Solving First Order Differential Equations

Solving a Differential Equation (General Approach)

Solving a differential equation is the process of finding the unknown function (or functions) that satisfy the given equation relating the function, its independent variable(s), and its derivatives. The goal is to find a relationship between the variables that does not involve derivatives. This is the reverse process of forming a differential equation from a family of curves.

Finding solutions to differential equations is a central task in differential calculus and is essential for analyzing models of dynamic systems. The process typically involves using integration, as integration is the inverse operation of differentiation.

Types of Solutions

As discussed in a previous section, differential equations can have different types of solutions:

• General Solution: This solution contains arbitrary constants, the number of which is equal to the order of the differential equation. It represents a family of functions that satisfy the differential equation. For an $n$-th order ODE, the general solution will contain $n$ independent arbitrary constants.
• Particular Solution: This is a specific solution obtained from the general solution by determining the exact values of the arbitrary constants. These values are determined by using additional information, such as initial conditions (the state of the system at a particular value of the independent variable) or boundary conditions (the state of the system at different values of the independent variable).

For instance, solving $\frac{dy}{dx} = \cos x$ gives the general solution $y = \sin x + C$. If we're given the initial condition $y(0)=1$, plugging in $x=0, y=1$ gives $1 = \sin(0) + C \implies 1 = 0 + C \implies C=1$. The particular solution is $y = \sin x + 1$.

General Approach for First-Order Ordinary Differential Equations

A first-order ordinary differential equation (ODE) involves only the first derivative of a single dependent variable with respect to a single independent variable. Such an equation can generally be written in the form $\frac{dy}{dx} = f(x, y)$ or $M(x, y) dx + N(x, y) dy = 0$. There is no single method that solves all first-order ODEs. Instead, different techniques are used depending on the specific structure of the equation.

The general process for solving a first-order ODE involves the following steps:

1. Examine and Identify the Type of the Equation: Look at the structure of the differential equation to determine which class it belongs to. Common types include:
   • Variable Separable Equations ($\frac{dy}{dx} = f(x)g(y)$)
   • Homogeneous Equations ($\frac{dy}{dx} = f(y/x)$)
   • Linear Equations ($\frac{dy}{dx} + P(x)y = Q(x)$)
   • Exact Equations ($M(x, y) dx + N(x, y) dy = 0$ where $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$)
   • Equations Reducible to Homogeneous or Linear Form
2. Apply the Appropriate Solution Method: Use the specific technique (such as separation of variables, substitution, multiplying by an integrating factor, etc.) that is designed for the identified type of equation. This step aims to transform the equation into a form that is directly integrable.
3. Perform Integration: Carry out the necessary integration(s) to eliminate the derivative(s). When finding indefinite integrals during this process, remember to include the constant of integration ($C$). For a first-order ODE, you will introduce one arbitrary constant $C$.
4. Express the General Solution: Algebraically manipulate the resulting equation to obtain the relationship between $y$ and $x$. This can be an explicit solution of the form $y = \phi(x)$ (if $y$ can be isolated) or an implicit solution of the form $G(x, y) = C$ or $G(x, y) = 0$. This represents the general solution, the family of curves that satisfy the differential equation.
5. Apply Initial/Boundary Conditions (if given): If specific conditions (e.g., $y(x_0)=y_0$) are provided, substitute the values of $x$ and $y$ from these conditions into the general solution. This allows you to solve for the unique value of the constant of integration $C$ that corresponds to the specific situation. Substituting this value of $C$ back into the general solution gives the particular solution.

It is important to note that not all differential equations have solutions that can be expressed in terms of elementary functions (like polynomials, trigonometric functions, exponentials, logarithms, etc.). In many cases, solutions might need to be expressed as infinite series or can only be approximated using numerical methods.

Variable Separable Method

The variable separable method is one of the simplest and most direct techniques for solving first-order Ordinary Differential Equations. It applies to differential equations that can be rearranged so that all terms involving the dependent variable $y$ and its differential $dy$ are on one side of the equation, and all terms involving the independent variable $x$ and its differential $dx$ are on the other side.

Recognizing the Form of a Variable Separable Equation

A first-order differential equation is said to be variable separable if it can be written in the form:

$g(y) dy = f(x) dx$

where $g(y)$ is a function of $y$ only, and $f(x)$ is a function of $x$ only.

This form is often obtained from an equation given as $\frac{dy}{dx} = h(x, y)$ if the function $h(x, y)$ can be factored into a product of a function of $x$ only and a function of $y$ only. That is, if the equation is of the form $\frac{dy}{dx} = f(x) \cdot K(y)$, where $K(y) = \frac{1}{g(y)}$ (assuming $g(y) \neq 0$). Then, separating variables gives $\frac{dy}{K(y)} = f(x) dx$, which is $g(y) dy = f(x) dx$.

