrand:

$\frac{y}{y + 2} = \frac{y + 2 - 2}{y + 2} = 1 - \frac{2}{y + 2}$

$\int \left(1 - \frac{2}{y + 2}\right) dy = \int 1 \, dy - \int \frac{2}{y + 2} dy = y - 2 \int \frac{1}{y + 2} dy = y - 2 \log|y + 2| + C_1$

For the right side integral, we rewrite the integrand:

$\frac{x + 2}{x} = \frac{x}{x} + \frac{2}{x} = 1 + \frac{2}{x}$

$\int \left(1 + \frac{2}{x}\right) dx = \int 1 \, dx + \int \frac{2}{x} dx = x + 2 \int \frac{1}{x} dx = x + 2 \log|x| + C_2$

Equating the results from both sides of the integration:

$y - 2 \log|y + 2| = x + 2 \log|x| + C'$

where $C' = C_2 - C_1$ is the arbitrary constant of integration.

Rearrange the terms:

$y - x = 2 \log|x| + 2 \log|y + 2| + C'$

$y - x = 2 (\log|x| + \log|y + 2|) + C'$

Using the logarithm property $\log a + \log b = \log(ab)$:

$y - x = 2 \log|x(y + 2)| + C'$

Now, apply the initial condition $y = -1$ when $x = 1$ to find the value of $C'$.

Substitute the value of $C'$ back into the general solution to get the particular solution:

$y - x = 2 \log|x(y + 2)| - 2$

We can rearrange this slightly:

$y - x + 2 = 2 \log|x(y + 2)|$

Or, by dividing by 2 and taking the exponential:

$\frac{y - x + 2}{2} = \log|x(y + 2)|$

$e^{\frac{y - x + 2}{2}} = |x(y + 2)|$

Since the point $(1, -1)$ is on the curve, $x=1 > 0$ and $y+2 = 1 > 0$, the term $x(y+2)$ is positive near this point, so we can remove the absolute value for the particular solution around this point:

$e^{\frac{y - x + 2}{2}} = x(y + 2)$

Given:

The curve passes through the point (0, –2).

At any point $(x, y)$ on the curve, the product of the slope of its tangent and the $y$ coordinate is equal to the $x$ coordinate.

To Find:

The equation of the curve passing through the point (0, –2).

Solution:

Let the equation of the curve be $y = f(x)$.

The slope of the tangent at any point $(x, y)$ on the curve is given by $\frac{dy}{dx}$.

According to the problem statement, the product of the slope of the tangent and the $y$ coordinate is equal to the $x$ coordinate.

This can be written as the differential equation:

$y \frac{dy}{dx} = x$

This is a variable separable differential equation. Separate the variables:

$y \, dy = x \, dx$

Integrate both sides to find the general solution:

$\int y \, dy = \int x \, dx$

Evaluate the integrals:

$\frac{y^2}{2} = \frac{x^2}{2} + C'$

where $C'$ is the constant of integration.

Multiply the entire equation by 2 to simplify the form:

$y^2 = x^2 + 2C'$

Let $C = 2C'$. Since $C'$ is an arbitrary constant, $C$ is also an arbitrary constant.

The general solution (the equation of the family of curves) is:

$y^2 = x^2 + C$

Now, we use the given condition that the curve passes through the point (0, –2). This means when $x = 0$, $y = -2$. Substitute these values into the general solution to find the value of $C$.

Substitute the value of $C$ back into the general solution to obtain the particular solution, which is the equation of the required curve:

$y^2 = x^2 + 4$

This equation can also be written as $y^2 - x^2 = 4$, which represents a hyperbola.

Given:

The curve passes through the point (–2, 1).

At any point $(x, y)$ on the curve, the slope of the tangent is twice the slope of the line segment joining $(x, y)$ to $(-4, -3)$.

To Find:

The equation of the curve passing through the point (–2, 1).

Solution:

Let the slope of the tangent at any point $(x, y)$ on the curve be $\frac{dy}{dx}$.

The slope of the line segment joining the point of contact $(x, y)$ to the point $(-4, -3)$ is given by the formula $\frac{y_2 - y_1}{x_2 - x_1}$.

Slope of the line segment = $\frac{-3 - y}{-4 - x} = \frac{-(y + 3)}{-(x + 4)} = \frac{y + 3}{x + 4}$, provided $x \neq -4$.

According to the problem statement, the slope of the tangent is twice the slope of this line segment:

$\frac{dy}{dx} = 2 \times \left( \frac{y + 3}{x + 4} \right)$

This is a variable separable differential equation. Separate the variables by dividing by $(y+3)$ and multiplying by $dx$, assuming $y \neq -3$ and $x \neq -4$:

$\frac{dy}{y + 3} = \frac{2}{x + 4} dx$

Integrate both sides:

$\int \frac{dy}{y + 3} = \int \frac{2}{x + 4} dx$

Evaluate the integrals:

$\int \frac{dy}{y + 3} = \log|y + 3| + C_1$

$\int \frac{2}{x + 4} dx = 2 \int \frac{1}{x + 4} dx = 2 \log|x + 4| + C_2$

Equating the results:

$\log|y + 3| = 2 \log|x + 4| + C'$

where $C' = C_2 - C_1$ is the arbitrary constant of integration.

Using logarithm properties, $2 \log|x + 4| = \log|(x + 4)^2|$:

$\log|y + 3| = \log|(x + 4)^2| + C'$

Rearrange the terms:

$\log|y + 3| - \log|(x + 4)^2| = C'$

Using the logarithm property $\log a - \log b = \log(a/b)$:

$\log\left|\frac{y + 3}{(x + 4)^2}\right| = C'$

Take the exponential of both sides:

$\left|\frac{y + 3}{(x + 4)^2}\right| = e^{C'}$

Let $C = \pm e^{C'}$. Since $e^{C'} > 0$, $C$ is a non-zero arbitrary constant.

$\frac{y + 3}{(x + 4)^2} = C$

$y + 3 = C(x + 4)^2$

This is the general solution (the equation of the family of curves).

Now, we use the given condition that the curve passes through the point (–2, 1). This means when $x = -2$, $y = 1$. Substitute these values into the general solution to find the value of $C$.

Substitute the value of $C = 1$ back into the general solution to obtain the particular solution, which is the equation of the required curve:

$y + 3 = 1 \cdot (x + 4)^2$

$y + 3 = (x + 4)^2$

This can also be written as $y = (x+4)^2 - 3 = x^2 + 8x + 16 - 3 = x^2 + 8x + 13$. This is a parabola.

Given:

The volume of the spherical balloon changes at a constant rate.

Initial condition: Radius $r = 3$ units when time $t = 0$ seconds.

After 3 seconds: Radius $r = 6$ units when time $t = 3$ seconds.

To Find:

The radius of the balloon after $t$ seconds, i.e., $r(t)$.

Solution:

Let $V$ be the volume of the spherical balloon and $r$ be its radius at time $t$.

The volume of a sphere is given by the formula $V = \frac{4}{3}\pi r^3$.

The problem states that the volume changes at a constant rate. This can be written as a differential equation:

We need to express $\frac{dV}{dt}$ in terms of the radius $r$. Differentiate the volume formula with respect to time $t$ using the chain rule:

Substitute this into the differential equation $\frac{dV}{dt} = k$:

This is a variable separable differential equation. Separate the variables:

Integrate both sides:

We use the given conditions to find the values of the constants $k$ and $C$.

Using the initial condition: $r = 3$ when $t = 0$. Substitute these values into equation (i):

Using the condition at $t=3$: $r = 6$ when $t = 3$. Substitute these values and $C = 36\pi$ into equation (i):

Now, substitute the values of $k = 84\pi$ and $C = 36\pi$ back into equation (i):

We need to find the radius $r$ as a function of $t$. Divide the entire equation by $\frac{4}{3}\pi$:

Take the cube root of both sides to find $r$:

$r = (63t + 27)^{1/3}$

Conclusion:

The radius of the balloon after $t$ seconds is given by the formula $r = (63t + 27)^{1/3}$ units.

Given:

Principal $P$ increases continuously at the rate of $r\%$ per year.

Initial condition: $P = \textsf{₹} 100$ when $t = 0$ years.

Second condition: $P = \textsf{₹} 200$ when $t = 10$ years.

Value of $\log_e 2 = 0.6931$.

To Find:

The value of the rate $r$ (in percent).

Solution:

Let $P(t)$ be the principal amount at time $t$. The statement "principal increases continuously at the rate of $r\%$ per year" translates to a differential equation where the rate of change of principal with respect to time is proportional to the current principal.

The rate of increase is $\frac{r}{100}$ times the principal $P$. So, we have:

This is a variable separable differential equation. Assuming $P > 0$, we can separate the variables:

Integrate both sides:

Since the principal $P$ is always positive, we have $\log|P| = \log P$.

(Note: The integral $\int \frac{1}{P} dP$ is $\log_e|P|$, which we denote as $\log|P|$ as per instruction).

Apply the initial condition: $P = 100$ when $t = 0$. Substitute these values into the general solution:

Substitute the value of $C'$ back into the general solution:

Rearrange the terms:

Using the logarithm property $\log a - \log b = \log(a/b)$:

Now, apply the second condition: $P = 200$ when $t = 10$. Substitute these values into equation (iii):

Solve for $r$:

The problem provides the value $\log_e 2 = 0.6931$. This means the logarithm base is $e$. We will use this numerical value for $\log 2$.

The rate is given as $r\%$ per year. Therefore, the value of $r$ is $6.931$.

Answer:

The value of $r$ is $6.931$.

Given:

Principal $P$ increases continuously at the rate of 5% per year.

Rate of interest $r = 5\% = 0.05$ per year.

Initial principal amount $P_0 = \textsf{₹} 1000$ when time $t = 0$.

Time period $t = 10$ years.

Value of $e^{0.5} = 1.648$.

To Find:

The amount the principal will be worth after 10 years ($P(10)$).

Solution:

Let $P(t)$ be the principal amount at time $t$. The principal increases continuously at the rate of 5% per year, which means the rate of change of principal with respect to time is proportional to the principal itself.

The differential equation describing this situation is:

This is a variable separable differential equation. Assuming $P > 0$, we can separate the variables:

Integrate both sides:

Since the principal $P$ is always positive, we have $\log|P| = \log P$.

Apply the initial condition: $P = 1000$ when $t = 0$. Substitute these values into the general solution:

Substitute the value of $C'$ back into the general solution:

Rearrange the terms:

Using the logarithm property $\log a - \log b = \log(a/b)$:

Take the exponential of both sides (assuming the logarithm base is $e$):

Now, find the amount after $t = 10$ years. Substitute $t=10$ into equation (iv):

The problem provides the value $e^{0.5} = 1.648$. Substitute this value:

Answer:

The amount will be worth $\textsf{₹} 1648$ after 10 years.

Given:

Initial bacteria count $N_0 = 1,00,000$ at $t=0$.

Bacteria count increased by 10% in 2 hours, so $N = 1,00,000 \times (1 + 0.10) = 1,10,000$ at $t=2$ hours.

Target bacteria count $N_{target} = 2,00,000$.

Rate of growth of bacteria is proportional to the number present.

$\log_e 2 = 0.6931$.

To Find:

The time $t$ in hours when the bacteria count reaches 2,00,000.

Solution:

Let $N(t)$ be the number of bacteria at time $t$. The statement that the rate of growth of bacteria is proportional to the number present translates to the differential equation:

This is a variable separable differential equation. Separate the variables:

Integrate both sides:

Since the bacteria count $N$ is positive, we can write $\log N = kt + C'$.

Exponentiating both sides (assuming $\log$ here represents the natural logarithm, as suggested by the integral form and the given $\log_e$ value):

Let $C = e^{C'}$. Then the general solution is $N(t) = C e^{kt}$.

Apply the initial condition: $N = 1,00,000$ at $t = 0$.

So the formula for the bacteria count is $N(t) = 1,00,000 e^{kt}$.

Use the second condition: $N = 1,10,000$ at $t = 2$ hours.

Divide by $1,00,000$:

We want to find the time $t$ when the count reaches $2,00,000$. Let this time be $T$.

Divide by $1,00,000$:

Take the natural logarithm (log base $e$) of equations (iii) and (iv).

From (iii): $\log_e(1.1) = \log_e(e^{2k}) = 2k$

From (iv): $\log_e 2 = \log_e(e^{kT}) = kT$

We have the system of equations:

From equation (v), $k = \frac{\log_e(1.1)}{2}$. Substitute this into equation (vi):

Solve for $T$:

Using the given value $\log_e 2 = 0.6931$, the time is:

Assuming the use of the standard $\log$ notation for natural logarithm in the final answer format:

Answer:

The count will reach 2,00,000 in $\frac{2 \log 2}{\log(1.1)}$ hours.

Using the provided value $\log_e 2 = 0.6931$, the time is $\frac{2 \times 0.6931}{\log_e(1.1)} = \frac{1.3862}{\log_e(1.1)}$ hours.

Given:

The differential equation is $\frac{dy}{dx} = e^{x+y}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation can be written as:

This is a variable separable differential equation. We can separate the variables by dividing by $e^y$ (or multiplying by $e^{-y}$) and multiplying by $dx$:

Integrate both sides of the separated equation:

Evaluate the integrals.

For the left side integral, $\int e^{-y} dy$, let $u = -y$, so $du = -dy$. Thus $dy = -du$.

For the right side integral, $\int e^x dx$:

Equating the results of the integration:

Rearrange the terms to find the general solution. Move the terms involving $x$ and $y$ to one side and the constants to the other side:

Multiply the entire equation by -1:

Let $C = C_1 - C_2$. Since $C_1$ and $C_2$ are arbitrary constants, $C$ is also an arbitrary constant.

Comparing this general solution with the given options:

(A) $e^x + e^{–y} = C$

(B) $e^x + e^{y} = C$

(C) $e^{–x} + e^{y} = C$

(D) $e^{–x} + e^{–y} = C$

The obtained solution matches option (A).

Answer:

The correct option is (A).

The given differential equation is:

We can rewrite this as:

To Show it is Homogeneous:

Let $f(x, y) = x + 2y$ and $g(x, y) = x – y$.

Then $\frac{dy}{dx} = \frac{f(x, y)}{g(x, y)}$.

Consider $f(\lambda x, \lambda y) = (\lambda x) + 2(\lambda y) = \lambda (x + 2y) = \lambda f(x, y)$.

Consider $g(\lambda x, \lambda y) = (\lambda x) – (\lambda y) = \lambda (x – y) = \lambda g(x, y)$.

Since $f(\lambda x, \lambda y) = \lambda^1 f(x, y)$ and $g(\lambda x, \lambda y) = \lambda^1 g(x, y)$, both $f(x, y)$ and $g(x, y)$ are homogeneous functions of degree 1.

Thus, the ratio $\frac{f(x, y)}{g(x, y)}$ is a homogeneous function of degree $1-1=0$.

Hence, the given differential equation is homogeneous.

To Solve the Differential Equation:

The equation is $\frac{dy}{dx} = \frac{x + 2y}{x – y}$.

Substitute $y = vx$.

Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting these into equation (i):

Separate the variables:

Separating the variables $v$ and $x$:

Integrate both sides:

For the integral on the left, we can write the numerator as a combination of the derivative of the denominator and a constant.

Derivative of $v^2 + v + 1$ is $2v + 1$.

We want $1 - v = A(2v + 1) + B$.

$1 - v = 2Av + A + B$.

Comparing coefficients of $v$: $-1 = 2A \implies A = -\frac{1}{2}$.

Comparing constant terms: $1 = A + B \implies 1 = -\frac{1}{2} + B \implies B = 1 + \frac{1}{2} = \frac{3}{2}$.

So, $1 - v = -\frac{1}{2}(2v + 1) + \frac{3}{2}$.

The integral becomes:

For the first integral on the left, $\int \frac{2v + 1}{v^2 + v + 1} dv = \log|v^2 + v + 1|$ (since the numerator is the derivative of the denominator).

For the second integral on the left, complete the square in the denominator: $v^2 + v + 1 = v^2 + v + (\frac{1}{2})^2 - (\frac{1}{2})^2 + 1 = (v + \frac{1}{2})^2 + \frac{3}{4} = (v + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$.

The integral $\int \frac{1}{(v + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} dv$ is of the form $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$, where $u = v + \frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$.

The integral on the right is $\int \frac{1}{x} dx = \log|x|$.

Combining the results:

Multiply by 2:

Let $C_1 = 2C$.

Substitute back $v = \frac{y}{x}$:

Let $C_2 = -C_1$.

The general solution is:

(Note: $x^2+xy+y^2 = (x+\frac{y}{2})^2 + \frac{3y^2}{4} > 0$ for real $x, y$ unless $x=y=0$, so $|x^2+xy+y^2|$ can be written as $(x^2+xy+y^2)$.)

Final Solution Summary:

The differential equation $(x – y) \frac{dy}{dx} = x + 2y$ was shown to be homogeneous by demonstrating that $\frac{dy}{dx}$ can be expressed as a function of $\frac{y}{x}$.

Using the substitution $y=vx$, the equation was transformed into a separable equation in terms of $v$ and $x$.

Integration of the separated equation led to the solution.

The general solution is:

where C is an arbitrary constant.

The given differential equation is:

We can rewrite this as:

Dividing the numerator and denominator by $x$:

This equation is in the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$, where $f\left(\frac{y}{x}\right) = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + 1}{\cos \left( \frac{y}{x} \right)}$.

To Show it is Homogeneous:

Let $F(x, y) = \frac{y \cos \left( \frac{y}{x} \right) + x}{x \cos \left( \frac{y}{x} \right)}$.

Consider $F(\lambda x, \lambda y) = \frac{(\lambda y) \cos \left( \frac{\lambda y}{\lambda x} \right) + (\lambda x)}{(\lambda x) \cos \left( \frac{\lambda y}{\lambda x} \right)}$.

Since $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$, the function $F(x, y)$ is a homogeneous function of degree 0.

Hence, the given differential equation is homogeneous.

To Solve the Differential Equation:

The equation is $\frac{dy}{dx} = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + 1}{\cos \left( \frac{y}{x} \right)}$.

Substitute $y = vx$.

Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting these into equation (i):

Subtract $v$ from both sides:

Separating the variables $v$ and $x$:

Integrate both sides:

Substitute back $v = \frac{y}{x}$:

where $C$ is the arbitrary constant of integration.

Final Solution Summary:

The differential equation $x \cos x \left( \frac{y}{x} \right) \frac{dy}{dx} = y \cos \left( \frac{y}{x} \right) + x$ was shown to be homogeneous as $\frac{dy}{dx}$ is a homogeneous function of degree 0.

Using the substitution $y=vx$, the equation was transformed into a separable equation in terms of $v$ and $x$.

Integration of the separated equation $\cos(v) dv = \frac{1}{x} dx$ led to the solution.

The general solution is:

where C is an arbitrary constant.

Given:

The differential equation is $2y\; e^{\frac{x}{y}} dx + \left( y − 2x\; e^{\frac{x}{y}} \right) dy = 0$.

The particular solution is required for the condition $x = 0$ when $y = 1$.

To Show:

The given differential equation is homogeneous.

To Find:

The particular solution of the differential equation.

Solution:

The given differential equation is:

We can rewrite this equation as:

$\left( y − 2x\; e^{\frac{x}{y}} \right) dy = -2y\; e^{\frac{x}{y}} dx$

$\frac{dy}{dx} = \frac{-2y\; e^{\frac{x}{y}}}{y − 2x\; e^{\frac{x}{y}}}$

$\frac{dy}{dx} = \frac{2y\; e^{\frac{x}{y}}}{2x\; e^{\frac{x}{y}} - y}$

Let $f(x, y) = \frac{2y\; e^{\frac{x}{y}}}{2x\; e^{\frac{x}{y}} - y}$.

To check for homogeneity, we evaluate $f(kx, ky)$:

$f(kx, ky) = \frac{2(ky)\; e^{\frac{kx}{ky}}}{2(kx)\; e^{\frac{kx}{ky}} - (ky)}$

$f(kx, ky) = \frac{2ky\; e^{\frac{x}{y}}}{2kx\; e^{\frac{x}{y}} - ky}$

Take $k$ common from the numerator and denominator:

$f(kx, ky) = \frac{k(2y\; e^{\frac{x}{y}})}{k(2x\; e^{\frac{x}{y}} - y)}$

$f(kx, ky) = \frac{2y\; e^{\frac{x}{y}}}{2x\; e^{\frac{x}{y}} - y}$

Since $f(kx, ky) = f(x, y)$, the given differential equation is a homogeneous differential equation.

To find the particular solution, we use the substitution $x = vy$.

Differentiating with respect to $y$, we get:

$\frac{dx}{dy} = v + y\frac{dv}{dy}$

From equation (1), let's write it as $\frac{dx}{dy}$:

$\frac{dx}{dy} = -\frac{y − 2x\; e^{\frac{x}{y}}}{2y\; e^{\frac{x}{y}}}$

Substitute $x = vy$ into this equation:

$\frac{dx}{dy} = -\frac{y − 2(vy)\; e^{\frac{vy}{y}}}{2y\; e^{\frac{vy}{y}}}$

$\frac{dx}{dy} = -\frac{y − 2vy\; e^{v}}{2y\; e^{v}}$

$\frac{dx}{dy} = -\frac{y(1 − 2v\; e^{v})}{y(2\; e^{v})}$

$\frac{dx}{dy} = -\frac{1 − 2v\; e^{v}}{2\; e^{v}} = \frac{2v\; e^{v} - 1}{2\; e^{v}} = v - \frac{1}{2e^{v}}$

Now equate the two expressions for $\frac{dx}{dy}$:

$v + y\frac{dv}{dy} = v - \frac{1}{2e^{v}}$

$y\frac{dv}{dy} = - \frac{1}{2e^{v}}$

Separate the variables:

$2e^{v}\; dv = -\frac{1}{y}\; dy$

Integrate both sides:

$\int 2e^{v}\; dv = \int -\frac{1}{y}\; dy$

$2\int e^{v}\; dv = -\int \frac{1}{y}\; dy$

$2e^{v} = -\log|y| + C$

Substitute back $v = \frac{x}{y}$:

This is the general solution.

Now, we use the given condition $x = 0$ when $y = 1$ to find the value of $C$. Substitute these values into equation (3):

$2e^{\frac{0}{1}} = -\log|1| + C$

$2e^{0} = -\log(1) + C$

$2(1) = -0 + C$

$2 = C$

Substitute the value of $C$ from equation (4) into equation (3) to get the particular solution:

$2e^{\frac{x}{y}} = -\log|y| + 2$

The particular solution can be written as:

$\mathbf{2e^{\frac{x}{y}} + \log|y| = 2}$

Given:

The slope of the tangent at any point $(x, y)$ on the curve is given by $\frac{x^2 + y^2}{2xy}$.

To Show:

The family of curves is given by $x^2 - y^2 = cx$.

Solution:

The slope of the tangent at any point $(x, y)$ on a curve is given by $\frac{dy}{dx}$.

So, the differential equation representing the family of curves is:

Let $f(x, y) = \frac{x^2 + y^2}{2xy}$.

To check if the differential equation is homogeneous, we evaluate $f(kx, ky)$:

$f(kx, ky) = \frac{(kx)^2 + (ky)^2}{2(kx)(ky)}$

$f(kx, ky) = \frac{k^2x^2 + k^2y^2}{2k^2xy}$

$f(kx, ky) = \frac{k^2(x^2 + y^2)}{k^2(2xy)}$

$f(kx, ky) = \frac{x^2 + y^2}{2xy}$

Since $f(kx, ky) = f(x, y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous differential equation, we use the substitution $y = vx$.

Differentiating $y = vx$ with respect to $x$, we get:

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into equation (1):

$v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)}$

$v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2vx^2}$

$v + x\frac{dv}{dx} = \frac{x^2(1 + v^2)}{x^2(2v)}$

$v + x\frac{dv}{dx} = \frac{1 + v^2}{2v}$

$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v$

$x\frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v}$

$x\frac{dv}{dx} = \frac{1 - v^2}{2v}$

Separate the variables:

$\frac{2v}{1 - v^2}\; dv = \frac{1}{x}\; dx$

Integrate both sides:

$\int \frac{2v}{1 - v^2}\; dv = \int \frac{1}{x}\; dx$

For the left side integral, let $u = 1 - v^2$, so $du = -2v\; dv$, which means $2v\; dv = -du$.

$\int \frac{-du}{u} = \int \frac{1}{x}\; dx$

$-\int \frac{1}{u}\; du = \int \frac{1}{x}\; dx$

$-\log|u| = \log|x| + \log|C|$

Substitute back $u = 1 - v^2$:

$-\log|1 - v^2| = \log|x| + \log|C|$

$\log|1 - v^2|^{-1} = \log|Cx|$

$\log\left|\frac{1}{1 - v^2}\right| = \log|Cx|$

$\left|\frac{1}{1 - v^2}\right| = |Cx|$

$\frac{1}{1 - v^2} = \pm Cx$

Let the constant $\pm C$ be denoted by $K$.

$\frac{1}{1 - v^2} = Kx$

$1 = Kx(1 - v^2)$

Substitute back $v = \frac{y}{x}$:

$1 = Kx\left(1 - \left(\frac{y}{x}\right)^2\right)$

$1 = Kx\left(1 - \frac{y^2}{x^2}\right)$

$1 = Kx\left(\frac{x^2 - y^2}{x^2}\right)$

$1 = K\frac{x^2 - y^2}{x}$

$x = K(x^2 - y^2)$

$\frac{x}{K} = x^2 - y^2$

Let $\frac{1}{K} = c$, where $c$ is an arbitrary constant.

$x^2 - y^2 = cx$

This is the required family of curves.

Given:

The differential equation is $(x^2 + xy) dy = (x^2 + y^2) dx$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation is $(x^2 + xy) dy = (x^2 + y^2) dx$.

We can rewrite this equation in the form $\frac{dy}{dx} = f(x,y)$:

$\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$

Let $f(x,y) = \frac{x^2 + y^2}{x^2 + xy}$.

To check if the differential equation is homogeneous, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:

$f(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{(\lambda x)^2 + (\lambda x)(\lambda y)}$

$f(\lambda x, \lambda y) = \frac{\lambda^2 x^2 + \lambda^2 y^2}{\lambda^2 x^2 + \lambda^2 xy}$

Factor out $\lambda^2$ from the numerator and the denominator:

$f(\lambda x, \lambda y) = \frac{\lambda^2 (x^2 + y^2)}{\lambda^2 (x^2 + xy)}$

Cancel $\lambda^2$ (since $\lambda \neq 0$):

$f(\lambda x, \lambda y) = \frac{x^2 + y^2}{x^2 + xy}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the function $f(x,y)$ is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.

To find the general solution of this homogeneous equation, we use the substitution:

$y = vx$

Where $v$ is a function of $x$.

