Integrating Factor Method (First-Order Linear Differential Equations)

Step-by-Step Derivation, Formulas, Worked Examples, and Practice Exercises

This tutorial explains how to solve first-order linear differential equations using the integrating factor method. You will find a clear derivation of the formula, followed by fully worked examples and practice exercises with answers.

General Form

A first-order linear differential equation has the form:

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

where \( P(x) \) and \( Q(x) \) are known functions of \( x \).

Derivation of the Integrating Factor

Multiply both sides by an unknown function \( u(x) \):

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

We want the left side to become the derivative of a product. Using the product rule:

\[ \frac{d(uy)}{dx} = y\frac{du}{dx} + u\frac{dy}{dx} \]

For this to match the previous expression, we require:

\[ y\frac{du}{dx} = uP(x)y \]

Dividing by \( y \):

\[ \frac{du}{dx} = uP(x) \]

or

\[ \frac{1}{u}\frac{du}{dx} = P(x) \]

Integrating:

\[ \ln u = \int P(x)\,dx \]

Hence the integrating factor is:

Integrating Factor Formula: \[ u(x) = e^{\int P(x)\,dx} \]

Multiplying the original equation by this factor gives:

\[ \frac{d(uy)}{dx} = u(x)Q(x) \]

Integrating both sides:

\[ u(x)y = \int u(x)Q(x)\,dx \]

Finally:

General Solution Formula: \[ y = \frac{1}{u(x)}\int u(x)Q(x)\,dx \]

Worked Examples

Step-by-Step Solved Examples

Example 1

Solve:

\[ \frac{dy}{dx} - 2xy = x \]

Here \( P(x) = -2x \), \( Q(x) = x \).

\[ u(x) = e^{\int -2x\,dx} = e^{-x^2} \] \[ e^{-x^2}y = \int xe^{-x^2}\,dx \] \[ e^{-x^2}y = -\frac{1}{2} e^{-x^2} + C \] \[ y = Ce^{x^2} - \frac{1}{2} \]

Example 2

Solve for \( x > 0 \):

\[ \frac{dy}{dx} + \frac{y}{x} = -2 \] \[ u(x) = e^{\int \frac{1}{x} dx} = x \] \[ xy = \int -2x\,dx = -x^2 + C \] \[ y = \frac{C}{x} - x \]

Example 3

\[ x\frac{dy}{dx} + y = -x^3,\quad x > 0 \]

Divide by \( x \):

\[ \frac{dy}{dx} + \frac{y}{x} = -x^2 \] \[ u(x) = x \] \[ xy = \int -x^3\,dx = -\frac{x^4}{4} + C \] \[ y = \frac{C}{x} - \frac{x^3}{4} \]

Exercises and Answers

Practice Problems & Solutions

Solve:

  1. \(\dfrac{dy}{dx} + y = 2x + 5\)
  2. \(\dfrac{dy}{dx} + y = x^4\)

Answers:

  1. \( y = 2x + 3 + Ce^{-x} \)
  2. \( y = x^4 - 4x^3 + 12x^2 - 24x + 24 + Ce^{-x} \)

Further Reading

Tip: The integrating factor method is fundamental in calculus and appears frequently in physics, engineering, and applied mathematics.