Recipe

Solve a first-order linear equation with an integrating factor

Use when: the equation can be arranged as dy/dx + P(x)y = Q(x), with y and y' appearing only to the first power and never multiplied by each other

Goal: the general solution, found without ever needing the equation to separate

Taught in First-order linear equations and integrating factors

1. 1

   Rearrange so the coefficient of dy/dx is exactly 1, then read off P(x) and Q(x).

   Example

   \dfrac{dy}{dx}+\dfrac1xy=x^2 (for x>0) is already in standard form, with P(x)=\dfrac1x and Q(x)=x^2.

   Why: P(x) has to be identified from an equation already scaled so the y' coefficient is 1; reading it off any other form gives the wrong integrating factor.

2. 2

   Compute \mu(x)=e^{\int P(x)\,dx}, dropping the constant of integration inside the exponent.

   Example

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

   Why: any nonzero multiple of \mu(x) collapses the equation's left side the same way, so the simplest choice, no constant, is standard.

3. 3

   Multiply every term of the standard-form equation by \mu(x). By construction, the left side becomes the derivative of a single product, \dfrac{d}{dx}\big[\mu(x)y\big].

   Example

   x\frac{dy}{dx}+y = x^3 \quad\Longrightarrow\quad \frac{d}{dx}\big[xy\big] = x^3

   Why: this is a product rule run in reverse, arranged by the specific choice of \mu(x) in Step 2 so the left side always collapses this way.

4. 4

   Integrate the right side with respect to x; the left side integrates instantly, since it's already a single derivative.

   Example

   xy = \int x^3\,dx = \frac{x^4}{4}+C

   Why: once the left side is one clean derivative, this is the only calculus step left in the whole method.

5. 5

   Divide both sides by \mu(x) to solve for y explicitly.

   Example

   \boxed{y = \frac{x^3}{4}+\frac{C}{x}}

   Why: this is the general solution, one member for every value of C, obtained without the equation ever needing to separate.