Differentiating both sides with respect to $x$ using the product rule, we get:

$\frac{dy}{dx} = \frac{d}{dx}(vx) = v \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(v)$

$\frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx}$

$\frac{dy}{dx} = v + x \frac{dv}{dx}$

Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$:

$v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{x^2 + x(vx)}$

$v + x \frac{dv}{dx} = \frac{x^2 + v^2 x^2}{x^2 + vx^2}$

Factor out $x^2$ from the numerator and the denominator on the right side:

$v + x \frac{dv}{dx} = \frac{x^2(1 + v^2)}{x^2(1 + v)}$

Cancel out $x^2$ (assuming $x \neq 0$):

$v + x \frac{dv}{dx} = \frac{1 + v^2}{1 + v}$

Now, we separate the variables $v$ and $x$. Move $v$ to the right side:

$x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v$

Combine the terms on the right side by finding a common denominator:

$x \frac{dv}{dx} = \frac{1 + v^2 - v(1 + v)}{1 + v}$

$x \frac{dv}{dx} = \frac{1 + v^2 - v - v^2}{1 + v}$

Simplify the numerator:

$x \frac{dv}{dx} = \frac{1 - v}{1 + v}$

Separate the variables $v$ and $x$ by moving terms involving $v$ to the left side and terms involving $x$ to the right side:

$\frac{1 + v}{1 - v} dv = \frac{1}{x} dx$

Integrate both sides of the separated equation:

$\int \frac{1 + v}{1 - v} dv = \int \frac{1}{x} dx$

Consider the integral on the left side: $\int \frac{1 + v}{1 - v} dv$. We can rewrite the integrand by performing polynomial division or algebraic manipulation:

$\frac{1 + v}{1 - v} = \frac{-(v + 1)}{-(v - 1)} = \frac{v + 1}{v - 1}$

We can write $v+1 = (v-1) + 2$. So,

$\frac{v + 1}{v - 1} = \frac{(v - 1) + 2}{v - 1} = 1 + \frac{2}{v - 1}$

Alternatively, from $\frac{1+v}{1-v}$, we can write $1+v = 1+v-1+1 = (1-v) + 2$. So,

$\frac{1+v}{1-v} = \frac{(1-v)+2}{1-v} = \frac{1-v}{1-v} + \frac{2}{1-v} = 1 + \frac{2}{1-v}$.

$\int \frac{1 + v}{1 - v} dv = \int \left(1 + \frac{2}{1 - v}\right) dv = \int 1 \, dv + \int \frac{2}{1 - v} dv$

$= v + 2 \int \frac{1}{1 - v} dv$

For the integral $\int \frac{1}{1 - v} dv$, let $w = 1 - v$. Then $dw = -dv$, so $dv = -dw$.

$\int \frac{1}{w} (-dw) = -\int \frac{1}{w} dw = -\log|w| = -\log|1 - v|$.

So the left side integral is $v + 2(-\log|1 - v|) + C_1 = v - 2\log|1 - v| + C_1$.

The integral on the right side is $\int \frac{1}{x} dx = \log|x| + C_2$.

Equating the results from both sides of the integration:

$v - 2\log|1 - v| = \log|x| + C$, where $C = C_2 - C_1$ is the arbitrary constant of integration.

Now, substitute back $v = \frac{y}{x}$ to express the solution in terms of $x$ and $y$:

$\frac{y}{x} - 2\log\left|1 - \frac{y}{x}\right| = \log|x| + C$

$\frac{y}{x} - 2\log\left|\frac{x - y}{x}\right| = \log|x| + C$

Using the logarithm property $\log(a/b) = \log a - \log b$:

$\frac{y}{x} - 2(\log|x - y| - \log|x|) = \log|x| + C$

$\frac{y}{x} - 2\log|x - y| + 2\log|x| = \log|x| + C$}

Rearrange the terms to match the desired form. Move the $\log|x|$ term to the right side and others to the left:

$\frac{y}{x} - 2\log|x - y| = \log|x| - 2\log|x| + C$

$\frac{y}{x} - 2\log|x - y| = -\log|x| + C$

Rearrange to group logarithmic terms:

$-\frac{y}{x} + 2\log|x - y| + \log|x| = -C$

Let $C' = -C$, which is also an arbitrary constant.

$2\log|x - y| + \log|x| - \frac{y}{x} = C'$

Using the property $n \log A = \log A^n$ and $\log A + \log B = \log(AB)$:

$\log|(x - y)^2| + \log|x| - \frac{y}{x} = C'$

$\log(|x| (x - y)^2) - \frac{y}{x} = C'$

$\log(|x| (x - y)^2) = C' + \frac{y}{x}$

Exponentiate both sides:

$|x| (x - y)^2 = e^{C' + y/x}$

$|x| (x - y)^2 = e^{C'} e^{y/x}$

Let $K = e^{C'}$, which is a positive arbitrary constant. We can absorb the absolute value of $x$ and the constant $K$ into a single arbitrary constant $C_{\text{new}}$.

$(x - y)^2 = \frac{K}{|x|} x e^{y/x}$ - This is not quite the target form.

Let's go back to $\log\left|\frac{x}{(x - y)^2}\right| - \frac{y}{x} = C$.

$\log\left|\frac{x}{(x - y)^2}\right| = C + \frac{y}{x}$

Exponentiate both sides:

$\left|\frac{x}{(x - y)^2}\right| = e^{C + y/x} = e^C e^{y/x}$

Let $A = e^C$. $A$ is a positive arbitrary constant.

$\frac{x}{(x - y)^2} = \pm A e^{y/x}$

Let $B = \pm A$. $B$ is a non-zero arbitrary constant.

$\frac{x}{(x - y)^2} = B e^{y/x}$

Take the reciprocal of both sides:

$\frac{(x - y)^2}{x} = \frac{1}{B} e^{-y/x}$

Let $C = \frac{1}{B}$. Since $B$ is a non-zero arbitrary constant, $C$ is also a non-zero arbitrary constant. Multiplying by $x$, we get:

$(x - y)^2 = C x e^{-y/x}$

This form requires $C \neq 0$. However, if we check the solution $y=x$ (which corresponds to $(x-y)^2=0$), substituting into the original DE shows it is a solution. This case $y=x$ is obtained from $(x-y)^2 = Cx e^{-y/x}$ when $C=0$. Thus, the arbitrary constant $C$ can be any real number.

General Solution:

The general solution of the differential equation is:

$(x - y)^2 = Cx e^{\frac{-y}{x}}$

where $C$ is an arbitrary constant.

Given:

The differential equation is $y’ = \frac{x + y}{x}$.

This can be written as $\frac{dy}{dx} = \frac{x + y}{x}$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation is $\frac{dy}{dx} = \frac{x + y}{x}$.

Let $f(x,y) = \frac{x + y}{x}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{\lambda (x + y)}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{x + y}{x}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$:

$v + x \frac{dv}{dx} = \frac{x + vx}{x}$

$v + x \frac{dv}{dx} = \frac{x(1 + v)}{x}$

$v + x \frac{dv}{dx} = 1 + v$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = 1 + v - v$

$x \frac{dv}{dx} = 1$

$\frac{dv}{dx} = \frac{1}{x}$

Separating variables, we get:

$dv = \frac{1}{x} dx$

Integrating both sides:

$\int dv = \int \frac{1}{x} dx$

$v = \log|x| + C$

Substitute back $v = y/x$:

$\frac{y}{x} = \log|x| + C$

Multiply by $x$ to solve for $y$:

$y = x (\log|x| + C)$

The general solution is given by:

$\boxed{\mathbf{y = x (\log|x| + C)}}$

where C is the arbitrary constant.

Given:

The differential equation is $(x – y) dy – (x + y) dx = 0$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$(x – y) dy = (x + y) dx$

$\frac{dy}{dx} = \frac{x + y}{x - y}$

Let $f(x,y) = \frac{x + y}{x - y}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x - \lambda y}$

$f(\lambda x, \lambda y) = \frac{\lambda (x + y)}{\lambda (x - y)}$

$f(\lambda x, \lambda y) = \frac{x + y}{x - y}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{x + y}{x - y}$:

$v + x \frac{dv}{dx} = \frac{x + vx}{x - vx}$

$v + x \frac{dv}{dx} = \frac{x(1 + v)}{x(1 - v)}$

$v + x \frac{dv}{dx} = \frac{1 + v}{1 - v}$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v$

$x \frac{dv}{dx} = \frac{1 + v - v(1 - v)}{1 - v}$

$x \frac{dv}{dx} = \frac{1 + v - v + v^2}{1 - v}$

$x \frac{dv}{dx} = \frac{1 + v^2}{1 - v}$

Separating variables, we get:

$\frac{1 - v}{1 + v^2} dv = \frac{dx}{x}$

Integrating both sides:

$\int \frac{1 - v}{1 + v^2} dv = \int \frac{dx}{x}$

$\int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \int \frac{dx}{x}$

Integrating term by term:

$\int \frac{1}{1 + v^2} dv = \tan^{-1}(v)$

$\int \frac{v}{1 + v^2} dv$. Let $u = 1 + v^2$, so $du = 2v \, dv$. Thus $v \, dv = \frac{1}{2} \, du$.

$\int \frac{1}{u} \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log|u| = \frac{1}{2} \log|1 + v^2|$. Since $1+v^2 \ge 1$, we can write $\frac{1}{2} \log(1 + v^2)$.

$\int \frac{dx}{x} = \log|x| + C$

Combining the results:

$\tan^{-1}(v) - \frac{1}{2} \log(1 + v^2) = \log|x| + C$

Substitute back $v = y/x$:

$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log(1 + (\frac{y}{x})^2) = \log|x| + C$

$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log(1 + \frac{y^2}{x^2}) = \log|x| + C$

$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log(\frac{x^2 + y^2}{x^2}) = \log|x| + C$

$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} (\log(x^2 + y^2) - \log(x^2)) = \log|x| + C$

$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} (\log(x^2 + y^2) - 2\log|x|) = \log|x| + C$

$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log(x^2 + y^2) + \log|x| = \log|x| + C$

$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log(x^2 + y^2) = C$

Multiplying by 2:

$2 \tan^{-1}(\frac{y}{x}) - \log(x^2 + y^2) = 2C$

Let $C_1 = 2C$ be the new arbitrary constant.

The general solution is given by:

$\boxed{\mathbf{2 \tan^{-1}\left(\frac{y}{x}\right) - \log(x^2 + y^2) = C_1}}$

where $C_1$ is the arbitrary constant.

Given:

The differential equation is $(x^2 – y^2) dx + 2xy\; dy = 0$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$2xy\; dy = -(x^2 – y^2) dx$

$2xy\; dy = (y^2 – x^2) dx$

$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$

Let $f(x,y) = \frac{y^2 - x^2}{2xy}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{(\lambda y)^2 - (\lambda x)^2}{2(\lambda x)(\lambda y)}$

$f(\lambda x, \lambda y) = \frac{\lambda^2 y^2 - \lambda^2 x^2}{2 \lambda^2 xy}$

$f(\lambda x, \lambda y) = \frac{\lambda^2 (y^2 - x^2)}{\lambda^2 (2xy)}$

$f(\lambda x, \lambda y) = \frac{y^2 - x^2}{2xy}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$:

$v + x \frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)}$

$v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2v x^2}$

$v + x \frac{dv}{dx} = \frac{x^2(v^2 - 1)}{2v x^2}$

$v + x \frac{dv}{dx} = \frac{v^2 - 1}{2v}$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v$

$x \frac{dv}{dx} = \frac{v^2 - 1 - v(2v)}{2v}$

$x \frac{dv}{dx} = \frac{v^2 - 1 - 2v^2}{2v}$

$x \frac{dv}{dx} = \frac{-v^2 - 1}{2v}$

$x \frac{dv}{dx} = -\frac{v^2 + 1}{2v}$

Separating variables, we get:

$\frac{2v}{v^2 + 1} dv = -\frac{1}{x} dx$

Integrating both sides:

$\int \frac{2v}{v^2 + 1} dv = \int -\frac{1}{x} dx$

On the left side, let $u = v^2 + 1$, then $du = 2v \, dv$. The integral becomes $\int \frac{1}{u} du = \log|u| = \log(v^2 + 1)$ (since $v^2 + 1 \ge 1$).

On the right side, the integral is $-\log|x| + C$.

So, the integrated equation is:

$\log(v^2 + 1) = -\log|x| + C$

We can move the logarithmic term to the left side:

$\log(v^2 + 1) + \log|x| = C$

Using the logarithm property $\log A + \log B = \log(AB)$:

$\log((v^2 + 1)|x|) = C$

Substitute back $v = y/x$:

$\log((\frac{y}{x})^2 + 1)|x|) = C$

$\log((\frac{y^2}{x^2} + 1)|x|) = C$

$\log((\frac{y^2 + x^2}{x^2})|x|) = C$

$\log\left(\frac{x^2 + y^2}{x^2} |x|\right) = C$

$\log\left(\frac{x^2 + y^2}{|x|}\right) = C$

Exponentiating both sides:

$\frac{x^2 + y^2}{|x|} = e^C$

$x^2 + y^2 = e^C |x|$

Let $K = e^C$, where $K$ is an arbitrary positive constant.

$x^2 + y^2 = K|x|$

Alternatively, if we used the constant as $\log|K|$ in the integration step:

$\log(v^2 + 1) = -\log|x| + \log|K|$

$\log(v^2 + 1) = \log\left|\frac{K}{x}\right|$

$v^2 + 1 = \frac{K}{x}$

Substitute back $v = y/x$:

$(\frac{y}{x})^2 + 1 = \frac{K}{x}$

$\frac{y^2}{x^2} + 1 = \frac{K}{x}$

$\frac{y^2 + x^2}{x^2} = \frac{K}{x}$

Multiply by $x^2$ (assuming $x \ne 0$):

$x^2 + y^2 = Kx$

This form $x^2 + y^2 = Kx$ (or $x^2 + y^2 = Cx$ using constant C) is a common way to express the general solution, where $K$ is an arbitrary non-zero constant. If $x=0$, the original equation becomes $-y^2 dx + 2xy dy = 0$, which is $2xy dy = y^2 dx$. If $y \ne 0$, $2x \frac{dy}{dx} = y$, $\frac{dy}{y} = \frac{dx}{2x}$, $\log|y| = \frac{1}{2} \log|x| + C'$, $|y| = e^{C'} \sqrt{|x|}$. This suggests the solution $x^2+y^2=Kx$ covers the general case for $x \ne 0$.

The general solution is given by:

$\boxed{\mathbf{x^2 + y^2 = Kx}}$

where $K$ is the arbitrary constant.

Given:

The differential equation is $x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$\frac{dy}{dx} = \frac{x^2 - 2y^2 + xy}{x^2}$

$\frac{dy}{dx} = \frac{x^2}{x^2} - \frac{2y^2}{x^2} + \frac{xy}{x^2}$

$\frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$

Let $f(x,y) = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = 1 - 2\left(\frac{\lambda y}{\lambda x}\right)^2 + \frac{\lambda y}{\lambda x}$

$f(\lambda x, \lambda y) = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$:

$v + x \frac{dv}{dx} = 1 - 2v^2 + v$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = 1 - 2v^2$

Separating variables, we get:

$\frac{dv}{1 - 2v^2} = \frac{dx}{x}$

Integrating both sides:

$\int \frac{dv}{1 - 2v^2} = \int \frac{dx}{x}$

Consider the integral on the left side: $\int \frac{dv}{1 - 2v^2}$. We can rewrite this as $\int \frac{dv}{1^2 - (\sqrt{2}v)^2}$.

Using the standard integral formula $\int \frac{du}{a^2 - u^2} = \frac{1}{2a} \log\left|\frac{a+u}{a-u}\right|$, with $a=1$ and $u=\sqrt{2}v$, and $du = \sqrt{2} dv$ (so $dv = \frac{1}{\sqrt{2}} du$):

$\int \frac{1}{\sqrt{2}} \frac{du}{1^2 - u^2} = \frac{1}{\sqrt{2}} \left( \frac{1}{2(1)} \log\left|\frac{1+u}{1-u}\right| \right) = \frac{1}{2\sqrt{2}} \log\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|$

The integral on the right side is $\int \frac{dx}{x} = \log|x| + C_1$.

Equating the two integrals:

$\frac{1}{2\sqrt{2}} \log\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = \log|x| + C_1$

Multiply by $2\sqrt{2}$:

$\log\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = 2\sqrt{2} \log|x| + 2\sqrt{2} C_1$

Let $C = 2\sqrt{2} C_1$ be the new arbitrary constant.

$\log\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| = \log(|x|^{2\sqrt{2}}) + C$

Rearranging the terms:

$\log\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right| - \log(|x|^{2\sqrt{2}}) = C$

Using the logarithm property $\log A - \log B = \log(A/B)$:

$\log\left|\frac{(1+\sqrt{2}v)/(1-\sqrt{2}v)}{x^{2\sqrt{2}}}\right| = C$

Substitute back $v = y/x$:

$\log\left|\frac{(1+\sqrt{2}y/x)/(1-\sqrt{2}y/x)}{x^{2\sqrt{2}}}\right| = C$

$\log\left|\frac{(x+\sqrt{2}y)/x}{(x-\sqrt{2}y)/x} \cdot \frac{1}{x^{2\sqrt{2}}}\right| = C$

$\log\left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y} \cdot \frac{1}{x^{2\sqrt{2}}}\right| = C$

$\log\left|\frac{x+\sqrt{2}y}{x^{2\sqrt{2}}(x-\sqrt{2}y)}\right| = C$

The general solution is given by:

$\boxed{\mathbf{\log\left|\frac{x+\sqrt{2}y}{x^{2\sqrt{2}}(x-\sqrt{2}y)}\right| = C}}$

where C is the arbitrary constant.

Given:

The differential equation is $x \;dy \;–\; y \;dx = \sqrt{x^2 + y^2}\; dx$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$x \;dy = y \;dx + \sqrt{x^2 + y^2}\; dx$

$x \;dy = (y + \sqrt{x^2 + y^2})\; dx$

$\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$

Let $f(x,y) = \frac{y + \sqrt{x^2 + y^2}}{x}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{(\lambda x)^2 + (\lambda y)^2}}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{\lambda^2 x^2 + \lambda^2 y^2}}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{\lambda y + \sqrt{\lambda^2 (x^2 + y^2)}}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{\lambda y + |\lambda| \sqrt{x^2 + y^2}}{\lambda x}$

Assuming $\lambda > 0$, $|\lambda| = \lambda$.

$f(\lambda x, \lambda y) = \frac{\lambda y + \lambda \sqrt{x^2 + y^2}}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{\lambda (y + \sqrt{x^2 + y^2})}{\lambda x}$

$f(\lambda x, \lambda y) = \frac{y + \sqrt{x^2 + y^2}}{x}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation of degree 1.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$:

$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + (vx)^2}}{x}$

$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2 x^2}}{x}$

$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2(1 + v^2)}}{x}$

$v + x \frac{dv}{dx} = \frac{vx + |x|\sqrt{1 + v^2}}{x}$

Assuming $x > 0$, $|x| = x$.

$v + x \frac{dv}{dx} = \frac{vx + x\sqrt{1 + v^2}}{x}$

$v + x \frac{dv}{dx} = v + \sqrt{1 + v^2}$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = \sqrt{1 + v^2}$

$\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$

Integrating both sides:

$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$

Using the standard integral formula $\int \frac{du}{\sqrt{a^2 + u^2}} = \log|u + \sqrt{a^2 + u^2}|$, with $a=1$ and $u=v$:

$\log|v + \sqrt{1 + v^2}| = \log|x| + C_1$

We can write the constant $C_1$ as $\log|C|$ for some non-zero constant $C$.

$\log|v + \sqrt{1 + v^2}| = \log|x| + \log|C|$

$\log|v + \sqrt{1 + v^2}| = \log|Cx|$

$|v + \sqrt{1 + v^2}| = |Cx|$

$v + \sqrt{1 + v^2} = \pm Cx$

Let $K = \pm C$, where $K$ is an arbitrary non-zero constant.

$v + \sqrt{1 + v^2} = Kx$

Substitute back $v = y/x$:

$\frac{y}{x} + \sqrt{1 + (\frac{y}{x})^2} = Kx$

$\frac{y}{x} + \sqrt{\frac{x^2 + y^2}{x^2}} = Kx$

$\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{|x|} = Kx$

Assuming $x>0$, $|x|=x$.

$\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = Kx$

Multiply by $x$ (for $x \ne 0$):

$y + \sqrt{x^2 + y^2} = Kx^2$

Note: If we assumed $x<0$ earlier, we would get $\frac{y}{x} - \frac{\sqrt{x^2 + y^2}}{x} = Kx$, leading to $y - \sqrt{x^2 + y^2} = Kx^2$. However, the form $y + \sqrt{x^2 + y^2} = Kx^2$ is generally accepted as the representation of the general solution for homogeneous equations of this type, absorbing the sign difference into the arbitrary constant $K$.

The general solution is given by:

$\boxed{\mathbf{y + \sqrt{x^2 + y^2} = Kx^2}}$

where $K$ is the arbitrary constant.

Given:

The differential equation is $\left\{ x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right\} y \;dx = \left\{ y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right\} x \;dy$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$\frac{dy}{dx} = \frac{\left\{ x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right\} y}{\left\{ y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right\} x}$

Let $f(x,y) = \frac{y \left( x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right)}{x \left( y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right)}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\lambda y \left( \lambda x \cos \left( \frac{\lambda y}{\lambda x} \right) + \lambda y \sin \left( \frac{\lambda y}{\lambda x} \right) \right)}{\lambda x \left( \lambda y \sin \left( \frac{\lambda y}{\lambda x} \right) − \lambda x \cos \left( \frac{\lambda y}{\lambda x} \right) \right)}$

$f(\lambda x, \lambda y) = \frac{\lambda y \cdot \lambda \left( x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right)}{\lambda x \cdot \lambda \left( y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right)}$

$f(\lambda x, \lambda y) = \frac{\lambda^2 y \left( x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right)}{\lambda^2 x \left( y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right)}$

$f(\lambda x, \lambda y) = \frac{y \left( x \cos \left( \frac{y}{x} \right) + y \sin \left( \frac{y}{x} \right) \right)}{x \left( y \sin \left( \frac{y}{x} \right) − x \cos \left( \frac{y}{x} \right) \right)}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation of degree 0.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation:

$v + x \frac{dv}{dx} = \frac{vx \left( x \cos \left( \frac{vx}{x} \right) + vx \sin \left( \frac{vx}{x} \right) \right)}{x \left( vx \sin \left( \frac{vx}{x} \right) − x \cos \left( \frac{vx}{x} \right) \right)}$

$v + x \frac{dv}{dx} = \frac{vx \left( x \cos (v) + vx \sin (v) \right)}{x \left( vx \sin (v) − x \cos (v) \right)}$

$v + x \frac{dv}{dx} = \frac{vx \cdot x (\cos v + v \sin v)}{x \cdot x (v \sin v - \cos v)}$

$v + x \frac{dv}{dx} = \frac{v (\cos v + v \sin v)}{v \sin v - \cos v}$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} - v$

$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v(v \sin v - \cos v)}{v \sin v - \cos v}$

$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v}$

$x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v}$

Separating variables, we get:

$\frac{v \sin v - \cos v}{2v \cos v} dv = \frac{dx}{x}$

$\frac{1}{2} \left( \frac{v \sin v}{v \cos v} - \frac{\cos v}{v \cos v} \right) dv = \frac{dx}{x}$

$\frac{1}{2} \left( \tan v - \frac{1}{v} \right) dv = \frac{dx}{x}$

Integrating both sides:

$\int \frac{1}{2} \left( \tan v - \frac{1}{v} \right) dv = \int \frac{dx}{x}$

$\frac{1}{2} \int \tan v \, dv - \frac{1}{2} \int \frac{1}{v} dv = \int \frac{dx}{x}$

Using the standard integral $\int \tan v \, dv = -\log|\cos v|$ and $\int \frac{1}{v} dv = \log|v|$ and $\int \frac{dx}{x} = \log|x|$:

$\frac{1}{2} (-\log|\cos v|) - \frac{1}{2} \log|v| = \log|x| + C_1$

$-\frac{1}{2} (\log|\cos v| + \log|v|) = \log|x| + C_1$

$-\frac{1}{2} \log|v \cos v| = \log|x| + C_1$

Multiply by -2:

$\log|v \cos v| = -2 \log|x| - 2C_1$

$\log|v \cos v| = \log(|x|^{-2}) + C_2$, where $C_2 = -2C_1$

$\log|v \cos v| = \log\left(\frac{1}{x^2}\right) + C_2$

Rearranging the terms:

$\log|v \cos v| - \log\left(\frac{1}{x^2}\right) = C_2$

Using the logarithm property $\log A - \log B = \log(A/B)$:

$\log\left|\frac{v \cos v}{1/x^2}\right| = C_2$

$\log|x^2 v \cos v| = C_2$

Substitute back $v = y/x$:

$\log\left|x^2 \frac{y}{x} \cos\left(\frac{y}{x}\right)\right| = C_2$

$\log\left|xy \cos\left(\frac{y}{x}\right)\right| = C_2$

Exponentiating both sides:

$\left|xy \cos\left(\frac{y}{x}\right)\right| = e^{C_2}$

$xy \cos\left(\frac{y}{x}\right) = \pm e^{C_2}$

Let $K = \pm e^{C_2}$, where $K$ is an arbitrary non-zero constant.

The general solution is given by:

$\boxed{\mathbf{xy \cos\left(\frac{y}{x}\right) = K}}$

where $K$ is the arbitrary constant.

Given:

The differential equation is $x \frac{dy}{dx} - y + x \sin \left( \frac{y}{x} \right) = 0$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$x \frac{dy}{dx} = y - x \sin \left( \frac{y}{x} \right)$

$\frac{dy}{dx} = \frac{y - x \sin \left( \frac{y}{x} \right)}{x}$

$\frac{dy}{dx} = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$

Let $f(x,y) = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \sin \left( \frac{\lambda y}{\lambda x} \right)$

$f(\lambda x, \lambda y) = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation of degree 0.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} - \sin \left( \frac{y}{x} \right)$:

$v + x \frac{dv}{dx} = \frac{vx}{x} - \sin \left( \frac{vx}{x} \right)$

$v + x \frac{dv}{dx} = v - \sin(v)$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = -\sin(v)$

$\frac{dv}{\sin(v)} = -\frac{dx}{x}$

$\text{cosec}(v) dv = -\frac{1}{x} dx$

Integrating both sides:

$\int \text{cosec}(v) dv = \int -\frac{1}{x} dx$

Using standard integrals:

$\log|\text{cosec}(v) - \cot(v)| = -\log|x| + \log|C|$

$\log|\text{cosec}(v) - \cot(v)| + \log|x| = \log|C|$

Using the logarithm property $\log A + \log B = \log(AB)$:

$\log|x (\text{cosec}(v) - \cot(v))| = \log|C|$

$|x (\text{cosec}(v) - \cot(v))| = |C|$

$x (\text{cosec}(v) - \cot(v)) = K$, where $K = \pm C$ is an arbitrary non-zero constant.

Substitute back $v = y/x$:

$x \left(\text{cosec}\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x}\right)\right) = K$

We can simplify the term in the parenthesis using trigonometric identities:

$\text{cosec}(A) - \cot(A) = \frac{1}{\sin A} - \frac{\cos A}{\sin A} = \frac{1 - \cos A}{\sin A}$

Using half-angle identities $1 - \cos A = 2 \sin^2(A/2)$ and $\sin A = 2 \sin(A/2) \cos(A/2)$:

$\frac{1 - \cos(y/x)}{\sin(y/x)} = \frac{2 \sin^2(y/(2x))}{2 \sin(y/(2x)) \cos(y/(2x))} = \tan\left(\frac{y}{2x}\right)$

So, the equation becomes:

$x \tan\left(\frac{y}{2x}\right) = K$

$\tan\left(\frac{y}{2x}\right) = \frac{K}{x}$

The general solution is given by:

$\boxed{\mathbf{\tan\left(\frac{y}{2x}\right) = \frac{K}{x}}}$

where $K$ is the arbitrary constant.

