Mathematics.

first order odes

Integrating Factors

Differential Equations30 minDifficulty6 out of 10

You should know: first order differential equation

Overview

An integrating factor is a function μ(x,y) that, when multiplied through a non-exact first-order equation M dx + N dy = 0, makes it exact — so it can then be solved by the exact-equation method. The most common use is on linear first-order equations y′ + P(x)y = Q(x), where μ(x) = e^{∫P dx} converts the left side into the derivative of a single product, (μy)′. Finding a general integrating factor for an arbitrary non-exact equation can be hard, but for the linear case the formula is explicit and always works. This technique is one of the two standard workhorses (alongside separation of variables) for solving first-order linear ODEs in closed form.

Intuition

The product rule says (μy)′ = μy′ + μ′y. If we choose μ so that μ′ = μP, then μy′ + μPy is exactly μy′ + μ′y = (μy)′ — so the messy left side y′ + Py collapses into the derivative of a single product once multiplied by the right μ. It's like finding the right substitution that turns a sum of two changing quantities into the change of one combined quantity, after which a single integration finishes the job.

Formal Definition

Definition

For the linear first-order equation, multiplying by μ(x) = e^{∫P(x)dx} makes the left side an exact derivative:

y′+P(x)y=Q(x)y' + P(x)y = Q(x)
Linear first-order ODE
μ(x)=e∫P(x) dx\mu(x) = e^{\int P(x)\,dx}
Integrating factor
ddx[μ(x)y]=μ(x)Q(x)\frac{d}{dx}\big[\mu(x) y\big] = \mu(x) Q(x)
Reduces to an exact derivative
y=1μ(x)(∫μ(x)Q(x) dx+C)y = \frac{1}{\mu(x)}\left(\int \mu(x) Q(x)\,dx + C\right)
General solution

Worked Examples

  1. Identify P(x) = 2 and compute the integrating factor.

    μ(x)=e∫2 dx=e2x\mu(x) = e^{\int 2\,dx} = e^{2x}

  2. Multiply through by μ; the left side becomes an exact derivative.

    ddx[e2xy]=e2x⋅e−x=ex\frac{d}{dx}[e^{2x}y] = e^{2x}\cdot e^{-x} = e^{x}

  3. Integrate both sides.

    e2xy=ex+Ce^{2x}y = e^{x} + C

  4. Divide by e^{2x} to isolate y.

    y=e−x+Ce−2xy = e^{-x} + Ce^{-2x}

Answer: y = e^(-x) + Ce^(-2x) (verify: y′ = -e^{-x} - 2Ce^{-2x}; y′+2y = -e^{-x}-2Ce^{-2x}+2e^{-x}+2Ce^{-2x} = e^{-x}, matching the right side).

Practice Problems

Difficulty 6/10

Solve y′ + 3y = 6 using an integrating factor.

Difficulty 7/10

Solve xy′ + y = x², for x>0 (first rewrite in standard linear form).

Difficulty 6/10

An RL circuit obeys L di/dt + Ri = V (constant V, L=R=1, V=10), with i(0)=0. Use an integrating factor to find i(t).

Quiz

For y′ + P(x)y = Q(x), the integrating factor is:
After multiplying by the correct integrating factor, the left side y′+Py becomes:
For y′ - (1/x)y = x (x>0), the integrating factor is:

Summary

  • An integrating factor μ multiplies a non-exact equation to make it exact, most simply for linear ODEs y′+Py=Q.
  • For the linear case, μ(x) = e^{∫P dx} turns the left side into the exact derivative (μy)′.
  • Integrating both sides and dividing by μ gives the closed-form general solution.

References

  1. WebsiteWikipedia — Integrating factor