Method of Solution for Variable Separable Equations

Once a differential equation is recognized as variable separable and written in the form $g(y) dy = f(x) dx$, the solution is found by integrating both sides independently.

1. Separate Variables: Manipulate the given differential equation algebraically to isolate all terms containing $y$ and $dy$ on one side of the equation, and all terms containing $x$ and $dx$ on the other side. Ensure that $dy$ and $dx$ appear in the numerator on their respective sides.

   Example: If the equation is $\frac{dy}{dx} = \frac{x^2}{y}$, multiply both sides by $y$ and $dx$: $y \frac{dy}{dx} dx = x^2 dx \implies y dy = x^2 dx$.

   Example: If the equation is $\frac{dy}{dx} = y \cos x$, divide by $y$ and multiply by $dx$: $\frac{dy}{y} = \cos x dx$.

2. Integrate Both Sides: Integrate the left side of the separated equation with respect to $y$ and the right side with respect to $x$.

   $\int g(y) dy = \int f(x) dx$

3. Evaluate Integrals and Add Constant of Integration: Evaluate the indefinite integrals on both sides. Let $G(y)$ be an antiderivative of $g(y)$ and $F(x)$ be an antiderivative of $f(x)$. The result of the integration is of the form:

   $G(y) + C_1 = F(x) + C_2$

   Since $C_1$ and $C_2$ are arbitrary constants, their difference $C_2 - C_1$ is also an arbitrary constant. We typically combine them into a single arbitrary constant $C$ and add it to only one side (conventionally, the side with the independent variable $x$).

   $G(y) = F(x) + C$

   This $C$ is the arbitrary constant of integration for the first-order ODE.

4. Solve for y Explicitly (Optional): The equation $G(y) = F(x) + C$ is the general solution in implicit form. If it is algebraically possible and desired, solve this equation for $y$ to express it explicitly as a function of $x$, $y = \phi(x)$. In many cases, solving for $y$ is difficult or impossible, and the implicit form is the accepted general solution.
5. Apply Initial Condition (if given): If an initial condition $y(x_0)=y_0$ is provided, substitute $x=x_0$ and $y=y_0$ into the general solution (either implicit or explicit form) and solve for the specific value of the constant $C$. Substituting this value back into the general solution gives the particular solution.

Answer:

Homogeneous Differential Equations

A first-order Ordinary Differential Equation is classified as homogeneous if the expression for $\frac{dy}{dx}$ can be written solely as a function of the ratio of the dependent variable to the independent variable, $\frac{y}{x}$. These equations have a specific structure that allows them to be transformed into a variable separable form using a suitable substitution.

Recognizing the Form of a Homogeneous Differential Equation

A first-order differential equation is homogeneous if it can be expressed in the form:

$\frac{dy}{dx} = F\left(\frac{y}{x}\right)$

where $F$ is some function.

Alternatively, a differential equation written in 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. A function $G(x, y)$ is defined as a homogeneous function of degree $n$ if for any non-zero scalar $\lambda$, $G(\lambda x, \lambda y) = \lambda^n G(x, y)$. If $M(x, y)$ and $N(x, y)$ are homogeneous of the same degree $n$, then the ratio $\frac{M(x, y)}{N(x, y)}$ is homogeneous of degree $0$, i.e., $\frac{M(\lambda x, \lambda y)}{N(\lambda x, \lambda y)} = \frac{\lambda^n M(x, y)}{\lambda^n N(x, y)} = \frac{M(x, y)}{N(x, y)}$. A function of degree 0 can always be written as a function of the ratio $y/x$. For example, divide numerator and denominator by $x^n$.

Example: Consider the equation $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$. Let's check if the right side is a function of $y/x$. Divide the numerator and denominator by $x^2$: $\frac{(x^2/x^2) + (y^2/x^2)}{(xy)/x^2} = \frac{1 + (y/x)^2}{y/x}$. This is indeed a function of $y/x$. $F(v) = \frac{1+v^2}{2v}$ where $v=y/x$. So the equation is homogeneous.

Method of Solution for Homogeneous Differential Equations

The standard method for solving a homogeneous differential equation involves a substitution that transforms it into a variable separable equation.

1. Confirm Homogeneity: Verify that the given differential equation can be written in the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ or that $M(x, y)$ and $N(x, y)$ in $M dx + N dy = 0$ are homogeneous functions of the same degree.
2. Introduce Substitution: Introduce a new dependent variable $v$ defined by the substitution $v = \frac{y}{x}$. This implies that $y = vx$.
3. Find $\frac{dy}{dx}$ in terms of $v$ and $\frac{dv}{dx}$: Differentiate the substitution equation $y = vx$ with respect to $x$ using the Product Rule. Remember that both $v$ and $x$ are treated as functions of $x$ (or $v$ depends on $x$).