Given:

The differential equation is $y \;dx + x \log \left( \frac{y}{x} \right) dy - 2x \;dy = 0$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$y \;dx = \left( 2x - x \log \left( \frac{y}{x} \right) \right) dy$

$\frac{dy}{dx} = \frac{y}{2x - x \log \left( \frac{y}{x} \right)}$

$\frac{dy}{dx} = \frac{y}{x \left( 2 - \log \left( \frac{y}{x} \right) \right)}$

$\frac{dy}{dx} = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)}$

Let $f(x,y) = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\frac{\lambda y}{\lambda x}}{2 - \log \left( \frac{\lambda y}{\lambda x} \right)}$

$f(\lambda x, \lambda y) = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)}$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation of degree 0.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)}$:

$v + x \frac{dv}{dx} = \frac{v}{2 - \log(v)}$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = \frac{v}{2 - \log(v)} - v$

$x \frac{dv}{dx} = \frac{v - v(2 - \log(v))}{2 - \log(v)}$

$x \frac{dv}{dx} = \frac{v - 2v + v \log(v)}{2 - \log(v)}$

$x \frac{dv}{dx} = \frac{-v + v \log(v)}{2 - \log(v)}$

$x \frac{dv}{dx} = \frac{v (\log(v) - 1)}{2 - \log(v)}$

Separating variables, we get:

$\frac{2 - \log(v)}{v (\log(v) - 1)} dv = \frac{dx}{x}$

Integrating both sides:

$\int \frac{2 - \log(v)}{v (\log(v) - 1)} dv = \int \frac{dx}{x}$

Consider the integral on the left side: $\int \frac{2 - \log(v)}{v (\log(v) - 1)} dv$.

Let $u = \log(v) - 1$. Then $du = \frac{1}{v} dv$.

Also, $\log(v) = u + 1$, so $2 - \log(v) = 2 - (u + 1) = 1 - u$.

The integral becomes $\int \frac{1 - u}{u} du = \int \left(\frac{1}{u} - 1\right) du$.

$\int \left(\frac{1}{u} - 1\right) du = \log|u| - u + C_1$.

Substitute back $u = \log(v) - 1$:

$\log|\log(v) - 1| - (\log(v) - 1) + C_1$

The integral on the right side is $\int \frac{dx}{x} = \log|x| + C_2$.

Equating the two integrals:

$\log|\log(v) - 1| - \log(v) + 1 = \log|x| + C$, where $C = C_2 - C_1$.

Rearrange the terms:

$\log|\log(v) - 1| - \log(v) - \log|x| = C - 1$

Using the logarithm property $\log A - \log B = \log(A/B)$:

$\log\left|\frac{\log(v) - 1}{v}\right| - \log|x| = C - 1$

$\log\left|\frac{(\log(v) - 1)/v}{x}\right| = C - 1$

$\log\left|\frac{\log(v) - 1}{vx}\right| = C - 1$

Substitute back $v = y/x$:

$\log\left|\frac{\log(y/x) - 1}{(y/x)x}\right| = C - 1$

$\log\left|\frac{\log(y/x) - 1}{y}\right| = C - 1$

Exponentiating both sides:

$\left|\frac{\log(y/x) - 1}{y}\right| = e^{C - 1}$

$\frac{\log(y/x) - 1}{y} = \pm e^{C - 1}$

Let $K = \pm e^{C - 1}$, where $K$ is an arbitrary non-zero constant.

$\log\left(\frac{y}{x}\right) - 1 = Ky$

$\log\left(\frac{y}{x}\right) = Ky + 1$

The general solution is given by:

$\boxed{\mathbf{\log\left(\frac{y}{x}\right) = Ky + 1}}$

where $K$ is the arbitrary constant.

Given:

The differential equation is $\left( 1 + e^{\frac{x}{y}} \right) dx + e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right) dy = 0$.

To Solve:

Show that the given differential equation is homogeneous and find its general solution.

Solution:

The given differential equation can be written as:

$\left( 1 + e^{\frac{x}{y}} \right) dx = - e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right) dy$

$\frac{dx}{dy} = \frac{- e^{\frac{x}{y}} \left( 1 − \frac{x}{y} \right)}{1 + e^{\frac{x}{y}}}$

$\frac{dx}{dy} = \frac{e^{\frac{x}{y}} \left( \frac{x}{y} - 1 \right)}{1 + e^{\frac{x}{y}}}$

Let $g(x,y) = \frac{e^{\frac{x}{y}} \left( \frac{x}{y} - 1 \right)}{1 + e^{\frac{x}{y}}}$.

To check for homogeneity, we evaluate $g(\lambda x, \lambda y)$:

$g(\lambda x, \lambda y) = \frac{e^{\frac{\lambda x}{\lambda y}} \left( \frac{\lambda x}{\lambda y} - 1 \right)}{1 + e^{\frac{\lambda x}{\lambda y}}}$

$g(\lambda x, \lambda y) = \frac{e^{\frac{x}{y}} \left( \frac{x}{y} - 1 \right)}{1 + e^{\frac{x}{y}}}$

$g(\lambda x, \lambda y) = g(x,y)$

Since $g(\lambda x, \lambda y) = g(x,y)$, the given differential equation is a homogeneous differential equation of degree 0.

To solve the homogeneous equation, since $\frac{dx}{dy}$ is a function of $\frac{x}{y}$, we substitute $x = vy$.

Differentiating with respect to $y$, we get $\frac{dx}{dy} = v + y \frac{dv}{dy}$.

Substituting $x=vy$ and $\frac{dx}{dy} = v + y \frac{dv}{dy}$ into the differential equation:

$v + y \frac{dv}{dy} = \frac{e^{\frac{vy}{y}} \left( \frac{vy}{y} - 1 \right)}{1 + e^{\frac{vy}{y}}}$

$v + y \frac{dv}{dy} = \frac{e^{v} (v - 1)}{1 + e^{v}}$

Now, we separate the variables $v$ and $y$:

$y \frac{dv}{dy} = \frac{e^{v} (v - 1)}{1 + e^{v}} - v$

$y \frac{dv}{dy} = \frac{ve^v - e^v - v(1 + e^v)}{1 + e^{v}}$

$y \frac{dv}{dy} = \frac{ve^v - e^v - v - ve^v}{1 + e^{v}}$

$y \frac{dv}{dy} = \frac{-e^v - v}{1 + e^{v}}$

$y \frac{dv}{dy} = -\frac{v + e^v}{1 + e^{v}}$

Separating variables, we get:

$\frac{1 + e^{v}}{v + e^{v}} dv = -\frac{1}{y} dy$

Integrating both sides:

$\int \frac{1 + e^{v}}{v + e^{v}} dv = \int -\frac{1}{y} dy$

For the integral on the left side, let $u = v + e^v$. Then $du = (1 + e^v) dv$.

$\int \frac{du}{u} = \log|u| + C_1 = \log|v + e^v| + C_1$.

For the integral on the right side: $\int -\frac{1}{y} dy = -\log|y| + C_2$.

Equating the two integrals:

$\log|v + e^v| = -\log|y| + C$, where $C = C_2 - C_1$ is an arbitrary constant.

Rearranging the terms:

$\log|v + e^v| + \log|y| = C$

Using the logarithm property $\log A + \log B = \log(AB)$:

$\log|(v + e^v)y| = C$

Exponentiating both sides:

$|(v + e^v)y| = e^C$

$(v + e^v)y = K$, where $K = \pm e^C$ is an arbitrary non-zero constant. Note that $K=0$ corresponds to $v+e^v=0$, which is not possible for real $v$.

Substitute back $v = x/y$:

$\left(\frac{x}{y} + e^{\frac{x}{y}}\right)y = K$

Multiplying by $y$ (assuming $y \ne 0$):

$\frac{x}{y} \cdot y + y \cdot e^{\frac{x}{y}} = K$

$x + y e^{\frac{x}{y}} = K$

The general solution is given by:

$\boxed{\mathbf{x + y e^{\frac{x}{y}} = K}}$

where $K$ is the arbitrary constant.

Given:

The differential equation is $(x + y) dy + (x – y) dx = 0$.

The initial condition is $y = 1$ when $x = 1$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation can be written as:

$(x + y) dy = -(x – y) dx$

$(x + y) dy = (y – x) dx$

$\frac{dy}{dx} = \frac{y – x}{x + y}$

Let $f(x,y) = \frac{y – x}{x + y}$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\lambda y – \lambda x}{\lambda x + \lambda y} = \frac{\lambda(y – x)}{\lambda(x + y)} = \frac{y – x}{x + y} = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y – x}{x + y}$:

$v + x \frac{dv}{dx} = \frac{vx – x}{x + vx} = \frac{x(v – 1)}{x(1 + v)} = \frac{v – 1}{1 + v}$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = \frac{v – 1}{1 + v} - v$

$x \frac{dv}{dx} = \frac{v – 1 - v(1 + v)}{1 + v}$

$x \frac{dv}{dx} = \frac{v – 1 - v - v^2}{1 + v}$

$x \frac{dv}{dx} = \frac{-1 - v^2}{1 + v} = -\frac{1 + v^2}{1 + v}$

Separating variables, we get:

$\frac{1 + v}{1 + v^2} dv = -\frac{dx}{x}$

Integrating both sides:

$\int \frac{1 + v}{1 + v^2} dv = \int -\frac{1}{x} dx$

$\int \left(\frac{1}{1 + v^2} + \frac{v}{1 + v^2}\right) dv = -\int \frac{1}{x} dx$

$\int \frac{1}{1 + v^2} dv + \int \frac{v}{1 + v^2} dv = -\int \frac{1}{x} dx$

Integrating each term:

$\tan^{-1}(v) + \frac{1}{2}\log(1+v^2) = -\log|x| + C$

Substitute back $v = y/x$:

$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\log\left(1+\left(\frac{y}{x}\right)^2\right) = -\log|x| + C$

$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\log\left(\frac{x^2+y^2}{x^2}\right) = -\log|x| + C$

$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}(\log(x^2+y^2) - \log(x^2)) = -\log|x| + C$

$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\log(x^2+y^2) - \log|x| = -\log|x| + C$

$\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\log(x^2+y^2) = C$

This is the general solution.

Now, we use the given condition $y = 1$ when $x = 1$ to find the value of $C$.

Substitute $x=1$ and $y=1$ into the general solution:

$\tan^{-1}\left(\frac{1}{1}\right) + \frac{1}{2}\log(1^2+1^2) = C$

$\tan^{-1}(1) + \frac{1}{2}\log(2) = C$

$\frac{\pi}{4} + \frac{1}{2}\log(2) = C$

Substitute this value of $C$ back into the general solution to get the particular solution:

$\boxed{\mathbf{\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\log(x^2+y^2) = \frac{\pi}{4} + \frac{1}{2}\log(2)}}$

Given:

The differential equation is $x^2 dy + (xy + y^2) dx = 0$.

The initial condition is $y = 1$ when $x = 1$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation can be written as:

$x^2 dy = -(xy + y^2) dx$

$\frac{dy}{dx} = \frac{-(xy + y^2)}{x^2}$

$\frac{dy}{dx} = \frac{-xy - y^2}{x^2}$

$\frac{dy}{dx} = -\frac{xy}{x^2} - \frac{y^2}{x^2}$

$\frac{dy}{dx} = -\frac{y}{x} - \left(\frac{y}{x}\right)^2$

Let $f(x,y) = -\frac{y}{x} - \left(\frac{y}{x}\right)^2$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = -\frac{\lambda y}{\lambda x} - \left(\frac{\lambda y}{\lambda x}\right)^2 = -\frac{y}{x} - \left(\frac{y}{x}\right)^2 = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = -\frac{y}{x} - \left(\frac{y}{x}\right)^2$:

$v + x \frac{dv}{dx} = -v - v^2$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = -v - v^2 - v$

$x \frac{dv}{dx} = -2v - v^2$

$x \frac{dv}{dx} = -(v^2 + 2v)$

Separating variables, we get:

$\frac{dv}{v^2 + 2v} = -\frac{dx}{x}$

Integrating both sides:

$\int \frac{dv}{v^2 + 2v} = \int -\frac{1}{x} dx$

Consider the integral on the left side: $\int \frac{dv}{v^2 + 2v}$. We can use partial fractions or complete the square in the denominator. Using partial fractions:

$\frac{1}{v^2 + 2v} = \frac{1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2}$

$1 = A(v + 2) + Bv$

If $v = 0$, $1 = 2A \implies A = 1/2$.

If $v = -2$, $1 = -2B \implies B = -1/2$.

So, $\int \left(\frac{1/2}{v} - \frac{1/2}{v + 2}\right) dv = \frac{1}{2} \int \frac{1}{v} dv - \frac{1}{2} \int \frac{1}{v + 2} dv$

$= \frac{1}{2} \log|v| - \frac{1}{2} \log|v + 2| + C_1$

$= \frac{1}{2} \log\left|\frac{v}{v + 2}\right| + C_1$

The integral on the right side is $\int -\frac{1}{x} dx = -\log|x| + C_2$.

Equating the two integrals:

$\frac{1}{2} \log\left|\frac{v}{v + 2}\right| = -\log|x| + C$, where $C = C_2 - C_1$

Multiply by 2:

$\log\left|\frac{v}{v + 2}\right| = -2\log|x| + 2C$

$\log\left|\frac{v}{v + 2}\right| = \log(|x|^{-2}) + C'$

$\log\left|\frac{v}{v + 2}\right| - \log\left(\frac{1}{x^2}\right) = C'$

$\log\left|\frac{v/(v + 2)}{1/x^2}\right| = C'$

$\log\left|\frac{vx^2}{v + 2}\right| = C'$

Substitute back $v = y/x$:

$\log\left|\frac{(y/x)x^2}{(y/x) + 2}\right| = C'$

$\log\left|\frac{xy}{(y + 2x)/x}\right| = C'$

$\log\left|\frac{x^2y}{y + 2x}\right| = C'$

Exponentiating both sides:

$\left|\frac{x^2y}{y + 2x}\right| = e^{C'}$

$\frac{x^2y}{y + 2x} = K$, where $K = \pm e^{C'}$ is an arbitrary non-zero constant.

$x^2y = K(y + 2x)$

This is the general solution.

Now, we use the given condition $y = 1$ when $x = 1$ to find the value of $K$.

Substitute $x=1$ and $y=1$ into the general solution:

$(1)^2(1) = K(1 + 2(1))$

$1 = K(1 + 2)$

$1 = 3K$

$K = \frac{1}{3}$

Substitute this value of $K$ back into the general solution to get the particular solution:

$x^2y = \frac{1}{3}(y + 2x)$

Multiply by 3:

$3x^2y = y + 2x$

$3x^2y - y = 2x$

$y(3x^2 - 1) = 2x$

$y = \frac{2x}{3x^2 - 1}$

The particular solution is given by:

$\boxed{\mathbf{y = \frac{2x}{3x^2 - 1}}}$

Given:

The differential equation is $\left[ x \sin^2 \left( \frac{y}{x} \right) − y \right] dx + x \;dy = 0$.

The initial condition is $y = \frac{π}{4}$ when $x = 1$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is $\left[ x \sin^2 \left( \frac{y}{x} \right) − y \right] dx + x \;dy = 0$.

We can rearrange this equation to express $\frac{dy}{dx}$. Move the first term to the right side:

$x \;dy = - \left[ x \sin^2 \left( \frac{y}{x} \right) − y \right] dx$

$x \;dy = \left[ y - x \sin^2 \left( \frac{y}{x} \right) \right] dx$

Divide both sides by $x \, dx$ (assuming $x \neq 0$):

$\frac{dy}{dx} = \frac{y - x \sin^2 \left( \frac{y}{x} \right)}{x}$

Separate the terms on the right side:

$\frac{dy}{dx} = \frac{y}{x} - \frac{x \sin^2 \left( \frac{y}{x} \right)}{x}$

$\frac{dy}{dx} = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$

This equation is in the form $\frac{dy}{dx} = f(x,y)$, where $f(x,y) = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$.

To check if the differential equation is homogeneous, we substitute $x$ with $\lambda x$ and $y$ with $\lambda y$ into $f(x,y)$:

$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \sin^2 \left( \frac{\lambda y}{\lambda x} \right)$

$f(\lambda x, \lambda y) = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we use the substitution:

$y = vx$

where $v$ is a function of $x$.

Differentiating both sides with respect to $x$ using the product rule, we get:

$\frac{dy}{dx} = v + x \frac{dv}{dx}$

Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} - \sin^2 \left( \frac{y}{x} \right)$:

$v + x \frac{dv}{dx} = \frac{vx}{x} - \sin^2 \left( \frac{vx}{x} \right)$

$v + x \frac{dv}{dx} = v - \sin^2(v)$

Subtract $v$ from both sides:

$x \frac{dv}{dx} = -\sin^2(v)$

Now, we separate the variables $v$ and $x$. Move terms involving $v$ to the left side and terms involving $x$ to the right side:

$\frac{dv}{\sin^2(v)} = -\frac{dx}{x}$

Using the reciprocal identity $\frac{1}{\sin^2 v} = \text{cosec}^2 v$, we get:

$\text{cosec}^2(v) dv = -\frac{1}{x} dx$

Integrate both sides of the separated equation:

$\int \text{cosec}^2(v) dv = \int -\frac{1}{x} dx$

The integral of $\text{cosec}^2(v)$ is $-\cot(v)$. The integral of $-\frac{1}{x}$ is $-\log|x|$.

$-\cot(v) = -\log|x| + C_{\text{gen}}$

Multiply by $-1$:

$\cot(v) = \log|x| - C_{\text{gen}}$

Let $C = -C_{\text{gen}}$. $C$ is an arbitrary constant.

$\cot(v) = \log|x| + C$

Substitute back $v = \frac{y}{x}$ to express the general solution in terms of $x$ and $y$:

$\cot\left(\frac{y}{x}\right) = \log|x| + C$

This is the general solution of the differential equation.

Now, we use the given initial condition $y = \frac{\pi}{4}$ when $x = 1$ to find the value of the constant $C$. Substitute these values into the general solution:

$\cot\left(\frac{\pi/4}{1}\right) = \log|1| + C$

$\cot\left(\frac{\pi}{4}\right) = \log(1) + C$

We know that $\cot\left(\frac{\pi}{4}\right) = 1$ and $\log(1) = 0$.

$1 = 0 + C$

$C = 1$

Substitute the value of $C=1$ back into the general solution $\cot\left(\frac{y}{x}\right) = \log|x| + C$ to obtain the particular solution:

$\cot\left(\frac{y}{x}\right) = \log|x| + 1$

We can rewrite $1$ as $\log e$, since $\log_e e = 1$.

$\cot\left(\frac{y}{x}\right) = \log|x| + \log e$

Using the property of logarithms $\log A + \log B = \log(AB)$:

So, the particular solution is $\cot\left(\frac{y}{x}\right) = \log|ex|$.

The particular solution is given by:

$\cot\left(\frac{y}{x}\right) = \log|ex|$

Given:

The differential equation is $\frac{dy}{dx} - \frac{y}{x} + \text{cosec} \left( \frac{y}{x} \right) = 0$.

The initial condition is $y = 0$ when $x = 1$.

To Find:

The particular solution satisfying the given initial condition.

Solution:

The given differential equation is $\frac{dy}{dx} - \frac{y}{x} + \text{cosec} \left( \frac{y}{x} \right) = 0$.

We can rewrite this equation as:

$\frac{dy}{dx} = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$

This equation is of the form $\frac{dy}{dx} = f(x,y)$, where $f(x,y) = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$.

To check if the differential equation is homogeneous, we evaluate $f(\lambda x, \lambda y)$ for any non-zero constant $\lambda$:

$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \text{cosec} \left( \frac{\lambda y}{\lambda x} \right)$

$f(\lambda x, \lambda y) = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we use the substitution:

$y = vx$

where $v$ is a function of $x$.

Differentiating both sides with respect to $x$ using the product rule, we get:

$\frac{dy}{dx} = v + x \frac{dv}{dx}$

Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} - \text{cosec} \left( \frac{y}{x} \right)$:

$v + x \frac{dv}{dx} = \frac{vx}{x} - \text{cosec} \left( \frac{vx}{x} \right)$

$v + x \frac{dv}{dx} = v - \text{cosec}(v)$

Subtract $v$ from both sides:

$x \frac{dv}{dx} = -\text{cosec}(v)$

Now, we separate the variables $v$ and $x$. Move terms involving $v$ to the left side and terms involving $x$ to the right side:

$\frac{dv}{\text{cosec}(v)} = -\frac{dx}{x}$

Using the reciprocal identity $\frac{1}{\text{cosec}(v)} = \sin(v)$, we get:

$\sin(v) dv = -\frac{1}{x} dx$}

Integrate both sides of the separated equation:

$\int \sin(v) dv = \int -\frac{1}{x} dx$

The integral of $\sin(v)$ with respect to $v$ is $-\cos(v)$. The integral of $-\frac{1}{x}$ with respect to $x$ is $-\log|x|$.

$-\cos(v) = -\log|x| + C_{\text{gen}}$

Multiply the entire equation by $-1$:

$\cos(v) = \log|x| - C_{\text{gen}}$

Let $C = -C_{\text{gen}}$. $C$ is an arbitrary constant of integration.

$\cos(v) = \log|x| + C$

Substitute back $v = \frac{y}{x}$ to express the general solution in terms of $x$ and $y$:

$\cos\left(\frac{y}{x}\right) = \log|x| + C$

This is the general solution of the differential equation.

Now, we use the given initial condition $y = 0$ when $x = 1$ to find the value of the constant $C$. Substitute these values into the general solution:

$\cos\left(\frac{0}{1}\right) = \log|1| + C$

$\cos(0) = \log(1) + C$

We know that $\cos(0) = 1$ and $\log(1) = 0$ (for the natural logarithm $\log$).

$1 = 0 + C$

$C = 1$

Substitute the value of $C=1$ back into the general solution $\cos\left(\frac{y}{x}\right) = \log|x| + C$ to obtain the particular solution:

$\cos\left(\frac{y}{x}\right) = \log|x| + 1$}

We can rewrite the constant $1$ as $\log e$, since the natural logarithm of $e$ is $1$ ($\log e = 1$).

$\cos\left(\frac{y}{x}\right) = \log|x| + \log e$}

Using the property of logarithms that $\log A + \log B = \log(AB)$, we can combine the terms on the right side:

$\cos\left(\frac{y}{x}\right) = \log|x \cdot e|$

$\cos\left(\frac{y}{x}\right) = \log|ex|$

The particular solution satisfying the given condition is:

$\cos\left(\frac{y}{x}\right) = \log|ex|$

Given:

The differential equation is $2xy + y^2 - 2x^2 \frac{dy}{dx} = 0$.

The initial condition is $y = 2$ when $x = 1$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation can be written as:

$2x^2 \frac{dy}{dx} = 2xy + y^2$

$\frac{dy}{dx} = \frac{2xy + y^2}{2x^2}$

$\frac{dy}{dx} = \frac{2xy}{2x^2} + \frac{y^2}{2x^2}$

$\frac{dy}{dx} = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$

Let $f(x,y) = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$.

To check for homogeneity, we evaluate $f(\lambda x, \lambda y)$:

$f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} + \frac{1}{2} \left(\frac{\lambda y}{\lambda x}\right)^2$

$f(\lambda x, \lambda y) = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$

$f(\lambda x, \lambda y) = f(x,y)$

Since $f(\lambda x, \lambda y) = f(x,y)$, the given differential equation is a homogeneous differential equation.

To solve the homogeneous equation, we substitute $y = vx$.

Differentiating with respect to $x$, we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{1}{2} \left(\frac{y}{x}\right)^2$:

$v + x \frac{dv}{dx} = v + \frac{1}{2} v^2$

Now, we separate the variables $v$ and $x$:

$x \frac{dv}{dx} = \frac{1}{2} v^2$

Separating variables, we get:

$\frac{dv}{v^2} = \frac{1}{2} \frac{dx}{x}$

$v^{-2} dv = \frac{1}{2} \frac{1}{x} dx$

Integrating both sides:

$\int v^{-2} dv = \int \frac{1}{2} \frac{1}{x} dx$

$\frac{v^{-1}}{-1} = \frac{1}{2} \log|x| + C$

$-\frac{1}{v} = \frac{1}{2} \log|x| + C$

Substitute back $v = y/x$:

$-\frac{1}{y/x} = \frac{1}{2} \log|x| + C$

$-\frac{x}{y} = \frac{1}{2} \log|x| + C$

This is the general solution.

Now, we use the given condition $y = 2$ when $x = 1$ to find the value of $C$.

Substitute $x=1$ and $y=2$ into the general solution:

$-\frac{1}{2} = \frac{1}{2} \log|1| + C$

$-\frac{1}{2} = \frac{1}{2} (0) + C$

$-\frac{1}{2} = C$

Substitute this value of $C$ back into the general solution to get the particular solution:

$-\frac{x}{y} = \frac{1}{2} \log|x| - \frac{1}{2}$

Multiply by $-2$:

$\frac{2x}{y} = -\log|x| + 1$

$\frac{2x}{y} = 1 - \log|x|$

Assuming $1 - \log|x| \ne 0$ and $y \ne 0$, we can write:

$y(1 - \log|x|) = 2x$

$y = \frac{2x}{1 - \log|x|}$

The particular solution is given by:

$\boxed{\mathbf{y = \frac{2x}{1 - \log|x|}}}$

Given:

A homogeneous differential equation of the form $\frac{dy}{dx} = h \left( \frac{x}{y} \right)$.

To Find:

The appropriate substitution to solve the given differential equation.

Solution:

The given differential equation is $\frac{dy}{dx} = h \left( \frac{x}{y} \right)$.

This equation is homogeneous because the right-hand side is a function of the ratio $\frac{x}{y}$.

Homogeneous differential equations can be solved by substituting either $y=vx$ or $x=vy$.

If we use the substitution $y=vx$, then $\frac{y}{x} = v$. Differentiating $y=vx$ with respect to $x$ gives $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting this into the given equation, we get $v + x \frac{dv}{dx} = h\left(\frac{x}{vx}\right) = h\left(\frac{1}{v}\right)$. This results in $x \frac{dv}{dx} = h\left(\frac{1}{v}\right) - v$, which is a separable equation in terms of $v$ and $x$.

Alternatively, we can rewrite the given equation by taking the reciprocal of both sides (assuming $h(\frac{x}{y}) \ne 0$):

$\frac{dx}{dy} = \frac{1}{h \left( \frac{x}{y} \right)}$

Let $G \left( \frac{x}{y} \right) = \frac{1}{h \left( \frac{x}{y} \right)}$. The equation becomes $\frac{dx}{dy} = G \left( \frac{x}{y} \right)$.

If we use the substitution $x=vy$, then $\frac{x}{y} = v$. Differentiating $x=vy$ with respect to $y$ gives $\frac{dx}{dy} = v \frac{dy}{dy} + y \frac{dv}{dy} = v + y \frac{dv}{dy}$. Substituting this into the rewritten equation, we get $v + y \frac{dv}{dy} = G(v)$. This results in $y \frac{dv}{dy} = G(v) - v$, which is a separable equation in terms of $v$ and $y$.

While both substitutions can lead to a solution, the standard practice is to use the substitution that directly introduces the ratio present in the function on the right-hand side as the new variable. Since the function $h$ is given in terms of $\frac{x}{y}$, the substitution $x=vy$ (which implies $v = \frac{x}{y}$) makes the right-hand side simpler, directly becoming $h(v)$, after rewriting the equation as $\frac{dx}{dy} = \frac{1}{h(x/y)}$.

Thus, the appropriate substitution for a homogeneous differential equation of the form $\frac{dy}{dx} = h \left( \frac{x}{y} \right)$ is $x = vy$.

Let's examine the given options:

(A) $y = vx$: This substitution is standard for $\frac{dy}{dx} = f(\frac{y}{x})$.

(B) $v = yx$: This is not a standard substitution for homogeneous equations.

(C) $x = vy$: This substitution is standard for $\frac{dx}{dy} = g(\frac{x}{y})$. The given equation can be rearranged to $\frac{dx}{dy} = \frac{1}{h(x/y)}$, which is of this form.

