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Lesson 2: First-Order Linear Equations and Integrating Factors

Estimated time: 35-40 minutes

Learning Objectives

Identify a first-order linear ODE in standard form
Compute the integrating factor μ(x) = e^∫P(x)dx
Multiply through by μ and recognize the left side as a product-rule derivative
Solve and apply initial conditions

Standard Form

First-Order Linear ODE (Standard Form):

dy/dx + P(x) y = Q(x)

The key: y and y' appear to the first power; P and Q depend only on x.

If the equation is not in standard form (e.g., has a coefficient on y'), divide through first.

The Integrating Factor Method

Integrating Factor:

μ(x) = e^{∫ P(x) dx}

Multiply both sides by μ(x). The left side becomes d/dx[μ(x) y], so:

μ(x) y = ∫ μ(x) Q(x) dx + C

Why Does This Work?

After multiplying by μ, the left side μy' + μPy equals d/dx[μy] by the product rule (since μ' = Pμ). This collapses the equation into a single derivative, which we can integrate directly.

Worked Examples

Example 1: Constant Coefficient

Solve y' + 2y = 6.

Step 1: P(x) = 2, Q(x) = 6. Already in standard form.

Step 2: μ = e^{∫2 dx} = e^2x.

Step 3: Multiply: e^2x y' + 2e^2x y = 6e^2x.

d/dx[e^2x y] = 6e^2x

Step 4: Integrate: e^2x y = 3e^2x + C.

Step 5: Solve: y = 3 + Ce^-2x.

Example 2: Variable Coefficient

Solve y' + (1/x)y = x, x > 0.

Step 1: P(x) = 1/x, Q(x) = x.

Step 2: μ = e^∫(1/x)dx = e^{ln x} = x.

Step 3: Multiply: xy' + y = x², i.e., d/dx[xy] = x².

Step 4: Integrate: xy = x³/3 + C.

Step 5: y = x²/3 + C/x.

Example 3: With IVP

Solve y' - 3y = e^2x, y(0) = 1.

Step 1: P(x) = -3, Q(x) = e^2x.

Step 2: μ = e^{∫-3 dx} = e^-3x.

Step 3: d/dx[e^-3x y] = e^-3x · e^2x = e^-x.

Step 4: Integrate: e^-3x y = -e^-x + C.

Step 5: y = -e^2x + Ce^3x.

Step 6: y(0) = 1: 1 = -1 + C, so C = 2.

Particular solution: y = -e^2x + 2e^3x.

Check Your Understanding

1. Solve y' + 5y = 10.

μ = e^5x. d/dx[e^5xy] = 10e^5x. Integrate: e^5xy = 2e^5x + C. y = 2 + Ce^-5x.

2. Solve y' + (2/x)y = x³, x > 0.

μ = e^∫(2/x)dx = x². d/dx[x²y] = x⁵. Integrate: x²y = x⁶/6 + C. y = x⁴/6 + C/x².

3. Solve y' - y = e^x, y(0) = 0.

μ = e^-x. d/dx[e^-xy] = 1. Integrate: e^-xy = x + C. y = (x+C)e^x. y(0)=0: C=0. y = xe^x.

4. What is the integrating factor for y' + (tan x)y = cos x?

μ = e^{∫tan x dx} = e^{-ln|cos x|} = 1/cos x = sec x.

5. Is the equation xy' + 2y = x³ linear? What is P(x) in standard form?

Yes, it is linear. Divide by x: y' + (2/x)y = x². So P(x) = 2/x.

Key Takeaways

Standard form: y' + P(x)y = Q(x). Always put the equation in this form first.
Integrating factor: μ = e^∫P(x)dx. No constant of integration needed here.
Multiply through by μ; the left side becomes d/dx[μy].
Integrate both sides, solve for y, then apply initial conditions.
This method works for every first-order linear equation -- it always produces the general solution.

Next Lesson

Learn to solve exact equations using the exactness condition.

Lesson 3

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