   "$\frac{dy}{dx} = \frac{d}{dx}(v \cdot x) = v \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(v)$"

   [Using Product Rule]

   "$ = v \cdot (1) + x \frac{dv}{dx} = v + x \frac{dv}{dx}$"

   So, $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

4. Substitute into the Original Differential Equation: Replace $\frac{y}{x}$ with $v$ and $\frac{dy}{dx}$ with $v + x \frac{dv}{dx}$ in the original differential equation (which should be in the form $\frac{dy}{dx} = F(\frac{y}{x})$).

   $v + x \frac{dv}{dx} = F(v)$

5. Separate Variables: Rearrange the equation obtained in Step 4 to separate the variables $v$ and $x$. The goal is to get all terms involving $v$ and $dv$ on one side and all terms involving $x$ and $dx$ on the other.

   Subtract $v$ from both sides: $x \frac{dv}{dx} = F(v) - v$.

   Assuming $F(v) - v \neq 0$, divide by $F(v) - v$ and multiply by $dx$: $\frac{dv}{F(v) - v} = \frac{dx}{x}$. This is now a variable separable equation in terms of $v$ and $x$.

6. Integrate Both Sides: Integrate the left side with respect to $v$ and the right side with respect to $x$.

   $\int \frac{dv}{F(v) - v} = \int \frac{dx}{x}$

   Evaluate the integrals. The integral on the right side is $\ln|x| + C_1$. The integral on the left side depends on the specific form of $F(v)$.

7. Substitute Back to the Original Variables: After evaluating the integrals and combining the constants of integration, the equation will be a relationship between $v$, $x$, and the constant $C$. Replace $v$ with its original expression in terms of $x$ and $y$, i.e., $v = \frac{y}{x}$.
8. Simplify: Algebraically manipulate the resulting equation to obtain the general solution, which is often in implicit form $G(x, y) = C$. If possible and required, express $y$ explicitly in terms of $x$. Apply initial conditions if given to find a particular solution.

Answer:

Equations Reducible to Homogeneous Form

Some first-order Ordinary Differential Equations that are not homogeneous can be transformed into homogeneous equations (which we know how to solve) using a clever substitution. A common type of such equation is of the form:

$\frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are constants. The method for reducing this equation to a homogeneous form depends on the relationship between the lines represented by the linear expressions in the numerator and denominator: $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$. These lines can either be parallel or intersect.

Case 1: The Lines are Parallel

The lines $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$ are parallel if the coefficients of $x$ and $y$ are proportional, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} = k$ for some constant $k$. In this case, $a_1 = k a_2$ and $b_1 = k b_2$. The numerator expression $a_1 x + b_1 y + c_1$ can be written as $k(a_2 x + b_2 y) + c_1$. The differential equation then takes the form $\frac{dy}{dx} = \frac{k(a_2 x + b_2 y) + c_1}{a_2 x + b_2 y + c_2}$.

Method when Lines are Parallel ($\frac{a_1}{a_2} = \frac{b_1}{b_2}$):

1. Identify Parallel Lines: Check if $\frac{a_1}{a_2} = \frac{b_1}{b_2}$. If so, the lines are parallel.
2. Introduce Substitution: Introduce a new dependent variable $z$ equal to the common linear expression in $x$ and $y$ (without the constant term). Let $z = a_2 x + b_2 y$ (or $z = a_1 x + b_1 y$).
3. Find $\frac{dz}{dx}$ in terms of $\frac{dy}{dx}$: Differentiate the substitution equation $z = a_2 x + b_2 y$ with respect to $x$:

   "$\frac{dz}{dx} = \frac{d}{dx}(a_2 x + b_2 y) = a_2 \frac{d}{dx}(x) + b_2 \frac{d}{dx}(y) = a_2(1) + b_2 \frac{dy}{dx}$"

   So, $\frac{dz}{dx} = a_2 + b_2 \frac{dy}{dx}$.

4. Solve for $\frac{dy}{dx}$: Rearrange the derivative of the substitution to express $\frac{dy}{dx}$ in terms of $z$ and $\frac{dz}{dx}$.

   $\frac{dz}{dx} - a_2 = b_2 \frac{dy}{dx}$

   Assuming $b_2 \neq 0$, $\frac{dy}{dx} = \frac{1}{b_2} \left(\frac{dz}{dx} - a_2\right)$. (If $b_2=0$, the original equation involves $\frac{dy}{dx}$ and only $x$ in the denominator, which is separable).

5. Substitute into the Original DE: Replace $\frac{dy}{dx}$ and the linear expressions in $x$ and $y$ with expressions involving $z$ and $\frac{dz}{dx}$. The original equation $\frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}$ will transform into a variable separable equation involving $z$ and $x$.
6. Solve the Separable Equation: Solve the resulting variable separable equation for $z$ in terms of $x$ using the integration method.
7. Substitute Back: Replace $z$ with its original expression in terms of $x$ and $y$ ($z = a_2 x + b_2 y$) to get the final solution in terms of $x$ and $y$.