(D) $x = v$: This is not a substitution for homogeneous equations involving two variables $x$ and $y$.

The correct substitution is $x = vy$.

The final answer is $\boxed{\text{(C) x = vy}}$.

Given:

Four differential equations are provided as options.

To Find:

Which of the given differential equations is homogeneous.

Solution:

A first-order differential equation of the form $\frac{dy}{dx} = f(x,y)$ is said to be homogeneous if the function $f(x,y)$ is a homogeneous function of degree zero, i.e., $f(\lambda x, \lambda y) = \lambda^0 f(x,y) = f(x,y)$ for any non-zero constant $\lambda$.

Alternatively, a differential equation of the form $M(x,y) dx + N(x,y) dy = 0$ is homogeneous if both $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree.

Let's check each option:

(A) $(4x + 6y + 5) dy – (3y + 2x + 4) dx = 0$

Rewrite as $\frac{dy}{dx} = \frac{3y + 2x + 4}{4x + 6y + 5}$.

Let $f(x,y) = \frac{3y + 2x + 4}{4x + 6y + 5}$.

$f(\lambda x, \lambda y) = \frac{3\lambda y + 2\lambda x + 4}{4\lambda x + 6\lambda y + 5} = \frac{\lambda(3y + 2x) + 4}{\lambda(4x + 6y) + 5}$.

This is not equal to $f(x,y)$ due to the constant terms (4 and 5). Thus, this equation is not homogeneous.

(B) $(xy) dx – (x^3 + y^3) dy = 0$

Rewrite as $\frac{dy}{dx} = \frac{xy}{x^3 + y^3}$.

Let $f(x,y) = \frac{xy}{x^3 + y^3}$.

$f(\lambda x, \lambda y) = \frac{(\lambda x)(\lambda y)}{(\lambda x)^3 + (\lambda y)^3} = \frac{\lambda^2 xy}{\lambda^3 x^3 + \lambda^3 y^3} = \frac{\lambda^2 xy}{\lambda^3 (x^3 + y^3)} = \lambda^{-1} \frac{xy}{x^3 + y^3} = \lambda^{-1} f(x,y)$.

This is not equal to $f(x,y)$ (unless $\lambda = 1$). Thus, this equation is not homogeneous (of degree 0).

Using the $M dx + N dy = 0$ form: $M(x,y) = xy$ (degree 2), $N(x,y) = -(x^3 + y^3)$ (degree 3). The degrees are different, so it's not homogeneous.

(C) $(x^3 + 2y^2) dx + 2xy dy = 0$

Rewrite as $\frac{dy}{dx} = -\frac{x^3 + 2y^2}{2xy}$.

Let $f(x,y) = -\frac{x^3 + 2y^2}{2xy}$.

$f(\lambda x, \lambda y) = -\frac{(\lambda x)^3 + 2(\lambda y)^2}{2(\lambda x)(\lambda y)} = -\frac{\lambda^3 x^3 + 2\lambda^2 y^2}{2\lambda^2 xy} = -\frac{\lambda^2(\lambda x^3 + 2y^2)}{\lambda^2(2xy)} = -\frac{\lambda x^3 + 2y^2}{2xy}$.

This is not equal to $f(x,y)$ (unless $\lambda = 1$). Thus, this equation is not homogeneous.

Using the $M dx + N dy = 0$ form: $M(x,y) = x^3 + 2y^2$. $M(\lambda x, \lambda y) = (\lambda x)^3 + 2(\lambda y)^2 = \lambda^3 x^3 + 2\lambda^2 y^2$. $M(x,y)$ is not a homogeneous function. Thus, it's not homogeneous.

(D) $y^2 dx + (x^2 – xy – y^2) dy = 0$

Rewrite as $\frac{dy}{dx} = -\frac{y^2}{x^2 – xy – y^2} = \frac{y^2}{xy + y^2 - x^2}$.

Let $f(x,y) = \frac{y^2}{xy + y^2 - x^2}$.

$f(\lambda x, \lambda y) = \frac{(\lambda y)^2}{(\lambda x)(\lambda y) + (\lambda y)^2 - (\lambda x)^2} = \frac{\lambda^2 y^2}{\lambda^2 xy + \lambda^2 y^2 - \lambda^2 x^2} = \frac{\lambda^2 y^2}{\lambda^2 (xy + y^2 - x^2)} = \frac{y^2}{xy + y^2 - x^2} = f(x,y)$.

Since $f(\lambda x, \lambda y) = f(x,y)$, this differential equation is homogeneous.

Using the $M dx + N dy = 0$ form: $M(x,y) = y^2$. $M(\lambda x, \lambda y) = (\lambda y)^2 = \lambda^2 y^2$. Homogeneous of degree 2.

$N(x,y) = x^2 – xy – y^2$. $N(\lambda x, \lambda y) = (\lambda x)^2 - (\lambda x)(\lambda y) - (\lambda y)^2 = \lambda^2 x^2 - \lambda^2 xy - \lambda^2 y^2 = \lambda^2 (x^2 - xy - y^2)$. Homogeneous of degree 2.

Since both $M$ and $N$ are homogeneous functions of the same degree (degree 2), the differential equation is homogeneous.

Therefore, the homogeneous differential equation is (D).

The final answer is $\boxed{(D) \ \ y^2 dx + (x^2 – xy – y^2 ) dy = 0}$.

Given:

The differential equation is $\frac{dy}{dx} - y = \cos x$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = -1$ and $Q(x) = \cos x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int (-1) dx = -x$.

So, the integrating factor is $IF = e^{-x}$.

The general solution of a linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y e^{-x} = \int (\cos x) e^{-x} dx + C$

We need to evaluate the integral $I = \int e^{-x} \cos x \, dx$. We can use integration by parts twice.

Let $u = \cos x$ and $dv = e^{-x} dx$. Then $du = -\sin x dx$ and $v = -e^{-x}$.

$I = uv - \int v du = -e^{-x} \cos x - \int (-e^{-x}) (-\sin x) dx = -e^{-x} \cos x - \int e^{-x} \sin x dx$

Now, evaluate $\int e^{-x} \sin x dx$. Let $u = \sin x$ and $dv = e^{-x} dx$. Then $du = \cos x dx$ and $v = -e^{-x}$.

$\int e^{-x} \sin x dx = -e^{-x} \sin x - \int (-e^{-x}) \cos x dx = -e^{-x} \sin x + \int e^{-x} \cos x dx$

Substitute this back into the expression for $I$:

$I = -e^{-x} \cos x - [-e^{-x} \sin x + I]$

$I = -e^{-x} \cos x + e^{-x} \sin x - I$

$2I = e^{-x} (\sin x - \cos x)$

$I = \frac{1}{2} e^{-x} (\sin x - \cos x)$

Now, substitute this back into the general solution formula:

$y e^{-x} = \frac{1}{2} e^{-x} (\sin x - \cos x) + C$

Multiply by $e^x$ to solve for $y$:

$y = \frac{1}{2} e^{-x} (\sin x - \cos x) e^x + C e^x$

$y = \frac{1}{2} (\sin x - \cos x) + C e^x$

The general solution is given by:

$\boxed{\mathbf{y = \frac{1}{2} (\sin x - \cos x) + C e^x}}$

where C is the arbitrary constant.

Given:

The differential equation is $x \frac{dy}{dx} + 2y = x^2$, for $x \ne 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.

Divide by $x$ to write it in the standard form of a linear first-order differential equation, which is $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} + \frac{2}{x}y = x$

Here, $P(x) = \frac{2}{x}$ and $Q(x) = x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \frac{2}{x} dx = 2 \log|x| = \log(x^2)$

So, the integrating factor is $IF = e^{\log(x^2)} = x^2$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y \cdot x^2 = \int x \cdot x^2 dx + C$

$yx^2 = \int x^3 dx + C$

Now, evaluate the integral:

$\int x^3 dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$

So, the equation becomes:

$yx^2 = \frac{x^4}{4} + C$

To express the solution for $y$, divide by $x^2$ (since $x \ne 0$):

$y = \frac{x^4}{4x^2} + \frac{C}{x^2}$

$y = \frac{x^2}{4} + \frac{C}{x^2}$

The general solution is given by:

$\boxed{\mathbf{y = \frac{x^2}{4} + \frac{C}{x^2}}}$

where C is the arbitrary constant.

Given:

The differential equation is $y dx – (x + 2y^2 ) dy = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $y dx – (x + 2y^2 ) dy = 0$.

Rearranging the terms, we get:

$y dx = (x + 2y^2) dy$

$\frac{dx}{dy} = \frac{x + 2y^2}{y}$

$\frac{dx}{dy} = \frac{x}{y} + \frac{2y^2}{y}$

$\frac{dx}{dy} = \frac{1}{y}x + 2y$

This equation is a linear first-order differential equation in the form $\frac{dx}{dy} + P(y)x = Q(y)$.

Here, $P(y) = -\frac{1}{y}$ and $Q(y) = 2y$.

The integrating factor (IF) is given by $e^{\int P(y) dy}$.

$\int P(y) dy = \int -\frac{1}{y} dy = -\log|y| = \log(|y|^{-1}) = \log\left(\frac{1}{|y|}\right)$

Assuming $y \ne 0$, we can take IF $= e^{\log(1/y)} = \frac{1}{y}$.

The general solution of a linear differential equation in $x$ is given by:

$x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$

Substitute the values of $IF$ and $Q(y)$:

$x \cdot \frac{1}{y} = \int (2y) \cdot \frac{1}{y} dy + C$

$\frac{x}{y} = \int 2 dy + C$

Now, evaluate the integral:

$\int 2 dy = 2y$

So, the equation becomes:

$\frac{x}{y} = 2y + C$

To express the solution for $x$, multiply by $y$ (assuming $y \ne 0$):

$x = y(2y + C)$

$x = 2y^2 + Cy$

The general solution is given by:

$\boxed{\mathbf{x = 2y^2 + Cy}}$

where C is the arbitrary constant.

Given:

The differential equation is $\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$, for $x \ne 0$.

The initial condition is $y = 0$ when $x = \frac{\pi}{2}$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = \cot x$ and $Q(x) = 2x + x^2 \cot x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \cot x dx = \log|\sin x|$

So, the integrating factor is $IF = e^{\log|\sin x|} = |\sin x|$.

For $x = \frac{\pi}{2}$, $\sin x = 1 > 0$. Let's consider an interval where $\sin x > 0$, e.g., $(0, \pi)$. Then $|\sin x| = \sin x$.

So, the integrating factor is $IF = \sin x$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y \sin x = \int (2x + x^2 \cot x) \sin x dx + C$

$y \sin x = \int (2x \sin x + x^2 \cot x \sin x) dx + C$

$y \sin x = \int \left(2x \sin x + x^2 \frac{\cos x}{\sin x} \sin x \right) dx + C$

$y \sin x = \int (2x \sin x + x^2 \cos x) dx + C$

Now, evaluate the integral $\int (2x \sin x + x^2 \cos x) dx$.

Notice that the integrand is of the form $\frac{d}{dx}(u \cdot w) = u'w + uw'$. Here, we can see a pattern related to the product rule of differentiation.

Consider the function $x^2 \sin x$. Its derivative is:

$\frac{d}{dx}(x^2 \sin x) = (2x)(\sin x) + (x^2)(\cos x) = 2x \sin x + x^2 \cos x$

So, the integral $\int (2x \sin x + x^2 \cos x) dx$ is simply $x^2 \sin x$.

Thus, the general solution is:

$y \sin x = x^2 \sin x + C$

Now, we use the given condition $y = 0$ when $x = \frac{\pi}{2}$ to find the value of $C$.

Substitute $x=\frac{\pi}{2}$ and $y=0$ into the general solution:

$0 \cdot \sin\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + C$

$0 \cdot 1 = \left(\frac{\pi^2}{4}\right) \cdot 1 + C$

$0 = \frac{\pi^2}{4} + C$

$C = -\frac{\pi^2}{4}$

Substitute this value of $C$ back into the general solution to get the particular solution:

$y \sin x = x^2 \sin x - \frac{\pi^2}{4}$

Divide by $\sin x$ (assuming $\sin x \ne 0$):

$y = x^2 - \frac{\pi^2}{4 \sin x}$

$y = x^2 - \frac{\pi^2}{4} \text{cosec } x$

The particular solution is given by:

$\boxed{\mathbf{y = x^2 - \frac{\pi^2}{4} \text{cosec } x}}$

Given:

The curve passes through the point (0, 1).

The slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the $x$ coordinate and the product of the $x$ coordinate and $y$ coordinate of that point.

To Find:

The equation of the curve.

Solution:

The slope of the tangent to a curve at any point $(x, y)$ is given by $\frac{dy}{dx}$.

According to the problem statement, the slope of the tangent at $(x, y)$ is equal to the sum of the $x$ coordinate and the product of the $x$ coordinate and $y$ coordinate.

This can be written as:

$\frac{dy}{dx} = x + xy$

$\frac{dy}{dx} = x(1 + y)$

This is a first-order differential equation. We can solve this by separating the variables $y$ and $x$.

Assuming $1 + y \ne 0$:

$\frac{dy}{1 + y} = x \, dx$

Integrate both sides:

$\int \frac{dy}{1 + y} = \int x \, dx$

$\log|1 + y| = \frac{x^2}{2} + C$

This is the general solution of the differential equation.

Now, we use the given condition that the curve passes through the point (0, 1) to find the value of the arbitrary constant $C$.

Substitute $x = 0$ and $y = 1$ into the general solution:

$\log|1 + 1| = \frac{0^2}{2} + C$

$\log|2| = 0 + C$

$\log 2 = C$

Substitute this value of $C$ back into the general solution to get the particular solution (the equation of the curve):

$\log|1 + y| = \frac{x^2}{2} + \log 2$

Rearranging the terms to isolate the $y$ term:

$\log|1 + y| - \log 2 = \frac{x^2}{2}$

Using the logarithm property $\log A - \log B = \log(A/B)$:

$\log\left|\frac{1 + y}{2}\right| = \frac{x^2}{2}$

Exponentiate both sides:

$\left|\frac{1 + y}{2}\right| = e^{\frac{x^2}{2}}$

$\frac{1 + y}{2} = \pm e^{\frac{x^2}{2}}$

$1 + y = \pm 2 e^{\frac{x^2}{2}}$

Let $A = \pm 2$, where $A$ is an arbitrary non-zero constant. Since the curve passes through (0, 1), $1+1 = A e^0$, $2 = A$. Thus, the positive sign is applicable.

$1 + y = 2 e^{\frac{x^2}{2}}$

$y = 2 e^{\frac{x^2}{2}} - 1$

Check if $y = 2e^{x^2/2} - 1$ satisfies the condition $1+y \ne 0$. $1 + (2e^{x^2/2} - 1) = 2e^{x^2/2}$. Since $e^{x^2/2}$ is always positive, $2e^{x^2/2} \ne 0$. So the assumption was valid.

The equation of the curve is given by:

$\boxed{\mathbf{y = 2 e^{\frac{x^2}{2}} - 1}}$

Given:

The differential equation is $\frac{dy}{dx} + 2y = \sin x$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = 2$ and $Q(x) = \sin x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int 2 dx = 2x$

So, the integrating factor is $IF = e^{2x}$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y e^{2x} = \int (\sin x) e^{2x} dx + C$

We need to evaluate the integral $I = \int e^{2x} \sin x \, dx$. We can use integration by parts twice.

Let $u = \sin x$ and $dv = e^{2x} dx$. Then $du = \cos x dx$ and $v = \frac{1}{2} e^{2x}$.

$I = uv - \int v du = \sin x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \cos x dx = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \int e^{2x} \cos x dx$

Now, evaluate $\int e^{2x} \cos x dx$. Let $u = \cos x$ and $dv = e^{2x} dx$. Then $du = -\sin x dx$ and $v = \frac{1}{2} e^{2x}$.

$\int e^{2x} \cos x dx = -e^{2x} (-\cos x) + \int e^{2x} \sin x dx = \frac{1}{2} e^{2x} \cos x - \int \frac{1}{2} e^{2x} (-\sin x) dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{2} \int e^{2x} \sin x dx = \frac{1}{2} e^{2x} \cos x + \frac{1}{2} I$

Substitute this back into the expression for $I$:

$I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left(\frac{1}{2} e^{2x} \cos x + \frac{1}{2} I\right)$

$I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x - \frac{1}{4} I$

Add $\frac{1}{4}I$ to both sides:

$I + \frac{1}{4} I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x$

$\frac{5}{4} I = \frac{1}{4} e^{2x} (2 \sin x - \cos x)$

Multiply by $\frac{4}{5}$:

$I = \frac{1}{5} e^{2x} (2 \sin x - \cos x)$

Now, substitute this integral result back into the general solution formula:

$y e^{2x} = \frac{1}{5} e^{2x} (2 \sin x - \cos x) + C$

Multiply by $e^{-2x}$ to solve for $y$:

$y = \frac{1}{5} e^{2x} (2 \sin x - \cos x) e^{-2x} + C e^{-2x}$

$y = \frac{1}{5} (2 \sin x - \cos x) + C e^{-2x}$

The general solution is given by:

$\boxed{\mathbf{y = \frac{1}{5} (2 \sin x - \cos x) + C e^{-2x}}}$

where C is the arbitrary constant.

Given:

The differential equation is $\frac{dy}{dx} + 3y = e^{−2x}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = 3$ and $Q(x) = e^{-2x}$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int 3 dx = 3x$

So, the integrating factor is $IF = e^{3x}$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y e^{3x} = \int e^{-2x} \cdot e^{3x} dx + C$

$y e^{3x} = \int e^{(-2x + 3x)} dx + C$

$y e^{3x} = \int e^x dx + C$

Now, evaluate the integral:

$\int e^x dx = e^x$

So, the equation becomes:

$y e^{3x} = e^x + C$

To express the solution for $y$, multiply by $e^{-3x}$:

$y = (e^x + C) e^{-3x}$

$y = e^x e^{-3x} + C e^{-3x}$

$y = e^{(x - 3x)} + C e^{-3x}$

$y = e^{-2x} + C e^{-3x}$

The general solution is given by:

$\boxed{\mathbf{y = e^{-2x} + C e^{-3x}}}$

where C is the arbitrary constant.

Given:

The differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = \frac{1}{x}$ and $Q(x) = x^2$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \frac{1}{x} dx = \log|x|$

So, the integrating factor is $IF = e^{\log|x|} = |x|$. We can take the positive value for the integrating factor, so $IF = x$ (assuming $x \ne 0$).

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y \cdot x = \int x^2 \cdot x \, dx + C$

$yx = \int x^3 dx + C$

Now, evaluate the integral:

$\int x^3 dx = \frac{x^{3+1}}{3+1} = \frac{x^4}{4}$

So, the equation becomes:

$yx = \frac{x^4}{4} + C$

To express the solution for $y$, divide by $x$ (since $x \ne 0$):

$y = \frac{x^4}{4x} + \frac{C}{x}$

$y = \frac{x^3}{4} + \frac{C}{x}$

The general solution is given by:

$\boxed{\mathbf{y = \frac{x^3}{4} + \frac{C}{x}}}$

where C is the arbitrary constant.

Given:

The differential equation is $\frac{dy}{dx} + (\sec x) y = \tan x$, for $0 \le x < \frac{\pi}{2}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = \sec x$ and $Q(x) = \tan x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \sec x dx$

Using the standard integral for $\sec x$:

$\int \sec x dx = \log|\sec x + \tan x|$

Since $0 \le x < \frac{\pi}{2}$, $\sec x \ge 1$ and $\tan x \ge 0$, so $\sec x + \tan x > 0$. Thus, we can write $\log(\sec x + \tan x)$.

So, the integrating factor is $IF = e^{\log(\sec x + \tan x)} = \sec x + \tan x$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y (\sec x + \tan x) = \int \tan x \cdot (\sec x + \tan x) dx + C$

$y (\sec x + \tan x) = \int (\tan x \sec x + \tan^2 x) dx + C$

Now, evaluate the integral $\int (\tan x \sec x + \tan^2 x) dx$. We can split this into two integrals:

$\int \tan x \sec x dx = \sec x$

For the second integral, use the trigonometric identity $\tan^2 x = \sec^2 x - 1$:

$\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \int \sec^2 x dx - \int 1 dx = \tan x - x$

Combining these results, the integral on the right side is $\sec x + \tan x - x$.

So, the general solution equation becomes:

$y (\sec x + \tan x) = \sec x + \tan x - x + C$

To express the solution for $y$, divide by $(\sec x + \tan x)$ (assuming $\sec x + \tan x \ne 0$, which is true for $0 \le x < \frac{\pi}{2}$):

$y = \frac{\sec x + \tan x - x + C}{\sec x + \tan x}$

$y = \frac{\sec x + \tan x}{\sec x + \tan x} - \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x}$

$y = 1 - \frac{x}{\sec x + \tan x} + \frac{C}{\sec x + \tan x}$

The general solution is given by:

$\boxed{\mathbf{y = 1 + \frac{C - x}{\sec x + \tan x}}}$

where C is the arbitrary constant.

Given:

The differential equation is $\cos^2 x \frac{dy}{dx} + y = \tan x$, for $0 \le x < \frac{\pi}{2}$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $\cos^2 x \frac{dy}{dx} + y = \tan x$.

Divide by $\cos^2 x$ (since $\cos x \ne 0$ for $0 \le x < \frac{\pi}{2}$) to write it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} + \frac{1}{\cos^2 x} y = \frac{\tan x}{\cos^2 x}$

$\frac{dy}{dx} + \sec^2 x \cdot y = \tan x \sec^2 x$

Here, $P(x) = \sec^2 x$ and $Q(x) = \tan x \sec^2 x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \sec^2 x dx = \tan x$

So, the integrating factor is $IF = e^{\tan x}$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y e^{\tan x} = \int (\tan x \sec^2 x) e^{\tan x} dx + C$

We need to evaluate the integral $\int \tan x \sec^2 x e^{\tan x} dx$.

Let $u = \tan x$. Then $du = \sec^2 x dx$.

The integral becomes $\int u e^u du$.

We integrate this by parts using $\int s \, dt = st - \int t \, ds$. Let $s = u$ and $dt = e^u du$. Then $ds = du$ and $t = e^u$.

$\int u e^u du = u e^u - \int e^u du = u e^u - e^u = e^u (u - 1)$

Substitute back $u = \tan x$:

$\int \tan x \sec^2 x e^{\tan x} dx = e^{\tan x}(\tan x - 1)$

So, the general solution equation becomes:

$y e^{\tan x} = e^{\tan x}(\tan x - 1) + C$

To express the solution for $y$, divide by $e^{\tan x}$ (which is never zero):

$y = \frac{e^{\tan x}(\tan x - 1)}{e^{\tan x}} + \frac{C}{e^{\tan x}}$

$y = \tan x - 1 + C e^{-\tan x}$

The general solution is given by:

$\boxed{\mathbf{y = \tan x - 1 + C e^{-\tan x}}}$

where C is the arbitrary constant.

Given:

The differential equation is $x \frac{dy}{dx} + 2y = x^2 \log x$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $x \frac{dy}{dx} + 2y = x^2 \log x$.

Divide by $x$ (assuming $x \ne 0$) to write it in the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} + \frac{2}{x}y = x \log x$

Here, $P(x) = \frac{2}{x}$ and $Q(x) = x \log x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \frac{2}{x} dx = 2 \log|x| = \log(x^2)$

So, the integrating factor is $IF = e^{\log(x^2)} = x^2$ (assuming $x \ne 0$).

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y \cdot x^2 = \int (x \log x) \cdot x^2 dx + C$

$yx^2 = \int x^3 \log x \, dx + C$

We need to evaluate the integral $\int x^3 \log x \, dx$. We use integration by parts, $\int u \, dv = uv - \int v \, du$.

Let $u = \log x$ and $dv = x^3 dx$.

Then $du = \frac{1}{x} dx$ and $v = \int x^3 dx = \frac{x^4}{4}$.

$\int x^3 \log x \, dx = (\log x) \left(\frac{x^4}{4}\right) - \int \frac{x^4}{4} \cdot \frac{1}{x} dx$

$= \frac{x^4}{4} \log x - \int \frac{x^3}{4} dx$

$= \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx$

$= \frac{x^4}{4} \log x - \frac{1}{4} \left(\frac{x^4}{4}\right)$

$= \frac{x^4}{4} \log x - \frac{x^4}{16}$

Now, substitute this integral result back into the general solution formula:

$yx^2 = \frac{x^4}{4} \log x - \frac{x^4}{16} + C$

To express the solution for $y$, divide by $x^2$ (since $x \ne 0$):

$y = \frac{1}{x^2} \left( \frac{x^4}{4} \log x - \frac{x^4}{16} + C \right)$

$y = \frac{x^4}{4x^2} \log x - \frac{x^4}{16x^2} + \frac{C}{x^2}$

$y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2}$

The general solution is given by:

$\boxed{\mathbf{y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2}}}$

where C is the arbitrary constant.

Given:

The differential equation is $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation involves $\log x$, so $x > 0$. Also, the term $x \log x$ multiplies $\frac{dy}{dx}$, so $x \log x \ne 0$, which means $x \ne 1$.

Divide the equation by $x \log x$ to write it in the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} + \frac{y}{x \log x} = \frac{\frac{2}{x} \log x}{x \log x}$

$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2}$

Here, $P(x) = \frac{1}{x \log x}$ and $Q(x) = \frac{2}{x^2}$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \frac{1}{x \log x} dx$

Let $u = \log x$, then $du = \frac{1}{x} dx$.

$\int \frac{1}{u} du = \log|u| = \log|\log x|$.

So, the integrating factor is $IF = e^{\log|\log x|} = |\log x|$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$. We consider two cases based on the sign of $\log x$. However, the constant of integration $C$ in the final solution typically absorbs the sign difference arising from the absolute value, so we can proceed with $\text{IF} = \log x$ and the understanding that the resulting general solution holds for $x > 0, x \ne 1$.

$y (\log x) = \int \frac{2}{x^2} (\log x) dx + C$

$y \log x = 2 \int x^{-2} \log x \, dx + C$

We need to evaluate the integral $\int x^{-2} \log x \, dx$. We use integration by parts, $\int u \, dv = uv - \int v \, du$.

Let $u = \log x$ and $dv = x^{-2} dx$.

Then $du = \frac{1}{x} dx$ and $v = \int x^{-2} dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$.

$\int x^{-2} \log x \, dx = (\log x) \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \frac{1}{x} dx$

$= -\frac{\log x}{x} + \int \frac{1}{x^2} dx$

$= -\frac{\log x}{x} + \int x^{-2} dx$

$= -\frac{\log x}{x} + \frac{x^{-1}}{-1}$

$= -\frac{\log x}{x} - \frac{1}{x}$

$= -\frac{\log x + 1}{x}$

Now, substitute this integral result back into the general solution formula:

$y \log x = 2 \left(-\frac{\log x + 1}{x}\right) + C$

$y \log x = -\frac{2 (\log x + 1)}{x} + C$

This can also be written as:

$y \log x + \frac{2 (\log x + 1)}{x} = C$

The general solution is given by:

$\boxed{\mathbf{y \log x + \frac{2 (\log x + 1)}{x} = C}}$

where C is the arbitrary constant.

Given:

The differential equation is $(1 + x^2) dy + 2xy dx = \cot x dx$, for $x \ne 0$.

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is $(1 + x^2) dy + 2xy dx = \cot x dx$.

We can rearrange this equation to write it in the standard form of a linear first-order differential equation, which is $\frac{dy}{dx} + P(x)y = Q(x)$.

