Separable Differential Equations

By the end of this lesson, you will be able to identify and solve separable differential equations, apply initial conditions to find particular solutions, and recognize implicit vs. explicit solutions.

Definition

A differential equation is separable if it can be written as:

$$\frac{dy}{dx} = g(x)h(y)$$

The variables can be "separated" — all $y$-terms on one side, all $x$-terms on the other.

Solving a Separable Equation

1. Write in the form $\dfrac{dy}{h(y)} = g(x),dx$
2. Integrate both sides
3. Solve for $y$ if possible (explicit solution)
4. Apply initial conditions to find $C$

Example 1 — Basic Separation

$\dfrac{dy}{dx} = \dfrac{x}{y}$

Separate: $y,dy = x,dx$

Integrate: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$

Simplify: $y^2 - x^2 = K$ (hyperbola family, $K = 2C$)

Explicit: $y = \pm\sqrt{x^2 + K}$

Example 2 — Exponential Growth

$\dfrac{dy}{dx} = ky$ ($k$ constant)

Separate: $\dfrac{dy}{y} = k,dx$

Integrate: $\ln|y| = kx + C$

Solve: $|y| = e^{kx+C} = e^C e^{kx}$

General solution: $y = Ae^{kx}$ where $A = \pm e^C$ (can be any nonzero constant; including $A=0$ as a trivial solution).

Example 3 — Initial Value Problem

$\dfrac{dy}{dx} = 2xy$, $y(0) = 3$

Separate: $\dfrac{dy}{y} = 2x,dx$

Integrate: $\ln|y| = x^2 + C$

$y = Ae^{x^2}$

Apply $y(0) = 3$: $A = 3$

Particular solution: $y = 3e^{x^2}$

Example 4 — Rational Separation

$\dfrac{dy}{dx} = \dfrac{x^2+1}{y^2}$, $y(0) = 2$

$y^2,dy = (x^2+1),dx$

$\dfrac{y^3}{3} = \dfrac{x^3}{3} + x + C$

Apply $y(0) = 2$: $8/3 = C$

$y^3 = x^3 + 3x + 8$

$y = \sqrt[3]{x^3 + 3x + 8}$ (one real cube root; explicitly solvable)

Example 5 — Logistic Equation

$\dfrac{dP}{dt} = rP\left(1-\dfrac{P}{K}\right)$ (logistic growth)

Separate: $\dfrac{dP}{P(1-P/K)} = r,dt$

Use partial fractions: $\dfrac{K}{P(K-P)} = \dfrac{1}{P} + \dfrac{1}{K-P}$