Example Structure: Consider $\frac{dy}{dx} = \frac{x + y + 1}{2x + 2y + 3}$. Here $a_1=1, b_1=1, a_2=2, b_2=2$. $\frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{1}{2}$, so they are equal. Let $z = x+y$. Then $\frac{dz}{dx} = 1 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dz}{dx} - 1$. The original DE becomes $\frac{dz}{dx} - 1 = \frac{z+1}{2z+3}$. Rearranging gives $\frac{dz}{dx} = 1 + \frac{z+1}{2z+3} = \frac{2z+3+z+1}{2z+3} = \frac{3z+4}{2z+3}$. This is separable: $\frac{2z+3}{3z+4} dz = dx$. The left integral might require polynomial long division in $z$ before integrating.

Case 2: The Lines are Intersecting

If the coefficients of $x$ and $y$ are not proportional, i.e., $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$ are intersecting lines. The intersection point is key to the substitution in this case.

Method when Lines are Intersecting ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$):

1. Find the Point of Intersection: Solve the system of linear equations:

   $a_1 x + b_1 y + c_1 = 0$

   $a_2 x + b_2 y + c_2 = 0$

   Find the unique point of intersection $(h, k)$.

2. Introduce New Variables: Introduce new independent and dependent variables, $X$ and $Y$, related to the original variables $x$ and $y$ by shifting the origin to the intersection point $(h, k)$.

   $x = X + h$

   $y = Y + k$

   Note that $X = x - h$ and $Y = y - k$.

3. Relate the Differentials: Differentiate the substitutions with respect to $X$: $\frac{dx}{dX} = \frac{d}{dX}(X+h) = 1$, so $dx = dX$. Differentiate $y=Y+k$ with respect to $X$ using Chain Rule: $\frac{dy}{dX} = \frac{dY}{dX}$. Also, $\frac{dy}{dx} = \frac{dY}{dX}$ (since $dx=dX$ and $dy=dY$).
4. Substitute into the Original DE: Replace $x, y,$ and $\frac{dy}{dx}$ in the original differential equation with expressions involving $X, Y,$ and $\frac{dY}{dX}$.

   "$\frac{dY}{dX} = \frac{a_1 (X + h) + b_1 (Y + k) + c_1}{a_2 (X + h) + b_2 (Y + k) + c_2}$"

   [Substitute $x, y, dy/dx$]

   "$\frac{dY}{dX} = \frac{a_1 X + b_1 Y + (a_1 h + b_1 k + c_1)}{a_2 X + b_2 Y + (a_2 h + b_2 k + c_2)}$"

   [Rearrange numerator and denominator]

5. Simplify using Intersection Point Property: Since $(h, k)$ is the point of intersection, it satisfies both linear equations: $a_1 h + b_1 k + c_1 = 0$ and $a_2 h + b_2 k + c_2 = 0$. The terms in parentheses in the numerator and denominator become zero. The equation simplifies to:

   $\frac{dY}{dX} = \frac{a_1 X + b_1 Y}{a_2 X + b_2 Y}$

6. Solve the Resulting Homogeneous Equation: The equation $\frac{dY}{dX} = \frac{a_1 X + b_1 Y}{a_2 X + b_2 Y}$ is now a homogeneous differential equation in the variables $Y$ and $X$. Solve this equation using the standard method for homogeneous equations, which involves the substitution $Y = vX$.
7. Substitute Back to the Original Variables: After finding the solution in terms of $Y$ and $X$ (usually of the form $G(Y/X) = \ln|X| + C$), substitute back $X = x - h$ and $Y = y - k$ to obtain the final solution in terms of the original variables $x$ and $y$.

Example Structure: Consider $\frac{dy}{dx} = \frac{x+y-1}{x-y-3}$. Lines $x+y-1=0$ and $x-y-3=0$. Adding them gives $(x+y-1)+(x-y-3)=0 \implies 2x-4=0 \implies x=2$. Substitute $x=2$ into $x+y-1=0 \implies 2+y-1=0 \implies y+1=0 \implies y=-1$. Intersection point $(h, k) = (2, -1)$. Substitute $x=X+2, y=Y-1$. The DE becomes $\frac{dY}{dX} = \frac{(X+2)+(Y-1)-1}{(X+2)-(Y-1)-3} = \frac{X+Y+2-1-1}{X-Y+2+1-3} = \frac{X+Y}{X-Y}$. This is homogeneous. Solve using $Y=vX$. After finding the solution in terms of $v$ and $X$, substitute back $v=Y/X = (y+1)/(x-2)$ and $X=x-2$.

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