First, move the term with $dx$ that does not have $dy$ to the right side, if necessary. In this case, we can rearrange the given equation as:

$(1 + x^2) dy = \cot x dx - 2xy dx$

$(1 + x^2) dy = (\cot x - 2xy) dx$

Now, divide both sides by $(1 + x^2) dx$ (assuming $1+x^2 \ne 0$, which is always true for real $x$, and $dx \ne 0$):

$\frac{(1 + x^2) dy}{(1 + x^2) dx} = \frac{(\cot x - 2xy) dx}{(1 + x^2) dx}$

$\frac{dy}{dx} = \frac{\cot x - 2xy}{1 + x^2}$

Separate the terms on the right side:

$\frac{dy}{dx} = \frac{\cot x}{1 + x^2} - \frac{2xy}{1 + x^2}$

Move the term involving $y$ to the left side to match the standard form:

$\frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{\cot x}{1 + x^2}$

$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cot x}{1 + x^2}$

This equation is in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing the equation with the standard form, we identify the functions $P(x)$ and $Q(x)$:

$P(x) = \frac{2x}{1 + x^2}$

$Q(x) = \frac{\cot x}{1 + x^2}$

To solve a linear differential equation, we first find the integrating factor (IF), which is given by the formula $e^{\int P(x) dx}$.

First, we evaluate the integral of $P(x)$ with respect to $x$:

$\int P(x) dx = \int \frac{2x}{1 + x^2} dx$}

We can evaluate this integral using substitution. Let $u = 1 + x^2$. Then the differential $du$ is the derivative of $1 + x^2$ with respect to $x$, multiplied by $dx$: $du = 2x \, dx$.

The integral becomes $\int \frac{1}{u} du$. The integral of $\frac{1}{u}$ is $\log|u|$.

$\int \frac{2x}{1 + x^2} dx = \int \frac{1}{u} du = \log|u|$

Substitute back $u = 1 + x^2$. Since $1 + x^2$ is always positive for real $x$, we can remove the absolute value.

$\int P(x) dx = \log(1 + x^2)$

Now, we find the integrating factor:

$IF = e^{\int P(x) dx} = e^{\log(1 + x^2)}$

Using the property $e^{\log a} = a$, we get:

$IF = 1 + x^2$

The general solution of a linear first-order differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $y$, $IF$, and $Q(x)$ into this formula:

$y (1 + x^2) = \int \left(\frac{\cot x}{1 + x^2}\right) \cdot (1 + x^2) dx + C$}

In the integral on the right side, the $(1 + x^2)$ terms cancel out (assuming $1+x^2 \ne 0$):

$y (1 + x^2) = \int \cot x \, dx + C$

Now, we need to evaluate the remaining integral $\int \cot x \, dx$.

$\int \cot x \, dx = \int \frac{\cos x}{\sin x} dx$

We can evaluate this integral using substitution. Let $w = \sin x$. Then $dw = \cos x \, dx$.

$\int \frac{dw}{w} = \log|w|$

Substitute back $w = \sin x$: $\int \cot x \, dx = \log|\sin x|$.

Substitute this result back into the general solution equation:

$y (1 + x^2) = \log|\sin x| + C$

To find the general solution for $y$, divide both sides by $(1 + x^2)$ (which is non-zero):

$y = \frac{\log|\sin x| + C}{1 + x^2}$

We can write this by splitting the fraction:

$y = \frac{1}{1 + x^2} \log|\sin x| + \frac{C}{1 + x^2}$

Using negative exponents, the solution can be written as:

$y = (1 + x^2)^{-1} \log|\sin x| + C(1 + x^2)^{-1}$

Where $C$ is the arbitrary constant of integration.

General Solution:

The general solution of the given differential equation is:

$y = (1 + x^2)^{-1} \log |\sin x| + C(1 + x^2)^{-1}$

where $C$ is an arbitrary constant.

Given:

The differential equation is $x \frac{dy}{dx} + y - x + xy\; \cot x = 0$, for $x \ne 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $x \frac{dy}{dx} + y - x + xy\; \cot x = 0$.

Rearrange the terms to write it in the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$:

$x \frac{dy}{dx} + y + xy\; \cot x = x$

$x \frac{dy}{dx} + y(1 + x \cot x) = x$

Divide by $x$ (since $x \ne 0$):

$\frac{dy}{dx} + \frac{1 + x \cot x}{x} y = 1$

$\frac{dy}{dx} + \left(\frac{1}{x} + \cot x\right) y = 1$

Here, $P(x) = \frac{1}{x} + \cot x$ and $Q(x) = 1$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \left(\frac{1}{x} + \cot x\right) dx$

$\int \left(\frac{1}{x} + \cot x\right) dx = \int \frac{1}{x} dx + \int \cot x dx = \log|x| + \log|\sin x| = \log|x \sin x|$

So, the integrating factor is $IF = e^{\log|x \sin x|} = |x \sin x|$. We can use $IF = x \sin x$ as the multiplying factor (assuming $x \sin x \ne 0$), and the absolute value will be absorbed into the constant of integration in the general solution.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y (x \sin x) = \int 1 \cdot (x \sin x) dx + C$

$yx \sin x = \int x \sin x \, dx + C$

We need to evaluate the integral $\int x \sin x \, dx$. We use integration by parts, $\int u \, dv = uv - \int v \, du$.

Let $u = x$ and $dv = \sin x dx$.

Then $du = dx$ and $v = \int \sin x dx = -\cos x$.

$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) dx$

$= -x \cos x + \int \cos x dx$

$= -x \cos x + \sin x$

Now, substitute this integral result back into the general solution formula:

$yx \sin x = -x \cos x + \sin x + C$

To express the solution for $y$, divide by $x \sin x$ (assuming $x \sin x \ne 0$):

$y = \frac{-x \cos x + \sin x + C}{x \sin x}$

$y = \frac{-x \cos x}{x \sin x} + \frac{\sin x}{x \sin x} + \frac{C}{x \sin x}$

$y = -\frac{\cos x}{\sin x} + \frac{1}{x} + \frac{C}{x \sin x}$

$y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}$

The general solution is given by:

$\boxed{\mathbf{y = \frac{1}{x} - \cot x + \frac{C}{x \sin x}}}$

where C is the arbitrary constant.

Given:

The differential equation is $(x + y) \frac{dy}{dx} = 1$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $(x + y) \frac{dy}{dx} = 1$.

We can rewrite this equation as:

$\frac{dx}{dy} = x + y$

Rearrange the terms to write it in the standard form of a linear first-order differential equation with $y$ as the independent variable and $x$ as the dependent variable, which is $\frac{dx}{dy} + P(y)x = Q(y)$:

$\frac{dx}{dy} - x = y$

Here, $P(y) = -1$ and $Q(y) = y$.

The integrating factor (IF) is given by $e^{\int P(y) dy}$.

$\int P(y) dy = \int (-1) dy = -y$

So, the integrating factor is $IF = e^{-y}$.

The general solution of a linear differential equation is given by the formula:

$x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$

Substitute the values of $IF$ and $Q(y)$:

$x e^{-y} = \int y e^{-y} dy + C$

We need to evaluate the integral $\int y e^{-y} dy$. We use integration by parts, $\int u \, dv = uv - \int v \, du$.

Let $u = y$ and $dv = e^{-y} dy$.

Then $du = dy$ and $v = \int e^{-y} dy = -e^{-y}$.

$\int y e^{-y} dy = y(-e^{-y}) - \int (-e^{-y}) dy$

$= -y e^{-y} + \int e^{-y} dy$

$= -y e^{-y} - e^{-y}$

Now, substitute this integral result back into the general solution formula:

$x e^{-y} = -y e^{-y} - e^{-y} + C$

Multiply the entire equation by $e^y$ to solve for $x$ (or to get rid of the exponential term):

$x e^{-y} \cdot e^y = (-y e^{-y} - e^{-y} + C) \cdot e^y$

$x = -y e^{-y} e^y - e^{-y} e^y + C e^y$

$x = -y - 1 + C e^y$

Rearranging the terms, we can write the solution as:

$x + y + 1 = C e^y$

The general solution is given by:

$\boxed{\mathbf{x + y + 1 = C e^y}}$

where C is the arbitrary constant.

Given:

The differential equation is $y \;dx + (x – y^2) \;dy = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $y \;dx + (x – y^2) \;dy = 0$.

Rearranging the terms, we get:

$y \;dx = -(x – y^2) \;dy$

$y \;dx = (y^2 - x) \;dy$

Divide by $dy$ (assuming $dy \ne 0$):

$y \frac{dx}{dy} = y^2 - x$

Divide by $y$ (assuming $y \ne 0$):

$\frac{dx}{dy} = \frac{y^2 - x}{y}$

$\frac{dx}{dy} = \frac{y^2}{y} - \frac{x}{y}$

$\frac{dx}{dy} = y - \frac{1}{y}x$

Rearrange into the standard form of a linear first-order differential equation, $\frac{dx}{dy} + P(y)x = Q(y)$:

$\frac{dx}{dy} + \frac{1}{y}x = y$

Here, $P(y) = \frac{1}{y}$ and $Q(y) = y$.

The integrating factor (IF) is given by $e^{\int P(y) dy}$.

$\int P(y) dy = \int \frac{1}{y} dy = \log|y|$

So, the integrating factor is $IF = e^{\log|y|} = |y|$. We can take $IF = y$ (assuming $y \ne 0$).

The general solution of a linear differential equation is given by the formula:

$x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$

Substitute the values of $IF$ and $Q(y)$:

$x \cdot y = \int y \cdot y \, dy + C$

$xy = \int y^2 dy + C$

Now, evaluate the integral:

$\int y^2 dy = \frac{y^{2+1}}{2+1} = \frac{y^3}{3}$

So, the equation becomes:

$xy = \frac{y^3}{3} + C$

To express the solution for $x$, divide by $y$ (assuming $y \ne 0$):

$x = \frac{y^3}{3y} + \frac{C}{y}$

$x = \frac{y^2}{3} + \frac{C}{y}$

The general solution is given by:

$\boxed{\mathbf{x = \frac{y^2}{3} + \frac{C}{y}}}$

where C is the arbitrary constant.

Given:

The differential equation is $(x + 3y^2) \frac{dy}{dx} = y$, with $y > 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is $(x + 3y^2) \frac{dy}{dx} = y$.

Since the equation involves terms of $y$ and $y^2$ multiplied by constants and $x$, and the term multiplied by $\frac{dy}{dx}$ contains $x$ and $y^2$, it is likely a linear differential equation with $y$ as the independent variable.

Rewrite the equation in terms of $\frac{dx}{dy}$ (assuming $y \ne 0$):

$\frac{dx}{dy} = \frac{x + 3y^2}{y}$

$\frac{dx}{dy} = \frac{x}{y} + \frac{3y^2}{y}$

$\frac{dx}{dy} = \frac{1}{y}x + 3y$

Rearrange into the standard form of a linear first-order differential equation, $\frac{dx}{dy} + P(y)x = Q(y)$:

$\frac{dx}{dy} - \frac{1}{y}x = 3y$

Here, $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.

The integrating factor (IF) is given by $e^{\int P(y) dy}$.

$\int P(y) dy = \int -\frac{1}{y} dy = -\log|y|$

Since $y > 0$ is given, $|y| = y$.

$\int P(y) dy = -\log y = \log(y^{-1})$

So, the integrating factor is $IF = e^{\log(y^{-1})} = y^{-1} = \frac{1}{y}$.

The general solution of a linear differential equation is given by the formula:

$x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$

Substitute the values of $IF$ and $Q(y)$:

$x \cdot \frac{1}{y} = \int (3y) \cdot \frac{1}{y} dy + C$

$\frac{x}{y} = \int 3 \, dy + C$

Now, evaluate the integral:

$\int 3 \, dy = 3y$

So, the equation becomes:

$\frac{x}{y} = 3y + C$

To express the solution for $x$, multiply by $y$:

$x = y(3y + C)$

$x = 3y^2 + Cy$

The general solution is given by:

$\boxed{\mathbf{x = 3y^2 + Cy}}$

where C is the arbitrary constant.

Given:

The differential equation is $\frac{dy}{dx} + 2y \tan x = \sin x$.

The initial condition is $y = 0$ when $x = \frac{\pi}{3}$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = 2 \tan x$ and $Q(x) = \sin x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int 2 \tan x dx = 2 \int \tan x dx$

$\int 2 \tan x dx = 2 (-\log|\cos x|) = -2 \log|\cos x| = \log(|\cos x|^{-2}) = \log(\sec^2 x)$

So, the integrating factor is $IF = e^{\log(\sec^2 x)} = \sec^2 x$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y \sec^2 x = \int (\sin x) (\sec^2 x) dx + C$

$y \sec^2 x = \int \sin x \frac{1}{\cos^2 x} dx + C$

$y \sec^2 x = \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} dx + C$

$y \sec^2 x = \int \tan x \sec x dx + C$

Now, evaluate the integral:

$\int \tan x \sec x dx = \sec x$

So, the equation becomes:

$y \sec^2 x = \sec x + C$

To express the solution for $y$, divide by $\sec^2 x$ (assuming $\sec^2 x \ne 0$):

$y = \frac{\sec x}{\sec^2 x} + \frac{C}{\sec^2 x}$

$y = \frac{1}{\sec x} + C \cos^2 x$

$y = \cos x + C \cos^2 x$

This is the general solution.

Now, we use the given condition $y = 0$ when $x = \frac{\pi}{3}$ to find the value of $C$.

Substitute $x = \frac{\pi}{3}$ and $y = 0$ into the general solution:

$0 = \cos\left(\frac{\pi}{3}\right) + C \cos^2\left(\frac{\pi}{3}\right)$

We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

$0 = \frac{1}{2} + C \left(\frac{1}{2}\right)^2$

$0 = \frac{1}{2} + C \left(\frac{1}{4}\right)$

$0 = \frac{1}{2} + \frac{C}{4}$

$-\frac{C}{4} = \frac{1}{2}$

$C = -2$

Substitute this value of $C$ back into the general solution to get the particular solution:

$y = \cos x + (-2) \cos^2 x$

$y = \cos x - 2 \cos^2 x$

The particular solution is given by:

$\boxed{\mathbf{y = \cos x - 2 \cos^2 x}}$

Given:

The differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$.

The initial condition is $y = 0$ when $x = 1$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$.

Divide by $(1 + x^2)$ (since $1 + x^2 \ne 0$ for real $x$) to write it in the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{1}{(1 + x^2)^2}$

Here, $P(x) = \frac{2x}{1 + x^2}$ and $Q(x) = \frac{1}{(1 + x^2)^2}$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int \frac{2x}{1 + x^2} dx$

Let $u = 1 + x^2$, then $du = 2x dx$.

$\int \frac{1}{u} du = \log|u| = \log(1 + x^2)$ (since $1 + x^2 > 0$).

So, the integrating factor is $IF = e^{\log(1 + x^2)} = 1 + x^2$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y (1 + x^2) = \int \frac{1}{(1 + x^2)^2} \cdot (1 + x^2) dx + C$

$y (1 + x^2) = \int \frac{1}{1 + x^2} dx + C$

Now, evaluate the integral:

$\int \frac{1}{1 + x^2} dx = \tan^{-1}(x)$

So, the equation becomes:

$y (1 + x^2) = \tan^{-1}(x) + C$

To express the solution for $y$, divide by $(1 + x^2)$:

$y = \frac{\tan^{-1}(x) + C}{1 + x^2}$

This is the general solution.

Now, we use the given condition $y = 0$ when $x = 1$ to find the value of $C$.

Substitute $x = 1$ and $y = 0$ into the general solution:

$0 = \frac{\tan^{-1}(1) + C}{1 + 1^2}$

$0 = \frac{\frac{\pi}{4} + C}{2}$

$0 = \frac{\pi}{4} + C$

$C = -\frac{\pi}{4}$

Substitute this value of $C$ back into the general solution to get the particular solution:

$y = \frac{\tan^{-1}(x) - \frac{\pi}{4}}{1 + x^2}$

The particular solution is given by:

$\boxed{\mathbf{y = \frac{\tan^{-1}(x) - \frac{\pi}{4}}{1 + x^2}}}$

Given:

The differential equation is $\frac{dy}{dx} - 3y \;\cot x = \sin 2x$.

The initial condition is $y = 2$ when $x = \frac{\pi}{2}$.

To Find:

The particular solution satisfying the given condition.

Solution:

The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a linear first-order differential equation.

Here, $P(x) = -3 \cot x$ and $Q(x) = \sin 2x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int -3 \cot x dx = -3 \int \cot x dx$

$\int -3 \cot x dx = -3 \log|\sin x| = \log(|\sin x|^{-3}) = \log(\text{cosec}^3 x)$

So, the integrating factor is $IF = e^{\log(\text{cosec}^3 x)} = \text{cosec}^3 x$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y \text{cosec}^3 x = \int (\sin 2x) (\text{cosec}^3 x) dx + C$

$y \text{cosec}^3 x = \int (2 \sin x \cos x) \left(\frac{1}{\sin^3 x}\right) dx + C$

$y \text{cosec}^3 x = \int \frac{2 \cos x}{\sin^2 x} dx + C$

$y \text{cosec}^3 x = 2 \int \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} dx + C$

$y \text{cosec}^3 x = 2 \int \cot x \text{cosec} x \, dx + C$

Now, evaluate the integral:

$\int \cot x \text{cosec} x \, dx = -\text{cosec } x$

So, the equation becomes:

$y \text{cosec}^3 x = 2 (-\text{cosec } x) + C$

$y \text{cosec}^3 x = -2 \text{cosec } x + C$

To express the solution for $y$, divide by $\text{cosec}^3 x$ (assuming $\text{cosec}^3 x \ne 0$):

$y = \frac{-2 \text{cosec } x}{\text{cosec}^3 x} + \frac{C}{\text{cosec}^3 x}$

$y = -2 \frac{1}{\text{cosec}^2 x} + C \sin^3 x$

$y = -2 \sin^2 x + C \sin^3 x$

This is the general solution.

Now, we use the given condition $y = 2$ when $x = \frac{\pi}{2}$ to find the value of $C$.

Substitute $x = \frac{\pi}{2}$ and $y = 2$ into the general solution:

$2 = -2 \sin^2\left(\frac{\pi}{2}\right) + C \sin^3\left(\frac{\pi}{2}\right)$

We know that $\sin\left(\frac{\pi}{2}\right) = 1$.

$2 = -2 (1)^2 + C (1)^3$

$2 = -2 (1) + C (1)$

$2 = -2 + C$

Add 2 to both sides:

$2 + 2 = C$

$C = 4$

Substitute this value of $C$ back into the general solution to get the particular solution:

$y = -2 \sin^2 x + 4 \sin^3 x$

The particular solution is given by:

$\boxed{\mathbf{y = -2 \sin^2 x + 4 \sin^3 x}}$

Given:

The curve passes through the origin (0, 0).

The slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.

To Find:

The equation of the curve.

Solution:

The slope of the tangent to a curve at any point $(x, y)$ is given by $\frac{dy}{dx}$.

According to the problem statement, the slope of the tangent at $(x, y)$ is equal to the sum of the coordinates $x$ and $y$.

This can be written as:

$\frac{dy}{dx} = x + y$

Rearrange the terms to write it in the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} - y = x$

Here, $P(x) = -1$ and $Q(x) = x$.

The integrating factor (IF) is given by $e^{\int P(x) dx}$.

$\int P(x) dx = \int (-1) dx = -x$

So, the integrating factor is $IF = e^{-x}$.

The general solution of a linear differential equation is given by the formula:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

Substitute the values of $IF$ and $Q(x)$:

$y e^{-x} = \int x e^{-x} dx + C$

We need to evaluate the integral $\int x e^{-x} dx$. We use integration by parts, $\int u \, dv = uv - \int v \, du$.

Let $u = x$ and $dv = e^{-x} dx$.

Then $du = dx$ and $v = \int e^{-x} dx = -e^{-x}$.

$\int x e^{-x} dx = x(-e^{-x}) - \int (-e^{-x}) dx$

$= -x e^{-x} + \int e^{-x} dx$

$= -x e^{-x} - e^{-x}$

Now, substitute this integral result back into the general solution formula:

$y e^{-x} = -x e^{-x} - e^{-x} + C$

Multiply the entire equation by $e^x$ to solve for $y$:

$y e^{-x} \cdot e^x = (-x e^{-x} - e^{-x} + C) \cdot e^x$

$y = -x e^{-x} e^x - e^{-x} e^x + C e^x$

$y = -x - 1 + C e^x$

This is the general solution of the differential equation.

Now, we use the given condition that the curve passes through the origin (0, 0) to find the value of the arbitrary constant $C$.

Substitute $x = 0$ and $y = 0$ into the general solution:

$0 = -0 - 1 + C e^0$

$0 = -1 + C \cdot 1$

$0 = -1 + C$

$C = 1$

Substitute this value of $C$ back into the general solution to get the particular solution (the equation of the curve):

$y = -x - 1 + 1 \cdot e^x$

$y = e^x - x - 1$

The equation of the curve is given by:

$\boxed{\mathbf{y = e^x - x - 1}}$

Given:

The curve passes through the point (0, 2).

The sum of the coordinates of any point $(x, y)$ on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

To Find:

The equation of the curve.

Solution:

Let the slope of the tangent to the curve at any point $(x, y)$ be $\frac{dy}{dx}$.

The sum of the coordinates is $x + y$.

The problem statement says that the sum of the coordinates exceeds the magnitude of the slope by 5. This can be interpreted as:

$(x + y) = \left|\frac{dy}{dx}\right| + 5$

However, checking with the initial condition $(0, 2)$: $0 + 2 = |\frac{dy}{dx}| + 5 \implies 2 = |\frac{dy}{dx}| + 5 \implies |\frac{dy}{dx}| = -3$, which is impossible for a real slope.

It is likely that the term "magnitude of the slope" is used in a less strict sense, or that the phrasing implies the difference between the sum of coordinates and the slope is 5. Based on typical problems of this type and to satisfy the initial condition, we interpret the condition as:

The sum of the coordinates minus the slope of the tangent is equal to 5.

$(x + y) - \frac{dy}{dx} = 5$

This can be rearranged to form a first-order linear differential equation:

$\frac{dy}{dx} = x + y - 5$

Rewrite in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$:

$\frac{dy}{dx} - y = x - 5$

Here, $P(x) = -1$ and $Q(x) = x - 5$.

The integrating factor (IF) is $e^{\int P(x) dx}$.

$\int P(x) dx = \int (-1) dx = -x$

So, the integrating factor is $IF = e^{-x}$.

The general solution of a linear differential equation is given by:

$y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$

$y e^{-x} = \int (x - 5) e^{-x} dx + C$

We evaluate the integral $\int (x - 5) e^{-x} dx$ using integration by parts $\int u \, dv = uv - \int v \, du$.

Let $u = x - 5$, so $du = dx$.

Let $dv = e^{-x} dx$, so $v = \int e^{-x} dx = -e^{-x}$.

$\int (x - 5) e^{-x} dx = (x - 5)(-e^{-x}) - \int (-e^{-x}) dx$

$= -(x - 5)e^{-x} + \int e^{-x} dx$

$= -(x - 5)e^{-x} - e^{-x}$

$= e^{-x} [-(x - 5) - 1]$

$= e^{-x} [-x + 5 - 1]$

$= e^{-x} (4 - x)$

Substitute this back into the general solution formula:

$y e^{-x} = e^{-x} (4 - x) + C$

Multiply both sides by $e^x$:

$y = (4 - x) + C e^x$

$y = 4 - x + C e^x$

This is the general solution.

Now, we use the given condition that the curve passes through the point (0, 2) to find the value of the arbitrary constant $C$.

Substitute $x = 0$ and $y = 2$ into the general solution:

$2 = 4 - 0 + C e^0$

$2 = 4 + C \cdot 1$

$2 = 4 + C$

$C = 2 - 4$

$C = -2$

Substitute this value of $C$ back into the general solution to get the particular solution (the equation of the curve):

$y = 4 - x + (-2) e^x$

$y = 4 - x - 2 e^x$

The equation of the curve is given by:

$\boxed{\mathbf{y = 4 - x - 2 e^x}}$

Given:

The differential equation is $x \frac{dy}{dx} - y = 2x^2$.

To Find:

The Integrating Factor of the given differential equation.

Solution:

The given differential equation is $x \frac{dy}{dx} - y = 2x^2$.

To find the integrating factor, we first need to write the differential equation in the standard form of a linear first-order differential equation:

$\frac{dy}{dx} + P(x)y = Q(x)$

Divide the given equation by $x$ (assuming $x \ne 0$):

$\frac{dy}{dx} - \frac{y}{x} = \frac{2x^2}{x}$

$\frac{dy}{dx} - \frac{1}{x} y = 2x$

Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify $P(x)$ and $Q(x)$.

Here, $P(x) = -\frac{1}{x}$ and $Q(x) = 2x$.

The integrating factor (IF) is given by the formula $e^{\int P(x) dx}$.

First, calculate the integral of $P(x)$:

$\int P(x) dx = \int -\frac{1}{x} dx = -\log|x|$

Now, calculate the integrating factor:

$IF = e^{\int P(x) dx} = e^{-\log|x|}$

Using the property of logarithms $a \log b = \log(b^a)$:

$IF = e^{\log(|x|^{-1})}$

Using the property $e^{\log u} = u$:

$IF = |x|^{-1} = \frac{1}{|x|}$

For the purpose of the integrating factor, we can take $\frac{1}{x}$ (or $-\frac{1}{x}$), as the constant multiple will be absorbed into the constant of integration when finding the general solution. Let's choose $\frac{1}{x}$.

Comparing this with the given options:

(A) $e^{–x}$

(B) $e^{–y}$

(C) $\frac{1}{x}$

(D) x

The calculated integrating factor is $\frac{1}{x}$.

The final answer is $\boxed{\text{(C) }\frac{1}{x}}$.

Given:

The differential equation is $(1 - y^2) \frac{dx}{dy} + yx = ay$, for $-1 < y < 1$.

To Find:

The Integrating Factor of the given differential equation.

Solution:

The given differential equation is $(1 - y^2) \frac{dx}{dy} + yx = ay$.

This is a first-order linear differential equation with $y$ as the independent variable and $x$ as the dependent variable.

To find the integrating factor, we first need to write the differential equation in the standard form:

$\frac{dx}{dy} + P(y)x = Q(y)$

Divide the given equation by $(1 - y^2)$. Since $-1 < y < 1$, $1 - y^2 > 0$, so we can safely divide by this term.

$\frac{dx}{dy} + \frac{yx}{1 - y^2} = \frac{ay}{1 - y^2}$

Rearrange the terms to match the standard form:

$\frac{dx}{dy} + \left(\frac{y}{1 - y^2}\right) x = \frac{ay}{1 - y^2}$

Comparing this with the standard form $\frac{dx}{dy} + P(y)x = Q(y)$, we identify $P(y)$.

Here, $P(y) = \frac{y}{1 - y^2}$.

The integrating factor (IF) is given by the formula $e^{\int P(y) dy}$.

First, calculate the integral of $P(y)$:

$\int P(y) dy = \int \frac{y}{1 - y^2} dy$

To evaluate this integral, we use a substitution. Let $u = 1 - y^2$.

Differentiating $u$ with respect to $y$, we get $du = -2y \, dy$.

So, $y \, dy = -\frac{1}{2} du$.

Substitute $u$ and $dy$ into the integral:

$\int \frac{-\frac{1}{2} du}{u} = -\frac{1}{2} \int \frac{1}{u} du$

Evaluating the integral of $\frac{1}{u}$:

$-\frac{1}{2} \log|u| + C_1$

Substitute back $u = 1 - y^2$:

$-\frac{1}{2} \log|1 - y^2|$

Since $-1 < y < 1$, $1 - y^2$ is always positive, so $|1 - y^2| = 1 - y^2$.

The integral is $-\frac{1}{2} \log(1 - y^2)$.

