In our previous discussions, we explored methods for solving first-order ordinary differential equations (ODEs) using direct integration and separable equations. These methods are effective for a variety of equations but are not always applicable to every type of first-order ODE.

In this article, we will focus on linear first-order ODEs.

1. Linear Equations

Linear first-order ordinary differential equations (ODEs) have the standard form:

\frac{dy}{dx} + P(x)y = Q(x)

Here, $P(x) \$ and $Q(x) \$ are functions of $x \$ .

When $Q(x) = 0 \$ the equation is called a homogeneous linear differential equation. If $Q(x) \$ is nonzero, the equation is referred to as a non-homogeneous linear differential equation.

To solve these equations, we use a method known as the integrating factor method. The integrating factor, denoted by $\mu(x) \$ , is defined as:

\mu(x) = e^{\int P(x) dx}

Once the integrating factor is determined, we multiply the entire ODE by $\ \mu(x) \$ to simplify and solve the equation.

\mu(x)\frac{dy}{dx} + \mu(x)P(x)y = \mu(x)Q(x)

The following outlines the steps to solve a linear first-order ordinary differential equation (ODE) of the general form:

$\frac{dy}{dx} + P(x)y = Q(x)\$

where $Px \$ and $Qx \$ are functions of $x \$ .

1. Identify the Integrating Factor:

The integrating factor $\mu x \$ is calculated as:

$\mu(x) = e^{\int P(x) \, dx} \$

2. Multiply the Differential Equation by the Integrating Factor:

Multiply every term of the original ODE by $\mu x \$ :

$\mu(x) \frac{dy}{dx} + \mu(x) P(x) y = \mu(x) Q(x) \$

3. Recognize the Left Side as a Derivative:

The left side of the equation can be expressed as the derivative of a product:

$\frac{d}{dx} \left( \mu(x) y \right) = \mu(x) Q(x)$

4. Integrate Both Sides:

Integrate both sides with respect to $x$ :

$\int \frac{d}{dx} \left( \mu(x) y \right) \, dx = \int \mu(x) Q(x) \, dx$

This simplifies to:

$\mu(x) y = \int \mu(x) Q(x) \, dx + C$

where $C$ is the constant of integration.

5. Solve for $C$ :

Finally, solve for $y$ by isolating it:

$\ y = \frac{1}{\mu(x)} (\int \mu(x) Q(x) \, dx + C )\$

This method provides a systematic approach to solving linear first-order ODEs by converting them into a more straightforward integral problem.

Let's look at examples to understand these steps.

2. Examples

Example 1:

Solve

\frac{dy}{dx} + y = 2

Solution:

1. Standard Form: $\frac{dy}{dx} + y = 2\$

2. Integrating Factor:

$\mu(x) = e^{\int 1 \, dx} = e^x \$

3. Multiply through by $\mu(x)\$ :

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

4. Recognize the left side as a derivative:

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

5. Integrate both sides:

$\ e^x y = \int 2e^x \, dx = 2e^x + C \$

6. Solve for $y$ :

$\ y = 2 + Ce^{-x} \$

Example 2:

Solve

\frac{dy}{dx} - 3y = 6

Solution:

1. Standard Form: $\frac{dy}{dx} - 3y = 6\$

2. Integrating Factor:

$\mu(x) = e^{\int -3 \, dx} = e^{-3x} \$

3. Multiply through by $\mu(x)\$ :

$\ e^{-3x} \frac{dy}{dx} - 3e^{-3x} y = 6e^{-3x} \$

4. Recognize the left side as a derivative:

$\frac{d}{dx}(e^{-3x} y) = 6e^{-3x} \$

5. Integrate both sides:

$\ e^{-3x} y = \int 6e^{-3x} \, dx = -2e^{-3x} + C \$

6. Solve for $y$ :

$\ y = -2 + Ce^{3x} \$

Example 3:

Solve

\frac{dy}{dx} + 4y = 8

Solution:

1. Standard Form: $\frac{dy}{dx} + 4y = 8\$

2. Integrating Factor:

$\mu(x) = e^{\int 4 \, dx} = e^{4x} \$

3. Multiply through by $\mu(x)\$ :

$\ e^{4x} \frac{dy}{dx} + 4e^{4x} y = 8e^{4x}\$

4. Recognize the left side as a derivative:

$\frac{d}{dx}(e^{4x} y) = 8e^{4x} \$

5. Integrate both sides:

$\ e^{4x} y = \int 8e^{4x} \, dx = 2e^{4x} + C \$

6. Solve for $y$ :

$\ y = 2 + Ce^{-4x} \$

Can you try on your own and check for the answer later?

