2. First-Order Linear

Video coming soon2 worked examples

Theory

A first-order linear equation has the standard form $\frac{dy}{dx} + P(x)\,y = Q(x)$ . Multiply through by the integrating factor:

I = e^{\int P\,dx}

The left side then becomes an exact derivative, $\frac{d}{dx}(Iy)$ , so integrating gives $Iy = \int IQ\,dx$ .

The Golden Rule: put the equation in standard form (coefficient of $\frac{dy}{dx}$ equal to 1) before reading off $P$ , then multiply by $e^{\int P\,dx}$ and recognise the left side as $\frac{d}{dx}(Iy)$ .

⚠️ Common Examiner Traps

Not in standard form: divide through so the $\frac{dy}{dx}$ coefficient is 1 before identifying $P$ .
Integrating factor: use $e^{\int P\,dx}$ with no separate constant.
The left side: after multiplying, it collapses to $\frac{d}{dx}(Iy)$ — use that directly.

Worked examples

Example 1

Solve $\dfrac{dy}{dx} + \dfrac{1}{x}y = 3x$ for $x > 0$ .

Step 1: Here $P = \frac{1}{x}$ , so the integrating factor is:

I = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x

Step 2: Multiply through; the left side becomes $\frac{d}{dx}(xy)$ :

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

Step 3: Integrate and solve for $y$ :

xy = x^3 + c \implies y = x^2 + \frac{c}{x}

Example 2

Solve $\dfrac{dy}{dx} - y = e^{2x}$ .

Step 1: Here $P = -1$ , so the integrating factor is $I = e^{\int -1\,dx} = e^{-x}$ .

Step 2: Multiply through; the left side becomes $\frac{d}{dx}(e^{-x}y)$ :

\frac{d}{dx}\bigl(e^{-x}y\bigr) = e^{-x}e^{2x} = e^{x}

Step 3: Integrate and solve:

e^{-x}y = e^{x} + c \implies y = e^{2x} + ce^{x}