Now, calculate the integrating factor:

$IF = e^{\int P(y) dy} = e^{-\frac{1}{2} \log(1 - y^2)}$

Using the property of logarithms $a \log b = \log(b^a)$:

$IF = e^{\log((1 - y^2)^{-1/2})}$

Using the property $e^{\log v} = v$:

$IF = (1 - y^2)^{-1/2}$

$IF = \frac{1}{(1 - y^2)^{1/2}}$

$IF = \frac{1}{\sqrt{1 - y^2}}$

Comparing this result with the given options:

(A) $\frac{1}{y^2 − 1}$

(B) $\frac{1}{\sqrt{y^2 − 1}}$

(C) $\frac{1}{1 − y^2}$

(D) $\frac{1}{\sqrt{1 − y^2}}$

The calculated integrating factor matches option (D).

The final answer is $\boxed{\text{(D) }\frac{1}{\sqrt{1 − y^2}}}$.

Given:

The function $y = c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx$, where $c_1$ and $c_2$ are arbitrary constants.

The differential equation is $\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y = 0$.

To Verify:

That the given function $y$ is a solution of the given differential equation.

Solution:

To verify that the function $y = c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx$ is a solution to the differential equation $\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y = 0$, we need to compute the first and second derivatives of $y$ with respect to $x$ and substitute them into the differential equation.

The function is $y = c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx$. We can write this as $y = e^{ax}(c_1 \cos bx + c_2 \sin bx)$.

Step 1: Find the first derivative $\frac{dy}{dx}$.

We differentiate $y$ with respect to $x$ using the product rule $\frac{d}{dx}(uv) = \frac{du}{dx} v + u \frac{dv}{dx}$.

Let $u = e^{ax}$ and $v = c_1 \cos bx + c_2 \sin bx$.

$\frac{du}{dx} = \frac{d}{dx}(e^{ax}) = a e^{ax}$

$\frac{dv}{dx} = \frac{d}{dx}(c_1 \cos bx + c_2 \sin bx) = c_1 (-b \sin bx) + c_2 (b \cos bx) = -bc_1 \sin bx + bc_2 \cos bx$

Now, apply the product rule:

$\frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx}$

$\frac{dy}{dx} = (a e^{ax})(c_1 \cos bx + c_2 \sin bx) + (e^{ax})(-bc_1 \sin bx + bc_2 \cos bx)$

Distribute $e^{ax}$ and $a$ in the first term and $e^{ax}$ in the second term:

$\frac{dy}{dx} = a c_1 e^{ax} \cos bx + a c_2 e^{ax} \sin bx - bc_1 e^{ax} \sin bx + bc_2 e^{ax} \cos bx$}

Group the terms with $\cos bx$ and $\sin bx$:

$\frac{dy}{dx} = (a c_1 + bc_2) e^{ax} \cos bx + (a c_2 - bc_1) e^{ax} \sin bx$}

Step 2: Find the second derivative $\frac{d^2y}{dx^2}$.

We differentiate $\frac{dy}{dx}$ with respect to $x$. We use the product rule again for each term in the expression for $\frac{dy}{dx}$.

Consider the first term: $(a c_1 + bc_2) e^{ax} \cos bx$. Let $u_1 = (a c_1 + bc_2) e^{ax}$ and $v_1 = \cos bx$.

$\frac{du_1}{dx} = (a c_1 + bc_2) (a e^{ax}) = a(a c_1 + bc_2) e^{ax}$}

$\frac{dv_1}{dx} = -b \sin bx$}

Derivative of the first term: $a(a c_1 + bc_2) e^{ax} \cos bx + (a c_1 + bc_2) e^{ax} (-b \sin bx)$

$= (a^2 c_1 + abc_2) e^{ax} \cos bx - (abc_1 + b^2c_2) e^{ax} \sin bx$}

Consider the second term: $(a c_2 - bc_1) e^{ax} \sin bx$. Let $u_2 = (a c_2 - bc_1) e^{ax}$ and $v_2 = \sin bx$.}

$\frac{du_2}{dx} = (a c_2 - bc_1) (a e^{ax}) = a(a c_2 - bc_1) e^{ax}$}

$\frac{dv_2}{dx} = b \cos bx$}

Derivative of the second term: $a(a c_2 - bc_1) e^{ax} \sin bx + (a c_2 - bc_1) e^{ax} (b \cos bx)$

$= (a^2 c_2 - abc_1) e^{ax} \sin bx + (abc_2 - b^2c_1) e^{ax} \cos bx$}

Now, add the derivatives of the two terms to get $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = (a^2 c_1 + abc_2) e^{ax} \cos bx - (abc_1 + b^2c_2) e^{ax} \sin bx$}

$+ (a^2 c_2 - abc_1) e^{ax} \sin bx + (abc_2 - b^2c_1) e^{ax} \cos bx$}

Group the terms with $e^{ax} \cos bx$ and $e^{ax} \sin bx$:

$\frac{d^2y}{dx^2} = [(a^2 c_1 + abc_2) + (abc_2 - b^2c_1)] e^{ax} \cos bx$}

$+ [-(abc_1 + b^2c_2) + (a^2 c_2 - abc_1)] e^{ax} \sin bx$}

$\frac{d^2y}{dx^2} = [a^2 c_1 - b^2c_1 + 2abc_2] e^{ax} \cos bx + [a^2 c_2 - b^2c_2 - 2abc_1] e^{ax} \sin bx$}

$\frac{d^2y}{dx^2} = [(a^2 - b^2) c_1 + 2abc_2] e^{ax} \cos bx + [(a^2 - b^2) c_2 - 2abc_1] e^{ax} \sin bx$}

Step 3: Substitute $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the differential equation.

The differential equation is $\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y = 0$.

Substitute the expressions for $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the left-hand side (LHS) of the equation:

LHS = $\left([(a^2 - b^2) c_1 + 2abc_2] e^{ax} \cos bx + [(a^2 - b^2) c_2 - 2abc_1] e^{ax} \sin bx\right)$}

$ - 2a \left([(a c_1 + bc_2) e^{ax} \cos bx + (a c_2 - bc_1) e^{ax} \sin bx]\right)$}

$ + (a^2 + b^2) \left[c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx\right]$

Factor out $e^{ax}$ from all terms:

LHS = $e^{ax} \left\{ [(a^2 - b^2) c_1 + 2abc_2] \cos bx + [(a^2 - b^2) c_2 - 2abc_1] \sin bx\right.$}

$ \left. - 2a [(a c_1 + bc_2) \cos bx + (a c_2 - bc_1) e^{ax} \sin bx]\right.$}

$ \left. + (a^2 + b^2) [c_1 \cos bx + c_2 \sin bx] \right\}$

Distribute the terms inside the curly brackets:

LHS = $e^{ax} \left\{ [(a^2 - b^2) c_1 + 2abc_2] \cos bx + [(a^2 - b^2) c_2 - 2abc_1] \sin bx\right.$}

$ \left. - [2a(a c_1 + bc_2)] \cos bx - [2a(a c_2 - bc_1)] \sin bx\right.$}

$ \left. + [(a^2 + b^2) c_1] \cos bx + [(a^2 + b^2) c_2] \sin bx \right\}$

Expand the terms being multiplied by $2a$ and $(a^2+b^2)$:

LHS = $e^{ax} \left\{ [(a^2 - b^2) c_1 + 2abc_2] \cos bx + [(a^2 - b^2) c_2 - 2abc_1] \sin bx\right.$}

$ \left. - [2a^2 c_1 + 2abc_2] \cos bx - [2a^2 c_2 - 2abc_1] \sin bx\right.$}

$ \left. + [a^2 c_1 + b^2 c_1] \cos bx + [a^2 c_2 + b^2 c_2] \sin bx \right\}$

Now, collect the coefficients of $\cos bx$ and $\sin bx$ inside the curly brackets.

Coefficient of $\cos bx$:

$(a^2 - b^2) c_1 + 2abc_2 - (2a^2 c_1 + 2abc_2) + (a^2 c_1 + b^2 c_1)$

$= a^2 c_1 - b^2 c_1 + 2abc_2 - 2a^2 c_1 - 2abc_2 + a^2 c_1 + b^2 c_1$}

Combine terms with $c_1$ and $c_2$:

$= (a^2 - b^2 - 2a^2 + a^2 + b^2) c_1 + (2ab - 2ab) c_2$}

$= (0) c_1 + (0) c_2 = 0$

Coefficient of $\sin bx$:

$(a^2 - b^2) c_2 - 2abc_1 - (2a^2 c_2 - 2abc_1) + (a^2 c_2 + b^2 c_2)$

$= a^2 c_2 - b^2 c_2 - 2abc_1 - 2a^2 c_2 + 2abc_1 + a^2 c_2 + b^2 c_2$}

Combine terms with $c_1$ and $c_2$:

$= (-2ab + 2ab) c_1 + (a^2 - b^2 - 2a^2 + a^2 + b^2) c_2$}

$= (0) c_1 + (0) c_2 = 0$

So, the expression inside the curly brackets is $0 \cdot \cos bx + 0 \cdot \sin bx = 0$.

LHS = $e^{ax} \{0\} = 0$

Step 4: Conclusion.

The right-hand side (RHS) of the given differential equation is 0.

We have shown that the LHS = 0.

Since LHS = RHS, the given function $y = c_1 e^{ax} \cos bx + c_2 e^{ax} \sin bx$ satisfies the differential equation $\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2) y = 0$.

Hence, the function is a solution of the given differential equation.

Given:

The differential equation is $ \log \left( \frac{dy}{dx} \right) = 3x + 4y $ .

The initial condition is $ y = 0 $ when $ x = 0 $ .

To Find:

The particular solution of the given differential equation.

Solution:

We have the differential equation:

Convert the equation from logarithmic form to exponential form:

$ \frac{dy}{dx} = e^{3x + 4y} $

Using the property of exponents $ e^{a+b} = e^a \cdot e^b $ , we can write:

$ \frac{dy}{dx} = e^{3x} e^{4y} $

This is a separable differential equation. Separate the variables $ y $ and $ x $ :

$ \frac{dy}{e^{4y}} = e^{3x} dx $

Rewrite the left side:

$ e^{-4y} dy = e^{3x} dx $

Integrate both sides:

$ \int e^{-4y} dy = \int e^{3x} dx $

Performing the integration:

where $ C $ is the constant of integration.

Finding the value of $ C $ using the initial condition:

Given that $ y = 0 $ when $ x = 0 $ . Substitute these values into equation (2):

$ -\frac{1}{4} e^{-4(0)} = \frac{1}{3} e^{3(0)} + C $

$ -\frac{1}{4} e^{0} = \frac{1}{3} e^{0} + C $

Since $ e^0 = 1 $ :

$ -\frac{1}{4} (1) = \frac{1}{3} (1) + C $

$ -\frac{1}{4} = \frac{1}{3} + C $

Solve for $ C $ :

$ C = -\frac{1}{4} - \frac{1}{3} $

Find a common denominator (12):

$ C = -\frac{3}{12} - \frac{4}{12} = -\frac{7}{12} $

Writing the particular solution:

Substitute the value of $ C = -\frac{7}{12} $ back into the general solution (equation 2):

$ -\frac{1}{4} e^{-4y} = \frac{1}{3} e^{3x} - \frac{7}{12} $

To clear the denominators, multiply the entire equation by 12:

$ 12 \left(-\frac{1}{4} e^{-4y}\right) = 12 \left(\frac{1}{3} e^{3x}\right) - 12 \left(\frac{7}{12}\right) $

$ -3 e^{-4y} = 4 e^{3x} - 7 $

Rearrange the terms:

This is the particular solution to the given differential equation that satisfies the initial condition.

Given:

The differential equation is $ (x \;dy \;–\; y \;dx) \;y \sin \left( \frac{y}{x} \right) = (y \;dx + x \;dy) \;x \cos \left( \frac{y}{x} \right) $ .

To Solve:

The given differential equation.

Solution:

The given differential equation is:

$ (x \;dy \;–\; y \;dx) \;y \sin \left( \frac{y}{x} \right) = (y \;dx + x \;dy) \;x \cos \left( \frac{y}{x} \right) $

First, expand the terms on both sides of the equation:

$ (xy \sin \left( \frac{y}{x} \right)) dy - (y^2 \sin \left( \frac{y}{x} \right)) dx = (xy \cos \left( \frac{y}{x} \right)) dx + (x^2 \cos \left( \frac{y}{x} \right)) dy $

Group the terms containing $dy$ on the left side and the terms containing $dx$ on the right side:

$ xy \sin \left( \frac{y}{x} \right) dy - x^2 \cos \left( \frac{y}{x} \right) dy = xy \cos \left( \frac{y}{x} \right) dx + y^2 \sin \left( \frac{y}{x} \right) dx $

Factor out $dy$ from the terms on the left and $dx$ from the terms on the right:

$ [xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)] dy = [xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right)] dx $

Now, express the equation in the form $ \frac{dy}{dx} = f(x,y) $ by dividing both sides by $ [xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)] dx $ (assuming the denominator is not zero):

$ \frac{dy}{dx} = \frac{xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right)}{xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)} $

To check if this is a homogeneous differential equation, we can divide the numerator and the denominator by $ x^2 $ :

$ \frac{dy}{dx} = \frac{\frac{xy}{x^2} \cos \left( \frac{y}{x} \right) + \frac{y^2}{x^2} \sin \left( \frac{y}{x} \right)}{\frac{xy}{x^2} \sin \left( \frac{y}{x} \right) - \frac{x^2}{x^2} \cos \left( \frac{y}{x} \right)} $

$ \frac{dy}{dx} = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + \left(\frac{y}{x}\right)^2 \sin \left( \frac{y}{x} \right)}{\frac{y}{x} \sin \left( \frac{y}{x} \right) - 1 \cdot \cos \left( \frac{y}{x} \right)} $

This shows that $ \frac{dy}{dx} $ can be expressed as a function of $ \frac{y}{x} $ alone. Let $ f(x,y) = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + \left(\frac{y}{x}\right)^2 \sin \left( \frac{y}{x} \right)}{\frac{y}{x} \sin \left( \frac{y}{x} \right) - \cos \left( \frac{y}{x} \right)} $. Then $ f(\lambda x, \lambda y) = f(x,y) $. Thus, the given differential equation is a homogeneous differential equation.

To solve a homogeneous equation, we use the substitution:

$ v = \frac{y}{x} $

This implies $ y = vx $. Differentiate both sides with respect to $ x $ using the product rule:

$ \frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx} = v + x \frac{dv}{dx} $

Substitute $ \frac{y}{x} = v $ and $ \frac{dy}{dx} = v + x \frac{dv}{dx} $ into the differential equation $ \frac{dy}{dx} = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + \left(\frac{y}{x}\right)^2 \sin \left( \frac{y}{x} \right)}{\frac{y}{x} \sin \left( \frac{y}{x} \right) - \cos \left( \frac{y}{x} \right)} $ :

$ v + x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} $

Separate the variables by moving $ v $ to the right side:

$ x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} - v $

Combine the terms on the right side by finding a common denominator:

$ x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v(v \sin v - \cos v)}{v \sin v - \cos v} $

$ x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} $

Simplify the numerator:

$ x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v} $

Separate the variables $ v $ and $ x $. Move terms involving $ v $ and $ dv $ to the left side and terms involving $ x $ and $ dx $ to the right side:

$ \frac{v \sin v - \cos v}{2v \cos v} dv = \frac{dx}{x} $

Split the fraction on the left side:

$ \frac{1}{2} \left( \frac{v \sin v}{v \cos v} - \frac{\cos v}{v \cos v} \right) dv = \frac{1}{x} dx $

$ \frac{1}{2} \left( \tan v - \frac{1}{v} \right) dv = \frac{1}{x} dx $

Integrate both sides of the separated equation:

$ \int \frac{1}{2} \left( \tan v - \frac{1}{v} \right) dv = \int \frac{1}{x} dx $

$ \frac{1}{2} \left( \int \tan v dv - \int \frac{1}{v} dv \right) = \int \frac{1}{x} dx $

The integrals are $\int \tan v dv = \log|\sec v|$ and $\int \frac{1}{v} dv = \log|v|$ and $\int \frac{1}{x} dx = \log|x|$.

$ \frac{1}{2} (\log |\sec v| - \log |v|) = \log |x| + C' $

Using the logarithm property $ \log A - \log B = \log(A/B) $ :

$ \frac{1}{2} \log \left| \frac{\sec v}{v} \right| = \log |x| + C' $

Multiply both sides by 2:

$ \log \left| \frac{\sec v}{v} \right| = 2 \log |x| + 2C' $

Using the logarithm property $ n \log A = \log A^n $ :

$ \log \left| \frac{\sec v}{v} \right| = \log |x^2| + 2C' $

Combine the logarithm terms on the right side, noting $ \log |x^2| = \log (x^2) $ since $ x^2 \ge 0 $:

$ \log \left| \frac{\sec v}{v} \right| = \log (x^2) + \log(e^{2C'}) $

Let $ K = e^{2C'} $. Since $ C' $ is arbitrary, $ K $ is an arbitrary positive constant.

$ \log \left| \frac{\sec v}{v} \right| = \log (K x^2) $

Exponentiate both sides to remove the logarithm:

$ \left| \frac{\sec v}{v} \right| = K x^2 $

$ \frac{\sec v}{v} = \pm K x^2 $

Let $ C = \pm K $. Since $ K $ is a positive arbitrary constant, $ C $ is a non-zero arbitrary constant. (The case $ C=0 $ needs separate consideration, but we will follow the standard form derived via separation). The solution $ y=0 $ gives $ 0=0 $, so $ y=0 $ is a solution. This is not included in $ \sec(y/x) = Cxy $ unless $ \sec(y/x) $ is undefined or $ xy=0 $. Similarly, $ y/x = (n + 1/2)\pi $ gives $ \cos(y/x)=0 $ and $\sec(y/x)$ undefined. These might be singular solutions.

$ \frac{\sec v}{v} = C x^2 $

Multiply by $ v $ :

$ \sec v = C v x^2 $

Substitute back $ v = \frac{y}{x} $ into the equation:

$ \sec \left(\frac{y}{x}\right) = C \left(\frac{y}{x}\right) x^2 $

Simplify the expression on the right side:

$ \sec \left(\frac{y}{x}\right) = C \frac{y}{x} x^2 = C yx $

$ \sec \left(\frac{y}{x}\right) = C xy $

General Solution:

The general solution to the differential equation is:

$ \sec \left(\frac{y}{x}\right) = C xy $

where $ C $ is an arbitrary constant.

Solution:

The given differential equation is:

$(\tan^{-1} y - x) dy = (1 + y^2) dx$

We can rewrite this equation in the form $\frac{dx}{dy} = f(x, y)$. Dividing both sides by $(1 + y^2) dy$, we get:

$\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1 + y^2}$

Separating the terms on the right side:

$\frac{dx}{dy} = \frac{\tan^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$

Rearranging this into the standard form of a linear differential equation $\frac{dx}{dy} + P(y)x = Q(y)$:

$\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{\tan^{-1} y}{1 + y^2}$

Comparing this to the standard form, we identify the coefficients:

$P(y) = \frac{1}{1 + y^2}$

$Q(y) = \frac{\tan^{-1} y}{1 + y^2}$

The Integrating Factor (IF) is calculated as $e^{\int P(y) dy}$:

$IF = e^{\int \frac{1}{1 + y^2} dy}$

Since the integral of $\frac{1}{1 + y^2}$ with respect to $y$ is $\tan^{-1} y$, the Integrating Factor is:

$IF = e^{\tan^{-1} y}$

The general solution of a linear differential equation is given by the formula:

$x \times IF = \int (Q(y) \times IF) dy + C$

Substituting the values of $Q(y)$ and $IF$ into this formula:

$x \times e^{\tan^{-1} y} = \int \left( \frac{\tan^{-1} y}{1 + y^2} \times e^{\tan^{-1} y} \right) dy + C$

To evaluate the integral $\int \left( \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} \right) dy$, we use a substitution. Let $u = \tan^{-1} y$. Then, the differential $du = \frac{1}{1 + y^2} dy$.

The integral transforms into:

$\int u e^u du$

We evaluate this integral using integration by parts, using the formula $\int v \, dw = vw - \int w \, dv$. Let $v = u$ and $dw = e^u du$. This gives $dv = du$ and $w = e^u$.

$\int u e^u du = u e^u - \int e^u du$

$\int u e^u du = u e^u - e^u$

Substituting back $u = \tan^{-1} y$, the evaluated integral is:

$\int \left( \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} \right) dy = (\tan^{-1} y) e^{\tan^{-1} y} - e^{\tan^{-1} y}$

Now, substitute this result back into the general solution equation:

$x \times e^{\tan^{-1} y} = (\tan^{-1} y) e^{\tan^{-1} y} - e^{\tan^{-1} y} + C$

To find the solution for $x$, divide the entire equation by $e^{\tan^{-1} y}$ (since $e^{\tan^{-1} y}$ is always positive and thus non-zero):

$x = \frac{(\tan^{-1} y) e^{\tan^{-1} y}}{e^{\tan^{-1} y}} - \frac{e^{\tan^{-1} y}}{e^{\tan^{-1} y}} + \frac{C}{e^{\tan^{-1} y}}$

Simplifying the expression gives the general solution:

$x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y}$

Solution:

We need to find the order and degree for each of the given differential equations.

(i) The given differential equation is: $\$ \frac{d^2y}{dx^2} + 5x \left( \frac{dy}{dx} \right)^2 - 6y = \log x$

The order of a differential equation is the order of the highest derivative present in the equation.

In this equation, the highest derivative is $\frac{d^2y}{dx^2}$, which is a derivative of order 2.

Therefore, the order of the equation is 2.

The degree of a differential equation is the highest power of the highest order derivative, provided that the equation is a polynomial in its derivatives.

The given equation is a polynomial in the derivatives $\frac{d^2y}{dx^2}$ and $\frac{dy}{dx}$. The highest order derivative is $\frac{d^2y}{dx^2}$, and its power is 1.

Therefore, the degree of the equation is 1.

(ii) The given differential equation is: $\$ \left( \frac{dy}{dx} \right)^3 - 4\left( \frac{dy}{dx} \right)^2 + 7y = \sin x$

The highest derivative present in this equation is $\frac{dy}{dx}$, which is a derivative of order 1.

Therefore, the order of the equation is 1.

The given equation is a polynomial in the derivative $\frac{dy}{dx}$. The highest order derivative is $\frac{dy}{dx}$, and its highest power is 3.

Therefore, the degree of the equation is 3.

(iii) The given differential equation is: $\$ \frac{d^4y}{dx^4} - \sin \left( \frac{d^3y}{dx^3} \right) = 0$

The highest derivative present in this equation is $\frac{d^4y}{dx^4}$, which is a derivative of order 4.

Therefore, the order of the equation is 4.

For the degree to be defined, the differential equation must be expressible as a polynomial in its derivatives. In this equation, the term $\sin \left( \frac{d^3y}{dx^3} \right)$ is a transcendental function (sine) of a derivative ($\frac{d^3y}{dx^3}$).

Since the equation is not a polynomial in its derivatives, the degree is undefined.

We need to verify for each of the given functions that it is a solution of the corresponding differential equation by substituting the function and its derivatives into the differential equation.

(i) Function: $xy = a e^x + b e^{-x} + x^2$

Differential Equation: $x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$

The given function is $xy = a e^x + b e^{-x} + x^2$.

To verify the differential equation, we need to find the first derivative $\frac{dy}{dx}$ and the second derivative $\frac{d^2y}{dx^2}$. We will use implicit differentiation with respect to $x$.

Differentiate the function with respect to $x$:

$\frac{d}{dx}(xy) = \frac{d}{dx}(a e^x + b e^{-x} + x^2)$

Using the product rule on the left side $\frac{d}{dx}(xy) = 1 \cdot y + x \frac{dy}{dx}$:

$y + x \frac{dy}{dx} = a e^x - b e^{-x} + 2x$

Now, differentiate this equation again with respect to $x$ to find the second derivative:

$\frac{d}{dx}\left(y + x \frac{dy}{dx}\right) = \frac{d}{dx}(a e^x - b e^{-x} + 2x)$

Using the sum and product rules:

$\frac{dy}{dx} + \left( \frac{d}{dx}(x) \cdot \frac{dy}{dx} + x \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right) \right) = a e^x - b (-e^{-x}) + 2$

$\frac{dy}{dx} + \left( 1 \cdot \frac{dy}{dx} + x \frac{d^2y}{dx^2} \right) = a e^x + b e^{-x} + 2$

$\frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = a e^x + b e^{-x} + 2$

$2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = a e^x + b e^{-x} + 2$

From the original function $xy = a e^x + b e^{-x} + x^2$, we can write $a e^x + b e^{-x} = xy - x^2$.

Substitute this expression for $a e^x + b e^{-x}$ into the equation for the second derivative:

$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} = (xy - x^2) + 2$

Rearrange the terms to match the differential equation:

$x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} - xy + x^2 - 2 = 0$

This matches the given differential equation. Thus, the function is a solution.

(ii) Function: $y = e^x (a \cos x + b \sin x)$

Differential Equation: $\frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$

The given function is $y = a e^x \cos x + b e^x \sin x$.

We need to find the first and second derivatives.

Differentiate $y$ with respect to $x$ using the product rule:

$\frac{dy}{dx} = \frac{d}{dx}(a e^x \cos x) + \frac{d}{dx}(b e^x \sin x)$

$\frac{dy}{dx} = a (e^x \cos x + e^x (-\sin x)) + b (e^x \sin x + e^x \cos x)$

$\frac{dy}{dx} = a e^x \cos x - a e^x \sin x + b e^x \sin x + b e^x \cos x$

$\frac{dy}{dx} = e^x [(a+b) \cos x + (b-a) \sin x]$

Now, differentiate $\frac{dy}{dx}$ with respect to $x$ to find the second derivative:

$\frac{d^2y}{dx^2} = \frac{d}{dx} [e^x ((a+b) \cos x + (b-a) \sin x)]$

Using the product rule:

$\frac{d^2y}{dx^2} = e^x [(a+b) \cos x + (b-a) \sin x] + e^x [(a+b)(-\sin x) + (b-a)(\cos x)]$}

$\frac{d^2y}{dx^2} = e^x [(a+b) \cos x + (b-a) \sin x - (a+b) \sin x + (b-a) \cos x]$

Combine the terms inside the bracket:

$\frac{d^2y}{dx^2} = e^x [((a+b) + (b-a)) \cos x + ((b-a) - (a+b)) \sin x]$

$\frac{d^2y}{dx^2} = e^x [(a+b+b-a) \cos x + (b-a-a-b) \sin x]$

$\frac{d^2y}{dx^2} = e^x [2b \cos x - 2a \sin x]$

Now, substitute $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ into the LHS of the differential equation $\frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$:

LHS = $(e^x [2b \cos x - 2a \sin x]) - 2 (e^x [(a+b) \cos x + (b-a) \sin x]) + 2 (e^x (a \cos x + b \sin x))$

Factor out $e^x$:

LHS = $e^x \{ [2b \cos x - 2a \sin x] - 2 [(a+b) \cos x + (b-a) \sin x] + 2 [a \cos x + b \sin x] \}$

LHS = $e^x \{ 2b \cos x - 2a \sin x - (2a+2b) \cos x - (2b-2a) \sin x + 2a \cos x + 2b \sin x \}$

Group the terms with $\cos x$ and $\sin x$:

Coefficient of $\cos x$: $2b - (2a+2b) + 2a = 2b - 2a - 2b + 2a = 0$

Coefficient of $\sin x$: $-2a - (2b-2a) + 2b = -2a - 2b + 2a + 2b = 0$

LHS = $e^x \{0 \cdot \cos x + 0 \cdot \sin x\} = e^x \cdot 0 = 0$

The RHS of the differential equation is 0. Since LHS = RHS, the function is a solution.