Example 4:

Example 4

Solve

\frac{dy}{dx} + \frac{2}{x}y = x

Answer

Solution:

1. Standard Form: $\frac{dy}{dx} + \frac{2}{x}y = x\$

2. Integrating Factor:

$\mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln x} = x^2 \$

3. Multiply through by $\mu(x)\$ :

$\ x^2 \frac{dy}{dx} + 2xy = x^3\$

4. Recognize the left side as a derivative:

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

5. Integrate both sides:

$\ x^2 y = \int x^3 \, dx = \frac{x^4}{4} + C \$

6. Solve for $y$ :

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

Example 5:

Example 5

Solve

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

Answer

Solution:

1. Standard Form: $\frac{dy}{dx} - \frac{1}{x}y = x^2\$

2. Integrating Factor:

$\mu(x) = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} \$

3. Multiply through by $\mu(x)\$ :

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

4. Recognize the left side as a derivative:

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

5. Integrate both sides:

$\frac{y}{x} = \int x \, dx = \frac{x^2}{2} + C \$

6. Solve for $y$ :

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

Example 6:

Example 6

Solve

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

Answer

Solution:

1. Standard Form: $\frac{dy}{dx} + \frac{2x}{x^2 + 1} y = 1\$

2. Integrating Factor:

$\mu(x) = e^{\int \frac{2x}{x^2 + 1} \, dx} \$

Using substitution $u = x^2 + 1, du = 2x \, dx\$ :

$\int \frac{2x}{x^2 + 1} \, dx = \int \frac{du}{u} = \ln|u| = \ln(x^2 + 1) \$

Thus:

$\mu(x) = e^{\ln(x^2 + 1)} = x^2 + 1 \$

3. Multiply through by $\mu(x)\$ :

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

4. Recognize the left side as a derivative:

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

5. Integrate both sides:

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

6. Solve for $y$ :

$\ y = \frac{x^3}{3(x^2 + 1)} + \frac{x}{x^2 + 1} + \frac{C}{x^2 + 1} \$

Example 7:

Example 7

Solve

\frac{dy}{dx} + \frac{2y}{x} = \ln(x)

Answer

Solution:

1. Standard Form: $\frac{dy}{dx} + \frac{2y}{x} = \ln(x)\$

2. Integrating Factor:

$\mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln x} = x^2 \$

3. Multiply through by $\mu(x)\$ :

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

4. Recognize the left side as a derivative:

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

5. Integrate both sides:

$\ x^2 y = \int x^2 \ln(x) \, dx \$

Use integration by parts: Let $u = \ln(x)\$ and $dv = x^2 \, dx\$ , then $du = \frac{1}{x} dx\$ and $v = \frac{x^3}{3}\$ .

$\int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3 \ln(x)}{3} - \frac{1}{3} \int x^2 \, dx \$

$\ = \frac{x^3 \ln(x)}{3} - \frac{x^3}{9} + C \$

6. Solve for $y$ :

$y = \frac{\frac{x^3 \ln(x)}{3} - \frac{x^3}{9} + C}{x^2} \$

$\ y = \frac{x \ln(x)}{3} - \frac{x}{9} + \frac{C}{x^2} \$

Example 8:

Example 8

Solve

\frac{dy}{dx} + \frac{2}{x} y = \frac{3}{x^2}

Answer

Solution:

1. Standard Form: $\frac{dy}{dx} + \frac{2}{x} y = \frac{3}{x^2}\$

2. Integrating Factor:

$\mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln x} = x^2 \$

3. Multiply through by $\mu(x)\$ :

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

4. Recognize the left side as a derivative:

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

5. Integrate both sides:

$\ x^2 y = \int 3 \, dx = 3x + C \$

6. Solve for $y$ :

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

Example 9:

Example 9

Solve

\frac{dy}{dx} - \frac{3y}{x} = x^5 e^x

Answer

Solution:

1. Rewrite the Equation in Standard Form

$\frac{dy}{dx} - \frac{3y}{x} = x^5 e^x\$

2. Find the Integrating Factor

$\mu(x) = e^{\int -\frac{3}{x} \, dx} = e^{-3 \ln x} = x^{-3}\$

3. Multiply Through by the Integrating Factor

$x^{-3} \frac{dy}{dx} - \frac{3y}{x} \cdot x^{-3} = x^{-3} \cdot x^5 e^x\$

This simplifies to:

$x^{-3} \frac{dy}{dx} - 3x^{-4} y = x^2 e^x\$

4. Recognize the Left Side as a Derivative

The left side of the equation can be written as the derivative of a product:

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

5. Integrate Both Sides

Integrate both sides with respect to $x$ :

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

The left side simplifies to:

$x^{-3} y\$

To integrate the right side, use integration by parts:

Let $u = x^2\$ and $dv = e^x \, dx\$ . Then $du = 2x \, dx\$ and $v = e^x\$ . Applying integration by parts:

$\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx\$

For $\int 2x e^x \, dx\$ , apply integration by parts again:

Let $u = 2x\$ and $dv = e^x \, dx$ Then $du = 2 \, dx\$ and $v = e^x\:$

$\int 2x e^x \, dx = 2x e^x - \int 2 e^x \, dx = 2x e^x - 2 e^x\$

Thus:

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

6. Solve for $y$

Combine the results:

$x^{-3} y = (x^2 - 2x + 2) e^x + C\$

Solve for $y$ y by multiplying through by $x^3\$ :

$y = x^3 \ (x^2 - 2x + 2) e^x + C \$

$y = x^3 (x^2 - 2x + 2) e^x + C \$

Example 10:

Example 10

Solve

\frac{dy}{dx} + y \sin x = \cos x

Answer

Solution:

1. Identify $P(x)\$ and $Q(x)\:$

$\ P(x) = \sin x, \quad Q(x) = \cos x \$

2. Find the Integrating Factor:

$\mu(x) = e^{\int \sin x \, dx} = e^{-\cos x} \$

3. Multiply Through by the Integrating Factor:

$\ e^{-\cos x} \frac{dy}{dx} + y e^{-\cos x} \sin x = \cos x \cdot e^{-\cos x} \$

The left side is the derivative of $y e^{-\cos x} \$ :

$\frac{d}{dx} \left( y e^{-\cos x} \right) = \cos x \cdot e^{-\cos x} \$

4. Integrate Both Sides:

$\int \frac{d}{dx} ( y e^{-\cos x} )\, dx = \int \cos x \cdot e^{-\cos x} \, dx \$

Integrating the right side:

$\int \cos x \cdot e^{-\cos x} \, dx = -e^{-\cos x} + C \$

Thus:

$\ y e^{-\cos x} = -e^{-\cos x} + C \$

5. Solve for $y$ :

$\ y = -1 + Ce^{\cos x} \$

3. Practice Questions set 1

Solve the following linear ODEs:

$\frac{dy}{dx} + y e^x = e^x\$
$\frac{dy}{dx} + 2y = 3e^{-2x}\$
$\frac{dy}{dx} + 3y = 6e^{-2x}\ y(0) = 5\$
$\frac{dI}{dt} + \frac{R}{L}I = \frac{V_0}{L}\ I(0) = 0\$

Not sure about your answers? Don't worry, click the link below to check the video where the answers are explained.

Practice Questions set 1 Discussion

4. Practice Questions set 2

Solve the following linear ODEs:

$x \frac{dy}{dx} - 3y = x^6 e^x\$
$\frac{dy}{dx} = \frac{x + y \sin x}{1 + \cos x}\$

Not sure about your answers? Don't worry, click the link below to check the video where the answers are explained.

Practice Questions set 2 Discussion

This article introduced the technique of using an integrating factor to solve linear first-order ODEs. We explored how this method simplifies linear differential equations and allows us to find solutions systematically. To reinforce your understanding, practice the problems provided, as solving more examples will help solidify your grasp of the integrating factor method.

Mastering differential equations requires consistent practice, so continue working on problems to build your confidence and proficiency.

In the next article, we will delve into more advanced techniques for solving ODEs. Stay tuned to enhance your skills and tackle more complex ODEs with confidence!

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4. Linear Equations

1. Linear Equations

2. Examples

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

3. Practice Questions set 1

4. Practice Questions set 2