(iii) Function: $y = x \sin 3x$

Differential Equation: $\frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$

The given function is $y = x \sin 3x$.

We need to find the first and second derivatives.

Differentiate $y$ with respect to $x$ using the product rule:

$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \sin 3x + x \cdot \frac{d}{dx}(\sin 3x)$

$\frac{dy}{dx} = 1 \cdot \sin 3x + x \cdot (\cos 3x \cdot 3)$

$\frac{dy}{dx} = \sin 3x + 3x \cos 3x$}

Now, differentiate $\frac{dy}{dx}$ with respect to $x$ to find the second derivative:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\sin 3x) + \frac{d}{dx}(3x \cos 3x)$

Using the chain rule and product rule:

$\frac{d^2y}{dx^2} = (\cos 3x \cdot 3) + \left( \frac{d}{dx}(3x) \cdot \cos 3x + 3x \cdot \frac{d}{dx}(\cos 3x) \right)$

$\frac{d^2y}{dx^2} = 3 \cos 3x + \left( 3 \cdot \cos 3x + 3x \cdot (-\sin 3x \cdot 3) \right)$

$\frac{d^2y}{dx^2} = 3 \cos 3x + 3 \cos 3x - 9x \sin 3x$}

$\frac{d^2y}{dx^2} = 6 \cos 3x - 9x \sin 3x$}

Now, substitute $y$ and $\frac{d^2y}{dx^2}$ into the LHS of the differential equation $\frac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$:

LHS = $(6 \cos 3x - 9x \sin 3x) + 9(x \sin 3x) - 6\cos 3x$

LHS = $6 \cos 3x - 9x \sin 3x + 9x \sin 3x - 6\cos 3x$}

Group like terms:

LHS = $(6 \cos 3x - 6 \cos 3x) + (-9x \sin 3x + 9x \sin 3x)$

LHS = $0 + 0 = 0$

The RHS of the differential equation is 0. Since LHS = RHS, the function is a solution.

(iv) Function: $x^2 = 2y^2 \log y$

Differential Equation: $(x^2 + y^2) \frac{dy}{dx} - xy = 0$

The given function is $x^2 = 2y^2 \log y$. We need to find the first derivative $\frac{dy}{dx}$. We will use implicit differentiation with respect to $x$. Remember that $y$ is a function of $x$.

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^2) = \frac{d}{dx}(2y^2 \log y)$

Using the chain rule on the left and the constant multiple and product rules on the right:

$2x = 2 \left[ \frac{d}{dx}(y^2) \cdot \log y + y^2 \cdot \frac{d}{dx}(\log y) \right]$

Using the chain rule $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$ and $\frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx}$:

$2x = 2 \left[ (2y \frac{dy}{dx}) \log y + y^2 \left(\frac{1}{y} \frac{dy}{dx}\right) \right]$

$2x = 2 \left[ 2y \log y \frac{dy}{dx} + y \frac{dy}{dx} \right]$

Divide both sides by 2:

$x = 2y \log y \frac{dy}{dx} + y \frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the terms on the right side:

$x = (2y \log y + y) \frac{dy}{dx}$

$x = y (2 \log y + 1) \frac{dy}{dx}$

From the original function $x^2 = 2y^2 \log y$, we can express $2 \log y$:

So, $2 \log y = \frac{x^2}{y^2}$. Substitute this into the equation for $x$:

$x = y \left( \frac{x^2}{y^2} + 1 \right) \frac{dy}{dx}$

Combine the terms inside the parenthesis:

$x = y \left( \frac{x^2 + y^2}{y^2} \right) \frac{dy}{dx}$

$x = \frac{x^2 + y^2}{y} \frac{dy}{dx}$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$

Now, substitute this expression for $\frac{dy}{dx}$ into the LHS of the differential equation $(x^2 + y^2) \frac{dy}{dx} - xy = 0$:

LHS = $(x^2 + y^2) \left( \frac{xy}{x^2 + y^2} \right) - xy$

Assuming $x^2 + y^2 \neq 0$, we can cancel the term $(x^2 + y^2)$:

LHS = $xy - xy$

LHS = $0$

The RHS of the differential equation is 0. Since LHS = RHS, the function is a solution.

Proof:

We are given the relation:

$x^2 - y^2 = c (x^2 + y^2)^2$

We need to show that this relation satisfies the differential equation:

$(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$

We can rewrite the differential equation in terms of $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y}$

Let's differentiate the given relation with respect to $x$. Treating $y$ as a function of $x$ and using implicit differentiation:

$\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(c (x^2 + y^2)^2)$

$2x - 2y \frac{dy}{dx} = c \cdot 2 (x^2 + y^2)^{2-1} \cdot \frac{d}{dx}(x^2 + y^2)$

$2x - 2y \frac{dy}{dx} = 2c (x^2 + y^2) (2x + 2y \frac{dy}{dx})$

$2x - 2y \frac{dy}{dx} = 4c (x^2 + y^2) (x + y \frac{dy}{dx})$

Divide both sides by 2:

$x - y \frac{dy}{dx} = 2c (x^2 + y^2) (x + y \frac{dy}{dx})$

Expand the right side:

$x - y \frac{dy}{dx} = 2cx (x^2 + y^2) + 2cy (x^2 + y^2) \frac{dy}{dx}$

Group terms with $\frac{dy}{dx}$ on one side and terms without $\frac{dy}{dx}$ on the other:

$x - 2cx (x^2 + y^2) = y \frac{dy}{dx} + 2cy (x^2 + y^2) \frac{dy}{dx}$

Factor out $x$ from the left side and $\frac{dy}{dx}$ and $y$ from the right side:

$x (1 - 2c (x^2 + y^2)) = \frac{dy}{dx} [y + 2cy (x^2 + y^2)]$

$x (1 - 2c (x^2 + y^2)) = \frac{dy}{dx} [y (1 + 2c (x^2 + y^2))]$

From the original relation $x^2 - y^2 = c (x^2 + y^2)^2$, we can express $c$ as $c = \frac{x^2 - y^2}{(x^2 + y^2)^2}$ (assuming $x^2+y^2 \neq 0$).

Substitute $c$ into the term $2c(x^2+y^2)$:

$2c(x^2+y^2) = 2 \frac{x^2 - y^2}{(x^2 + y^2)^2} (x^2 + y^2) = \frac{2(x^2 - y^2)}{x^2 + y^2}$

Now substitute this into the coefficients in the differentiated equation:

$1 - 2c(x^2 + y^2) = 1 - \frac{2(x^2 - y^2)}{x^2 + y^2} = \frac{(x^2 + y^2) - 2(x^2 - y^2)}{x^2 + y^2} = \frac{x^2 + y^2 - 2x^2 + 2y^2}{x^2 + y^2} = \frac{3y^2 - x^2}{x^2 + y^2}$

$1 + 2c(x^2 + y^2) = 1 + \frac{2(x^2 - y^2)}{x^2 + y^2} = \frac{(x^2 + y^2) + 2(x^2 - y^2)}{x^2 + y^2} = \frac{x^2 + y^2 + 2x^2 - 2y^2}{x^2 + y^2} = \frac{3x^2 - y^2}{x^2 + y^2}$

Substitute these expressions back into the differentiated equation:

$x \left( \frac{3y^2 - x^2}{x^2 + y^2} \right) = \frac{dy}{dx} \left[ y \left( \frac{3x^2 - y^2}{x^2 + y^2} \right) \right]$

Assuming $x^2+y^2 \neq 0$, we can multiply both sides by $x^2+y^2$:

$x (3y^2 - x^2) = \frac{dy}{dx} y (3x^2 - y^2)$

$3xy^2 - x^3 = \frac{dy}{dx} (3x^2y - y^3)$

Now, rearrange this equation to match the differential form $(...) dx = (...) dy$:

$(3xy^2 - x^3) dx = (3x^2y - y^3) dy$

This derived differential equation is the negative of the given differential equation $(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$. Multiplying our derived equation by -1:

$-(3xy^2 - x^3) dx = -(3x^2y - y^3) dy$

$(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$

This matches the given differential equation. Since the given relation $x^2 - y^2 = c (x^2 + y^2)^2$ contains an arbitrary constant $c$ and differentiating it yields the given differential equation, it is indeed the general solution.

Solution:

The given differential equation is:

$\frac{dy}{dx} + \sqrt{\frac{1 - y^2}{1 - x^2}} = 0$

We can rewrite the equation as:

$\frac{dy}{dx} = - \sqrt{\frac{1 - y^2}{1 - x^2}}$

$\frac{dy}{dx} = - \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}$

This is a variable separable differential equation. We can separate the variables $y$ and $x$ by moving all terms involving $y$ to one side and all terms involving $x$ to the other side:

$\frac{dy}{\sqrt{1 - y^2}} = - \frac{dx}{\sqrt{1 - x^2}}$

Now, integrate both sides of the equation:

$\int \frac{dy}{\sqrt{1 - y^2}} = \int - \frac{dx}{\sqrt{1 - x^2}}$

$\int \frac{dy}{\sqrt{1 - y^2}} = - \int \frac{dx}{\sqrt{1 - x^2}}$

We know the standard integral formulas $\int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1} u + C'$. Applying this to both sides:

$\sin^{-1} y = - \sin^{-1} x + C$

where $C$ is the constant of integration.

Rearranging the terms, we get the general solution:

$\sin^{-1} y + \sin^{-1} x = C$

Alternatively, using the identity $\sin^{-1} A + \sin^{-1} B = \sin^{-1} (A \sqrt{1-B^2} + B \sqrt{1-A^2})$, we can write:

$\sin^{-1} (y \sqrt{1 - x^2} + x \sqrt{1 - y^2}) = C$

Taking the sine of both sides:

$y \sqrt{1 - x^2} + x \sqrt{1 - y^2} = \sin C$

Let $K = \sin C$, where $K$ is another arbitrary constant. Then the solution can also be written as:

$y \sqrt{1 - x^2} + x \sqrt{1 - y^2} = K$

Both forms represent the general solution.

Proof:

The given differential equation is:

$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$

We can rewrite the equation as:

$\frac{dy}{dx} = - \frac{y^2 + y + 1}{x^2 + x + 1}$

This is a variable separable differential equation. We can separate the variables by moving terms involving $y$ to the left side and terms involving $x$ to the right side:

$\frac{dy}{y^2 + y + 1} = - \frac{dx}{x^2 + x + 1}$

Now, integrate both sides of the equation:

$\int \frac{dy}{y^2 + y + 1} = - \int \frac{dx}{x^2 + x + 1}$

To evaluate the integrals, we complete the square in the denominators:

$y^2 + y + 1 = y^2 + 2 \cdot \frac{1}{2}y + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1 = \left(y + \frac{1}{2}\right)^2 + \frac{3}{4} = \left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$

$x^2 + x + 1 = x^2 + 2 \cdot \frac{1}{2}x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$

The integrals are of the form $\int \frac{du}{(u+a)^2 + b^2} = \frac{1}{b} \tan^{-1}\left(\frac{u+a}{b}\right)$. Here $a = \frac{1}{2}$ and $b = \frac{\sqrt{3}}{2}$.

So, $\int \frac{dy}{\left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right)$

And $\int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)$

Applying these results to the integrated differential equation:

$\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) = - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C'$

where $C'$ is the constant of integration.

Divide both sides by $\frac{2}{\sqrt{3}}$:

$\tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) = - \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C$

where $C = \frac{\sqrt{3}}{2}C''$ (let's redefine $C$ to be the combined constant).

Rearrange the terms:

$\tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) = C$

Using the tangent addition formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A + B}{1 - AB}\right)$:

$\tan^{-1}\left(\frac{\frac{2y + 1}{\sqrt{3}} + \frac{2x + 1}{\sqrt{3}}}{1 - \left(\frac{2y + 1}{\sqrt{3}}\right) \left(\frac{2x + 1}{\sqrt{3}}\right)}\right) = C$

$\tan^{-1}\left(\frac{\frac{2y + 1 + 2x + 1}{\sqrt{3}}}{1 - \frac{(2y + 1)(2x + 1)}{3}}\right) = C$

$\tan^{-1}\left(\frac{\frac{2x + 2y + 2}{\sqrt{3}}}{\frac{3 - (4xy + 2x + 2y + 1)}{3}}\right) = C$

$\tan^{-1}\left(\frac{\frac{2(x + y + 1)}{\sqrt{3}}}{\frac{2 - 2x - 2y - 4xy}{3}}\right) = C$

$\tan^{-1}\left(\frac{2(x + y + 1)}{\sqrt{3}} \cdot \frac{3}{2(1 - x - y - 2xy)}\right) = C$

$\tan^{-1}\left(\frac{\sqrt{3}(x + y + 1)}{1 - x - y - 2xy}\right) = C$

Taking the tangent of both sides:

$\frac{\sqrt{3}(x + y + 1)}{1 - x - y - 2xy} = \tan C$

$\sqrt{3}(x + y + 1) = \tan C (1 - x - y - 2xy)$

Rearranging to match the given form $(x + y + 1) = A (1 – x – y – 2xy)$:

$x + y + 1 = \frac{\tan C}{\sqrt{3}} (1 - x - y - 2xy)$

Let $A = \frac{\tan C}{\sqrt{3}}$. Since $C$ is an arbitrary constant, $\tan C$ is an arbitrary constant (assuming $\tan C$ is defined), and thus $A$ is also an arbitrary constant (parameter).

So, we get the solution:

$(x + y + 1) = A (1 - x - y - 2xy)$

This matches the given general solution form, where A is an arbitrary parameter. Therefore, the given relation is the general solution of the differential equation.

Alternatively, we could start with the given solution and differentiate it implicitly to obtain the differential equation. Let's show this alternative verification.

Alternative Verification:

Given relation: $\$ (x + y + 1) = A (1 - x - y - 2xy)$

Differentiate implicitly with respect to $x$:

$\frac{d}{dx}(x + y + 1) = \frac{d}{dx}(A (1 - x - y - 2xy))$

$1 + \frac{dy}{dx} + 0 = A \left( 0 - 1 - \frac{dy}{dx} - (2 \cdot y + 2x \frac{dy}{dx}) \right)$

$1 + \frac{dy}{dx} = A \left( -1 - \frac{dy}{dx} - 2y - 2x \frac{dy}{dx} \right)$

$1 + \frac{dy}{dx} = -A - A \frac{dy}{dx} - 2Ay - 2Ax \frac{dy}{dx}$

Group terms with $\frac{dy}{dx}$ on one side:

$\frac{dy}{dx} + A \frac{dy}{dx} + 2Ax \frac{dy}{dx} = -A - 2Ay - 1$

$\frac{dy}{dx} (1 + A + 2Ax) = -(A + 2Ay + 1)$

$\frac{dy}{dx} = - \frac{A + 2Ay + 1}{1 + A + 2Ax}$

From the given relation, $A = \frac{x + y + 1}{1 - x - y - 2xy}$. Substitute this expression for $A$ into the equation for $\frac{dy}{dx}$:

$\frac{dy}{dx} = - \frac{\frac{x + y + 1}{1 - x - y - 2xy} + 2y \left(\frac{x + y + 1}{1 - x - y - 2xy}\right) + 1}{1 + \frac{x + y + 1}{1 - x - y - 2xy} + 2x \left(\frac{x + y + 1}{1 - x - y - 2xy}\right)}$

Combine terms in the numerator and denominator using a common denominator:

$\frac{dy}{dx} = - \frac{\frac{(x + y + 1) + 2y(x + y + 1) + (1 - x - y - 2xy)}{1 - x - y - 2xy}}{\frac{(1 - x - y - 2xy) + (x + y + 1) + 2x(x + y + 1)}{1 - x - y - 2xy}}$

Cancel the common denominator $(1 - x - y - 2xy)$ (assuming it's non-zero):

$\frac{dy}{dx} = - \frac{(x + y + 1)(1 + 2y) + (1 - x - y - 2xy)}{(1 - x - y - 2xy) + (x + y + 1) + 2x(x + y + 1)}$

Expand the terms in the numerator:

Numerator $= (x + y + 1 + 2xy + 2y^2 + 2y) + (1 - x - y - 2xy)$

Numerator $= x + y + 1 + 2xy + 2y^2 + 2y + 1 - x - y - 2xy$

Numerator $= (x - x) + (y - y + 2y) + (1 + 1) + (2xy - 2xy) + 2y^2$

Numerator $= 0 + 2y + 2 + 0 + 2y^2 = 2y^2 + 2y + 2 = 2(y^2 + y + 1)$

Expand the terms in the denominator:

Denominator $= (1 - x - y - 2xy) + (x + y + 1) + (2x^2 + 2xy + 2x)$

Denominator $= 1 - x - y - 2xy + x + y + 1 + 2x^2 + 2xy + 2x$

Denominator $= (1 + 1) + (-x + x + 2x) + (-y + y) + (-2xy + 2xy) + 2x^2$

Denominator $= 2 + 2x + 0 + 0 + 2x^2 = 2x^2 + 2x + 2 = 2(x^2 + x + 1)$

Substitute the simplified numerator and denominator back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = - \frac{2(y^2 + y + 1)}{2(x^2 + x + 1)}$

$\frac{dy}{dx} = - \frac{y^2 + y + 1}{x^2 + x + 1}$

Rearrange this to match the given differential equation:

$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$

Since the given relation, which contains one arbitrary parameter $A$, satisfies the differential equation, it is the general solution.

Solution:

The given differential equation is:

$\sin x \cos y dx + \cos x \sin y dy = 0$

We can rewrite the equation to separate the variables. Divide the entire equation by $\cos x \cos y$ (assuming $\cos x \neq 0$ and $\cos y \neq 0$):

$\frac{\sin x \cos y}{\cos x \cos y} dx + \frac{\cos x \sin y}{\cos x \cos y} dy = 0$

$\frac{\sin x}{\cos x} dx + \frac{\sin y}{\cos y} dy = 0$

$\tan x dx + \tan y dy = 0$

Now, integrate both sides of the separated equation:

$\int \tan x dx + \int \tan y dy = \int 0 \, dz$

We know that $\int \tan u \, du = -\log |\cos u| + C'$. Applying this formula:

$-\log |\cos x| - \log |\cos y| = C$

where $C$ is the constant of integration.

Rearranging the terms using logarithm properties ($\log A + \log B = \log (AB)$):

$\log |\cos x| + \log |\cos y| = -C$

$\log |\cos x \cos y| = -C$

Exponentiating both sides:

$|\cos x \cos y| = e^{-C}$

$\cos x \cos y = \pm e^{-C}$

Let $K = \pm e^{-C}$. Since $e^{-C}$ is a positive constant, $K$ is a non-zero arbitrary constant.

The general solution of the differential equation is:

$\cos x \cos y = K$

We are given that the curve passes through the point $\left( 0, \frac{\pi}{4} \right)$. We use this point to find the value of the constant $K$. Substitute $x = 0$ and $y = \frac{\pi}{4}$ into the general solution:

$\cos(0) \cos\left(\frac{\pi}{4}\right) = K$

$1 \cdot \frac{1}{\sqrt{2}} = K$

$K = \frac{1}{\sqrt{2}}$

Substitute the value of $K$ back into the general solution to get the equation of the curve passing through the given point:

$\cos x \cos y = \frac{1}{\sqrt{2}}$

This is the required equation of the curve.

Solution:

The given differential equation is:

$(1 + e^{2x}) dy + (1 + y^2) e^x dx = 0$

This is a variable separable differential equation. We can separate the variables by moving terms involving $y$ to the left side and terms involving $x$ to the right side. Divide the equation by $(1 + e^{2x})(1 + y^2)$:

$\frac{(1 + e^{2x}) dy}{(1 + e^{2x})(1 + y^2)} + \frac{(1 + y^2) e^x dx}{(1 + e^{2x})(1 + y^2)} = 0$

$\frac{dy}{1 + y^2} + \frac{e^x}{1 + e^{2x}} dx = 0$

Rearrange the terms:

$\frac{dy}{1 + y^2} = - \frac{e^x}{1 + e^{2x}} dx$

Now, integrate both sides of the equation:

$\int \frac{dy}{1 + y^2} = \int - \frac{e^x}{1 + e^{2x}} dx$

$\int \frac{dy}{1 + y^2} = - \int \frac{e^x}{1 + (e^x)^2} dx$

The integral on the left side is a standard integral:

$\int \frac{dy}{1 + y^2} = \tan^{-1} y$

For the integral on the right side, we can use a substitution. Let $u = e^x$. Then $du = e^x dx$. The integral becomes:

$\int \frac{du}{1 + u^2} = \tan^{-1} u$

Substituting back $u = e^x$, the integral on the right side is $\tan^{-1} (e^x)$.

Applying the integrals to the separated equation:

$\tan^{-1} y = - \tan^{-1} (e^x) + C$

where $C$ is the constant of integration.

Rearranging the terms gives the general solution:

$\tan^{-1} y + \tan^{-1} (e^x) = C$

We are given the initial condition that $y = 1$ when $x = 0$. Substitute these values into the general solution to find the value of $C$:

$\tan^{-1} (1) + \tan^{-1} (e^0) = C$

$\tan^{-1} (1) + \tan^{-1} (1) = C$

We know that $\tan^{-1}(1) = \frac{\pi}{4}$.

$\frac{\pi}{4} + \frac{\pi}{4} = C$

$C = \frac{2\pi}{4} = \frac{\pi}{2}$

Substitute the value of $C$ back into the general solution to obtain the particular solution:

$\tan^{-1} y + \tan^{-1} (e^x) = \frac{\pi}{2}$

Using the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)$ (if $AB < 1$) or alternative forms. A simpler approach is to use $\tan^{-1} A = \frac{\pi}{2} - \tan^{-1} B = \cot^{-1} B$.

$\tan^{-1} y = \frac{\pi}{2} - \tan^{-1} (e^x)$

$\tan^{-1} y = \cot^{-1} (e^x)$

We know that $\cot^{-1} z = \tan^{-1} (1/z)$. So,

$\tan^{-1} y = \tan^{-1} \left(\frac{1}{e^x}\right)$

Taking the tangent of both sides:

$y = \frac{1}{e^x}$

Which can also be written as:

$y = e^{-x}$

The particular solution is $y = e^{-x}$. Let's quickly check if this satisfies the initial condition: when $x=0$, $y = e^{-0} = e^0 = 1$. This matches the given condition $y=1$ when $x=0$. The relation $\tan^{-1} y + \tan^{-1} (e^x) = \frac{\pi}{2}$ is valid for all $x$ as long as $y$ is positive (since $e^x$ is always positive, $\tan^{-1}(e^x)$ is between 0 and $\pi/2$, $\tan^{-1} y$ needs to be between 0 and $\pi/2$ as well for the sum to be $\pi/2$, implying $y > 0$). Our solution $y=e^{-x}$ is always positive.

Solution:

The given differential equation is:

$y \;e^{\frac{x}{y}} \;dx = \left( x \;e^{\frac{x}{y}} + y^2 \right) \;dy$

Given that $y \neq 0$.

We can rewrite the equation in the form $\frac{dx}{dy} = f(x,y)$. Divide both sides by $y \;e^{\frac{x}{y}} \;dy$:

$\frac{dx}{dy} = \frac{x \;e^{\frac{x}{y}} + y^2}{y \;e^{\frac{x}{y}}}$

Now, divide the numerator and denominator by $y \;e^{\frac{x}{y}}$:

$\frac{dx}{dy} = \frac{x \;e^{\frac{x}{y}}}{y \;e^{\frac{x}{y}}} + \frac{y^2}{y \;e^{\frac{x}{y}}}$

$\frac{dx}{dy} = \frac{x}{y} + \frac{y}{e^{\frac{x}{y}}}$

This differential equation is of the form $\frac{dx}{dy} = g\left(\frac{x}{y}\right)$, which indicates it is a homogeneous differential equation.

To solve this, we use the substitution $x = vy$, where $v$ is a function of $y$. Differentiating $x = vy$ with respect to $y$ using the product rule:

$\frac{dx}{dy} = v \cdot \frac{dy}{dy} + y \frac{dv}{dy}$

$\frac{dx}{dy} = v + y \frac{dv}{dy}$

Substitute $x = vy$ and $\frac{dx}{dy} = v + y \frac{dv}{dy}$ into the differential equation:

$v + y \frac{dv}{dy} = v + \frac{y}{e^{\frac{vy}{y}}}$

$v + y \frac{dv}{dy} = v + \frac{y}{e^v}$

Subtract $v$ from both sides:

$y \frac{dv}{dy} = \frac{y}{e^v}$

Since $y \neq 0$, we can divide both sides by $y$:

$\frac{dv}{dy} = \frac{1}{e^v}$

This is a variable separable differential equation. Separate the variables $v$ and $y$:

$e^v dv = dy$

Integrate both sides of the equation:

$\int e^v dv = \int dy$

$e^v = y + C'$

where $C'$ is the constant of integration.

Now, substitute back $v = \frac{x}{y}$ to express the solution in terms of $x$ and $y$:

$e^{\frac{x}{y}} = y + C'$

This is the general solution of the given differential equation.

Sometimes, the constant is written on the other side, e.g., $e^{x/y} - y = C'$. Both forms are equivalent.

Alternatively, from the equation $y \frac{dv}{dy} = \frac{y}{e^v}$, if we had not divided by $y$, we would get $y e^v dv = y dy$. Integrating this gives $\int y e^v dv = \int y dy$. This requires $v$ to be a function of $y$. Let's return to $y \frac{dv}{dy} = \frac{1}{e^v}$ derived from the correct form $v + y \frac{dv}{dy} = v + \frac{1}{e^v}$.

$y \frac{dv}{dy} = \frac{1}{e^v}$

$e^v dv = \frac{1}{y} dy$

Integrating correctly gives:

$\int e^v dv = \int \frac{1}{y} dy$

$e^v = \log |y| + C$

Substitute back $v = \frac{x}{y}$:

$e^{\frac{x}{y}} = \log |y| + C$

This form of the solution is obtained when $\frac{dx}{dy} = \frac{x}{y} + \frac{y}{e^{x/y}}$. Let's re-examine the division by $y \;e^{x/y}$ at the start.

$y \;e^{\frac{x}{y}} \;dx = \left( x \;e^{\frac{x}{y}} + y^2 \right) \;dy$

Divide by $y^2 dx$ to get $\frac{dy}{dx}$ form:

$\frac{y \;e^{\frac{x}{y}}}{y^2} = \frac{x \;e^{\frac{x}{y}} + y^2}{y^2} \frac{dy}{dx}$

$\frac{1}{y} e^{\frac{x}{y}} = \left( \frac{x}{y^2} e^{\frac{x}{y}} + 1 \right) \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{\frac{1}{y} e^{\frac{x}{y}}}{\frac{x}{y^2} e^{\frac{x}{y}} + 1} = \frac{\frac{1}{y} e^{\frac{x}{y}}}{\frac{x e^{x/y} + y^2}{y^2}} = \frac{y^2}{y(x e^{x/y} + y^2)} e^{\frac{x}{y}} = \frac{y e^{x/y}}{x e^{x/y} + y^2}$

Let $v = \frac{y}{x}$. Then $y = vx$. $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

$v + x \frac{dv}{dx} = \frac{vx e^{1/v}}{x e^{1/v} + (vx)^2} = \frac{vx e^{1/v}}{x(e^{1/v} + v^2x)}$ This is not simplifying to a function of $v$.

The first approach $\frac{dx}{dy} = g(\frac{x}{y})$ is correct because the equation $\frac{dx}{dy} = \frac{x}{y} + \frac{y}{e^{x/y}}$ is indeed a function of $\frac{x}{y}$ and $y$. It's not strictly $g(x/y)$. Let's check the form $M(x,y) dx + N(x,y) dy = 0$. Here $M(x,y) = y e^{x/y}$ and $N(x,y) = -(x e^{x/y} + y^2)$.

Let's check homogeneity of degree. $M(\lambda x, \lambda y) = \lambda y e^{\lambda x / \lambda y} = \lambda y e^{x/y} = \lambda M(x,y)$. Degree 1.

$N(\lambda x, \lambda y) = -(\lambda x e^{\lambda x / \lambda y} + (\lambda y)^2) = -(\lambda x e^{x/y} + \lambda^2 y^2)$. This is not homogeneous of degree 1.

So the equation is not homogeneous in the standard sense $f(\lambda x, \lambda y) = \lambda^n f(x,y)$.

However, the substitution $x = vy$ or $y=vx$ is often useful when terms like $x/y$ or $y/x$ appear. Let's check the form of the equation again.

$y \;e^{\frac{x}{y}} \;dx = x \;e^{\frac{x}{y}} \;dy + y^2 \;dy$

$y \;e^{\frac{x}{y}} \;dx - x \;e^{\frac{x}{y}} \;dy = y^2 \;dy$

$e^{\frac{x}{y}} (y \;dx - x \;dy) = y^2 \;dy$

Recall the differential of $\frac{x}{y}$: $d\left(\frac{x}{y}\right) = \frac{y \;dx - x \;dy}{y^2}$.

So, $y \;dx - x \;dy = y^2 \;d\left(\frac{x}{y}\right)$.

Substitute this into the equation:

$e^{\frac{x}{y}} \left( y^2 \;d\left(\frac{x}{y}\right) \right) = y^2 \;dy$

Divide both sides by $y^2$ (since $y \neq 0$):

$e^{\frac{x}{y}} \;d\left(\frac{x}{y}\right) = dy$

This is a variable separable equation where the variable is $\frac{x}{y}$. Let $u = \frac{x}{y}$. The equation becomes:

$e^u du = dy$

Integrate both sides:

$\int e^u du = \int dy$

$e^u = y + C$

where $C$ is the constant of integration.

Substitute back $u = \frac{x}{y}$:

$e^{\frac{x}{y}} = y + C$

This matches the solution found using the $dx/dy$ form in the first attempt, but the derivation is much cleaner recognizing the differential form.

The general solution is:

$e^{\frac{x}{y}} - y = C$

Solution:

The given differential equation is:

$(x - y) (dx + dy) = dx - dy$

The initial condition is $y = -1$ when $x = 0$.

We use the suggested substitution:

$t = x - y$

Differentiating this substitution, we get:

$dt = d(x - y) = dx - dy$

Substitute $t = x - y$ and $dt = dx - dy$ into the given differential equation:

$t (dx + dy) = dt$

We need to express $dx + dy$ in terms of $dt$ and either $dx$ or $dy$. From $dt = dx - dy$, we can write $dx = dy + dt$.

Substitute this expression for $dx$ into $dx + dy$:

$dx + dy = (dy + dt) + dy = 2dy + dt$

Now substitute this into the equation $t (dx + dy) = dt$:

$t (2dy + dt) = dt$

$2t dy + t dt = dt$

Rearrange the terms to separate the variables $y$ and $t$:

$2t dy = dt - t dt$

$2t dy = (1 - t) dt$

If $t \neq 0$, we can write:

$dy = \frac{1 - t}{2t} dt$

$dy = \frac{1}{2} \left( \frac{1}{t} - \frac{t}{t} \right) dt$

$dy = \frac{1}{2} \left( \frac{1}{t} - 1 \right) dt$

Now, integrate both sides of the equation:

$\int dy = \int \frac{1}{2} \left( \frac{1}{t} - 1 \right) dt$

$y = \frac{1}{2} \left( \int \frac{1}{t} dt - \int 1 dt \right)$

$y = \frac{1}{2} (\log |t| - t) + C'$

where $C'$ is the constant of integration.

Substitute back $t = x - y$ into the general solution:

$y = \frac{1}{2} (\log |x - y| - (x - y)) + C'$

Multiply by 2:

$2y = \log |x - y| - (x - y) + 2C'$

$2y = \log |x - y| - x + y + 2C'$

Rearrange the terms:

$x + 2y - y - \log |x - y| = 2C'$

$x + y - \log |x - y| = 2C'$

Let $C = 2C'$. The general solution is:

$x + y - \log |x - y| = C$

Now, we use the initial condition $y = -1$ when $x = 0$ to find the value of the constant $C$. Substitute $x = 0$ and $y = -1$ into the general solution:

$0 + (-1) - \log |0 - (-1)| = C$

$-1 - \log |1| = C$

We know that $\log |1| = \log 1 = 0$ (for the natural logarithm).

$-1 - 0 = C$

$C = -1$

Substitute the value of $C$ back into the general solution to obtain the particular solution:

$x + y - \log |x - y| = -1$

We can rearrange this equation as:

$x + y + 1 = \log |x - y|$

Or, taking the exponential of both sides:

$e^{x + y + 1} = |x - y|$

The particular solution can be expressed as $x + y + 1 = \log |x - y|$.

Given:

The differential equation is $\left[ \frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}} \right] \frac{dy}{dx} = 1$, where $x ≠ 0$.

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is:

$\left[ \frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}} \right] \frac{dy}{dx} = 1$

Since $x \neq 0$, we can multiply both sides by $\sqrt{x}$:

$(e^{−2\sqrt{x}} − y) \frac{dy}{dx} = \sqrt{x}$

Expanding the left side:

$e^{−2\sqrt{x}} \frac{dy}{dx} - y \frac{dy}{dx} = \sqrt{x}$

This form does not directly match standard types like linear, exact, etc. However, observing the terms involving $\frac{1}{\sqrt{x}}$ and $\frac{dy}{dx}$, we consider if the equation leads to a linear first-order differential equation in the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Let us rearrange the terms to try and obtain a standard form. If we rearrange the equation to make $\frac{dy}{dx}$ the subject:

$\frac{dy}{dx} = \frac{1}{\frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}}} = \frac{\sqrt{x}}{e^{−2\sqrt{x}} - y}$

This does not easily simplify to the desired linear form $\frac{dy}{dx} + P(x)y = Q(x)$. However, given the structure of the terms, it is highly probable that the intended solvable form is linear in $y(x)$. Let's consider the equation in the form:

$\frac{dy}{dx} + \frac{y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$

This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{\sqrt{x}}$ and $Q(x) = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$.

To solve this linear equation, we find the integrating factor (IF):

$IF = e^{\int P(x) dx}$

$P(x) = \frac{1}{\sqrt{x}} = x^{-1/2}$.

The integral of $P(x)$ is:

$\int P(x) dx = \int x^{-1/2} dx = \frac{x^{-1/2+1}}{-1/2+1} = \frac{x^{1/2}}{1/2} = 2\sqrt{x}$.

The integrating factor is:

$IF = e^{2\sqrt{x}}$

Now, multiply the linear differential equation $\frac{dy}{dx} + \frac{y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$ by the integrating factor $e^{2\sqrt{x}}$:

$e^{2\sqrt{x}} \frac{dy}{dx} + e^{2\sqrt{x}} \frac{y}{\sqrt{x}} = e^{2\sqrt{x}} \left( \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \right)$

$e^{2\sqrt{x}} \frac{dy}{dx} + \frac{e^{2\sqrt{x}}}{\sqrt{x}} y = \frac{1}{\sqrt{x}}$

The left side of this equation is the exact derivative of the product of $y$ and the integrating factor, i.e., $\frac{d}{dx}(y \cdot IF)$. Let's verify:

$\frac{d}{dx} \left( y e^{2\sqrt{x}} \right) = e^{2\sqrt{x}} \frac{dy}{dx} + y \frac{d}{dx}(e^{2\sqrt{x}})$

$= e^{2\sqrt{x}} \frac{dy}{dx} + y \left( e^{2\sqrt{x}} \cdot \frac{d}{dx}(2\sqrt{x}) \right)$

$= e^{2\sqrt{x}} \frac{dy}{dx} + y \left( e^{2\sqrt{x}} \cdot 2 \cdot \frac{1}{2\sqrt{x}} \right)$

$= e^{2\sqrt{x}} \frac{dy}{dx} + y \frac{e^{2\sqrt{x}}}{\sqrt{x}}$

This matches the left side of the multiplied equation. So, the equation becomes:

$\frac{d}{dx} \left( y e^{2\sqrt{x}} \right) = \frac{1}{\sqrt{x}}$

Now, integrate both sides with respect to $x$:

$\int \frac{d}{dx} \left( y e^{2\sqrt{x}} \right) dx = \int \frac{1}{\sqrt{x}} dx$

$y e^{2\sqrt{x}} = \int x^{-1/2} dx$

$y e^{2\sqrt{x}} = \frac{x^{1/2}}{1/2} + C$

$y e^{2\sqrt{x}} = 2\sqrt{x} + C$

Where $C$ is the constant of integration.

Solving for $y$, we get the general solution:

$y = \frac{2\sqrt{x} + C}{e^{2\sqrt{x}}}$

or

$y = e^{-2\sqrt{x}}(2\sqrt{x} + C)$

Note on Interpretation:

The original equation as stated $\left[ \frac{e^{−2\sqrt{x}}}{\sqrt{x}} − \frac{y}{\sqrt{x}} \right] \frac{dy}{dx} = 1$ implies $\frac{dx}{dy} = \frac{e^{-2\sqrt{x}} - y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}$, which can be written as $\frac{dx}{dy} + \frac{1}{\sqrt{x}}y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$. This is not a standard linear ODE in $y(x)$ or $x(y)$ because the coefficient of $y$ is a function of $x$. The solution presented above is based on the assumption that the equation was intended to be equivalent to the standard linear form $\frac{dy}{dx} + \frac{y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$, which is a common solvable type involving these terms.

Given:

The differential equation is $\frac{dy}{dx} + y \;\cot x = 4x \;cosec \;x$, where $x ≠ 0$.

The initial condition is $y = 0$ when $x = \frac{\pi}{2}$.

To Find:

A particular solution of the given differential equation satisfying the initial condition.

Solution:

The given differential equation is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Comparing the given equation with the standard form, we have:

$P(x) = \cot x$

$Q(x) = 4x \;\text{cosec}\; x$

To find the general solution, we first calculate the integrating factor (IF), which is given by $IF = e^{\int P(x) dx}$.

Calculate $\int P(x) dx$:

$\int P(x) dx = \int \cot x \; dx = \log|\sin x|$

The integrating factor is:

$IF = e^{\int \cot x \; dx} = e^{\log|\sin x|} = |\sin x|$

Since the initial condition is given at $x = \frac{\pi}{2}$, and for $x$ in the neighborhood of $\frac{\pi}{2}$, $\sin x > 0$, we can take $IF = \sin x$.

The general solution is given by $y \cdot (IF) = \int Q(x) \cdot (IF) \; dx + C$, where C is the constant of integration.

Substitute the values of $Q(x)$ and $IF$:

$y \cdot \sin x = \int (4x \;\text{cosec}\; x) \cdot (\sin x) \; dx + C$

$y \sin x = \int 4x \cdot \frac{1}{\sin x} \cdot \sin x \; dx + C$

$y \sin x = \int 4x \; dx + C$

Now, evaluate the integral on the right side:

$\int 4x \; dx = 4 \int x \; dx = 4 \cdot \frac{x^2}{2} = 2x^2$

So, the general solution is:

$y \sin x = 2x^2 + C$

Now, we use the initial condition $y = 0$ when $x = \frac{\pi}{2}$ to find the value of $C$.

Substitute $y = 0$ and $x = \frac{\pi}{2}$ into the general solution:

$0 \cdot \sin \left( \frac{\pi}{2} \right) = 2 \left( \frac{\pi}{2} \right)^2 + C$

$0 \cdot 1 = 2 \cdot \frac{\pi^2}{4} + C$

$0 = \frac{\pi^2}{2} + C$

Solving for $C$:

$C = - \frac{\pi^2}{2}$

Substitute the value of $C$ back into the general solution to obtain the particular solution:

$y \sin x = 2x^2 - \frac{\pi^2}{2}$

The particular solution can also be written as:

$y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$

or

$y = (2x^2 - \frac{\pi^2}{2}) \;\text{cosec}\; x$

The particular solution is:

$y = (2x^2 - \frac{\pi^2}{2}) \;\text{cosec}\; x$

Given:

The differential equation is $(x + 1) \frac{dy}{dx} = 2 \;e^{-y} - 1$.

The initial condition is $y = 0$ when $x = 0$.

To Find:

A particular solution of the given differential equation satisfying the initial condition.

Solution:

The given differential equation is:

$(x + 1) \frac{dy}{dx} = 2 \;e^{-y} - 1$

This is a first-order separable differential equation. We can separate the variables $y$ and $x$ on different sides of the equation.

Provided that $x+1 \neq 0$ (i.e., $x \neq -1$) and $2e^{-y} - 1 \neq 0$, we can rewrite the equation as:

$\frac{dy}{2e^{-y} - 1} = \frac{dx}{x + 1}$

Now, integrate both sides of the equation:

$\int \frac{dy}{2e^{-y} - 1} = \int \frac{dx}{x + 1}$

Let's evaluate the integral on the left side. We can rewrite the denominator $2e^{-y} - 1$ as $\frac{2}{e^y} - 1 = \frac{2 - e^y}{e^y}$.

So, the integrand becomes $\frac{1}{\frac{2 - e^y}{e^y}} = \frac{e^y}{2 - e^y}$.

The left integral is $\int \frac{e^y}{2 - e^y} dy$.

Let $u = 2 - e^y$. Then, the differential $du = \frac{d}{dy}(2 - e^y) dy = -e^y dy$. This means $e^y dy = -du$.

Substituting $u$ and $du$ into the integral:

$\int \frac{-du}{u} = -\int \frac{du}{u} = -\log|u| + C_1$

Substitute back $u = 2 - e^y$:

$-\log|2 - e^y| + C_1$

Now, let's evaluate the integral on the right side:

$\int \frac{dx}{x + 1}$

Let $v = x+1$. Then, $dv = dx$.

The integral becomes $\int \frac{dv}{v} = \log|v| + C_2$.

Substitute back $v = x+1$:

$\log|x + 1| + C_2$

Equating the results of the integrals:

$-\log|2 - e^y| = \log|x + 1| + C'$, where $C' = C_2 - C_1$ is the constant of integration.

Rearrange the equation:

$\log|x + 1| + \log|2 - e^y| = -C'$

Using the logarithm property $\log A + \log B = \log(AB)$:

$\log(|x + 1| |2 - e^y|) = -C'$

Exponentiate both sides with base $e$:

$e^{\log(|x + 1| |2 - e^y|)} = e^{-C'}$

$|x + 1| |2 - e^y| = e^{-C'}$

Let $K = e^{-C'}$. Since $e^{anything}$ is always positive, $K$ is a positive constant. The general solution is:

$|x + 1| |2 - e^y| = K$

Now, we use the initial condition $y = 0$ when $x = 0$ to find the value of the constant $K$.

Substitute $x = 0$ and $y = 0$ into the general solution:

$|0 + 1| |2 - e^0| = K$

$|1| |2 - 1| = K$

$1 \cdot 1 = K$

$K = 1$

Substitute $K=1$ back into the general solution to get the particular solution:

$|x + 1| |2 - e^y| = 1$

Since the initial condition is at $(0, 0)$, where $x+1 = 1 > 0$ and $2-e^y = 2-e^0 = 2-1 = 1 > 0$, we can remove the absolute value signs in the neighborhood of the initial condition.

$(x + 1) (2 - e^y) = 1$

We need to express the particular solution by solving for $y$ in terms of $x$.

Divide both sides by $(x+1)$ (assuming $x \neq -1$):

$2 - e^y = \frac{1}{x + 1}$

Rearrange the terms to isolate $e^y$:

$e^y = 2 - \frac{1}{x + 1}$

Combine the terms on the right side:

$e^y = \frac{2(x + 1) - 1}{x + 1}$

$e^y = \frac{2x + 2 - 1}{x + 1}$

$e^y = \frac{2x + 1}{x + 1}$

Take the natural logarithm (log) of both sides to solve for $y$. This step requires $\frac{2x+1}{x+1} > 0$.

$y = \log\left(\frac{2x + 1}{x + 1}\right)$

This particular solution is valid for $x$ such that $x+1 \neq 0$ and $\frac{2x+1}{x+1} > 0$. The condition $\frac{2x+1}{x+1} > 0$ holds when $x < -1$ or $x > -\frac{1}{2}$. Since the initial condition is at $x=0$, which is in the interval $(-\frac{1}{2}, \infty)$, the solution is valid in this interval.

The particular solution is:

$y = \log\left(\frac{2x + 1}{x + 1}\right)$

Given:

The differential equation is $\frac{y \;dx − x \;dy}{y} = 0$.

To Find:

The general solution of the differential equation.

Solution:

The given differential equation is:

$\frac{y \;dx − x \;dy}{y} = 0$

Assuming $y \neq 0$, we can multiply both sides by $y$:

$y \;dx − x \;dy = 0$

This is a first-order differential equation. We can solve this by separating the variables $x$ and $y$.

Move the term containing $dy$ to the right side:

$y \;dx = x \;dy$}

Now, divide both sides by $xy$ to separate the variables (assuming $x \neq 0$ and $y \neq 0$):

$\frac{y \;dx}{xy} = \frac{x \;dy}{xy}$

Simplify both sides:

$\frac{dx}{x} = \frac{dy}{y}$

Now, integrate both sides of the separated equation:

$\int \frac{dx}{x} = \int \frac{dy}{y}$

The integral of $\frac{1}{z}$ with respect to $z$ is $\log|z|$. So, we have:

$\log|x| = \log|y| + C'$, where $C'$ is the constant of integration.

Rearrange the terms to find the general solution. Move $\log|y|$ to the left side:

$\log|x| - \log|y| = C'$}

Using the property of logarithms $\log A - \log B = \log(A/B)$:

$\log\left|\frac{x}{y}\right| = C'$}

To eliminate the logarithm, exponentiate both sides (using base $e$):

$\left|\frac{x}{y}\right| = e^{C'}$

Let $K = e^{C'}$. Since $C'$ is an arbitrary constant, $K$ is a positive arbitrary constant ($K > 0$).

$\frac{x}{y} = \pm K$

Let $C = \pm K$. Since $K$ is a positive arbitrary constant, $C$ is a non-zero arbitrary constant. Thus, we have:

$\frac{x}{y} = C$}

This can be rewritten as $x = Cy$. Alternatively, we can write $y = \frac{1}{C} x$. If we let $C_{\text{new}} = \frac{1}{C}$, then $y = C_{\text{new}} x$. Since $C$ is a non-zero arbitrary constant, $C_{\text{new}}$ is also a non-zero arbitrary constant.

However, standard practice often absorbs the constant differently. Let's start from $\log|x| = \log|y| + C'$. We can write $C'$ as $\log|K|$ where $K$ is an arbitrary positive constant.

$\log|x| = \log|y| + \log|K|$

$\log|x| = \log|Ky|$

Exponentiate both sides:

$|x| = |Ky| = |K||y|$

$|x| = K|y|$ (since $K > 0$)

This implies $x = \pm K y$. Let $C = \pm K$. Since $K$ is a positive constant, $C$ is a non-zero constant.

$x = Cy$ or $y = \frac{1}{C} x$. Let $C_{\text{final}} = \frac{1}{C}$. Then $y = C_{\text{final}} x$. This form covers all lines through the origin except the y-axis ($x=0$).

Alternatively, if we write $\log|y| = \log|x| - C'$. Let $C'' = -C'$. Then $\log|y| = \log|x| + C''$. Let $C'' = \log|A|$.

$\log|y| = \log|x| + \log|A| = \log|Ax|$

$|y| = |Ax| = |A||x|$

Let $C = \pm A$. Then $y = Cx$. If $A$ is a positive arbitrary constant, then $C$ is a non-zero arbitrary constant. This form covers all lines through the origin except the x-axis ($y=0$).

The solution $y=Cx$ for $C \neq 0$ represents lines passing through the origin, excluding the axes. The original equation $\frac{y \;dx − x \;dy}{y} = 0$ is equivalent to $y \, dx - x \, dy = 0$ for $y \neq 0$. The solution $y=Cx$ also includes the line $y=0$ when $C=0$. Although the original equation is undefined for $y=0$, the form $y=Cx$ is generally accepted as the general solution covering most cases and the limiting case $C=0$.

Comparing our solution with the given options:

(A) $xy = C$

(B) $x = Cy^2$

(C) $y = Cx$

(D) $y = Cx^2$}

Our solution $y = Cx$ matches option (C).

The general solution is $\mathbf{y = Cx}$.

The correct option is (C) y = Cx.

Given:

The question asks for the general solution of a differential equation of the type $\frac{dy}{dx} + P_1x = Q_1$. The options provide potential solution formulas.

To Find:

The correct general solution formula among the given options.

Solution:

The standard form of a linear first-order differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$. The general solution for this type is given by the formula $y \cdot (\text{Integrating Factor}) = \int Q(x) \cdot (\text{Integrating Factor}) dx + C$, where the Integrating Factor is $e^{\int P(x) dx}$. This corresponds to the structure of option (B) if $P_1$ and $Q_1$ were functions of $x$ and the equation was $\frac{dy}{dx} + P_1 y = Q_1$.

Another type of linear first-order differential equation is when $x$ is the dependent variable and $y$ is the independent variable. The standard form is $\frac{dx}{dy} + P(y)x = Q(y)$. The general solution for this type is given by the formula $x \cdot (\text{Integrating Factor}) = \int Q(y) \cdot (\text{Integrating Factor}) dy + C$, where the Integrating Factor is $e^{\int P(y) dy}$. This corresponds to the structure of option (C) if $P_1$ and $Q_1$ were functions of $y$ and the equation was $\frac{dx}{dy} + P_1 x = Q_1$.

The differential equation given in the question is $\frac{dy}{dx} + P_1x = Q_1$. The presence of the term $P_1x$ on the left side, where $x$ is multiplied by $P_1$, and the derivative $\frac{dy}{dx}$ suggests a form that does not directly match the standard linear forms unless $P_1$ and $Q_1$ have specific structures or there is a typo in the question.

However, looking at the options, option (C) provides a solution formula for a linear differential equation where the dependent variable is $x$ and the independent variable is $y$, i.e., $\frac{dx}{dy} + P(y)x = Q(y)$. The formula is $x \;e^{\int P(y) \;dy} = \int \left( Q(y) e^{\int P(y)\;dy} \right) dy + C$.

If we assume that the question intended to present a linear differential equation in $x$ with respect to $y$, of the form $\frac{dx}{dy} + P_1(y)x = Q_1(y)$, then $P(y) = P_1(y)$ and $Q(y) = Q_1(y)$. In this case, the integrating factor is $e^{\int P_1(y) dy}$ (or $e^{\int P_1 dy}$ using the given notation for $P_1$ which is assumed to be a function of $y$), and the general solution is $x \cdot e^{\int P_1 \;dy} = \int \left( Q_1 e^{\int P_1\;dy} \right) dy + C$.

This formula exactly matches option (C).

Given the options provided, it is most likely that the type of differential equation intended in the question was $\frac{dx}{dy} + P_1 x = Q_1$, where $P_1$ and $Q_1$ are functions of $y$, and the formula for its general solution is requested.

Let's verify the derivation for the standard form $\frac{dx}{dy} + P(y)x = Q(y)$:

The integrating factor (IF) is $e^{\int P(y) dy}$.

Multiply the differential equation by the integrating factor:

$e^{\int P(y) dy} \frac{dx}{dy} + P(y) e^{\int P(y) dy} x = Q(y) e^{\int P(y) dy}$

The left side is the derivative of the product $x \cdot e^{\int P(y) dy}$ with respect to $y$ (using the product rule and chain rule):

$\frac{d}{dy} \left( x \;e^{\int P(y) \;dy} \right) = x \cdot \frac{d}{dy}(e^{\int P(y) dy}) + e^{\int P(y) dy} \cdot \frac{dx}{dy}$

$\frac{d}{dy}(e^{\int P(y) dy}) = e^{\int P(y) dy} \cdot \frac{d}{dy}(\int P(y) dy) = e^{\int P(y) dy} \cdot P(y)$

So, $\frac{d}{dy} \left( x \;e^{\int P(y) \;dy} \right) = x \cdot P(y) e^{\int P(y) dy} + e^{\int P(y) dy} \frac{dx}{dy}$, which matches the left side of the multiplied equation.

Therefore, the equation becomes:

$\frac{d}{dy} \left( x \;e^{\int P(y) \;dy} \right) = Q(y) e^{\int P(y) dy}$

Integrate both sides with respect to $y$:

$\int \frac{d}{dy} \left( x \;e^{\int P(y) \;dy} \right) dy = \int \left( Q(y) e^{\int P(y) \;dy} \right) dy$

$x \;e^{\int P(y) \;dy} = \int \left( Q(y) e^{\int P(y) \;dy} \right) dy + C$

Using the notation from the options ($P_1$ and $Q_1$, assumed to be functions of $y$), the general solution is:

$x \;e^{\int P_1 \;dy} = \int \left( Q_1 e^{\int P_1\;dy} \right) dy + C$

This derivation matches the formula given in option (C).

The correct option is (C) $x \;e^{\int P_1 \;dy} = \int \left( Q_1 e^{\int P_1\;dy} \right) dy + C$.

Given:

The differential equation is $e^x dy + (y e^x + 2x) dx = 0$.

The options for the general solution are:

(A) $x e^y + x^2 = C$

(B) $x e^y + y^2 = C$

(C) $y e^x + x^2 = C$

(D) $y e^y + x^2 = C$

To Find:

The general solution of the given differential equation.

Solution:

The given differential equation is $e^x dy + (y e^x + 2x) dx = 0$.

This equation can be written in the form $M(x, y) dx + N(x, y) dy = 0$, where:

$M(x, y) = y e^x + 2x$

$N(x, y) = e^x$

We check if the differential equation is exact by verifying if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

Calculate $\frac{\partial M}{\partial y}$:

$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y e^x + 2x)$

Treating $x$ as a constant:

$\frac{\partial M}{\partial y} = e^x + 0 = e^x$

Calculate $\frac{\partial N}{\partial x}$:

$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(e^x)$

Treating $y$ as a constant (although $N$ does not contain $y$):

$\frac{\partial N}{\partial x} = e^x$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = e^x$, the given differential equation is exact.

For an exact differential equation, the general solution is given by $\int M(x, y) \;dx + \int (\text{terms in } N(x, y) \text{ independent of } x) \;dy = C$, where the first integral is taken with respect to $x$ keeping $y$ constant, and the second integral is taken with respect to $y$ considering only those terms in $N(x, y)$ that do not contain $x$.

Integrate $M(x, y)$ with respect to $x$ (keeping $y$ constant):

$\int (y e^x + 2x) \;dx = y \int e^x \;dx + \int 2x \;dx$

$= y e^x + x^2$

Identify terms in $N(x, y) = e^x$ that are independent of $x$. There are no terms in $e^x$ that do not contain $x$. Thus, the terms in $N$ independent of $x$ are $0$.

Integrate the terms in $N$ independent of $x$ with respect to $y$:

$\int 0 \;dy = 0$

Combine the results from the two integrals and add the constant of integration $C$:

$(y e^x + x^2) + 0 = C$

$y e^x + x^2 = C$

This is the general solution of the given differential equation.

Now, compare this solution with the given options:

(A) $x e^y + x^2 = C$

(B) $x e^y + y^2 = C$

(C) $y e^x + x^2 = C$

(D) $y e^y + x^2 = C$

The derived general solution matches option (C).

The final answer is:

(C) $y e^x + x^2 = C$