Solving Separable Differential Equations

Separable Differential Equations

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

What Is a Separable DE?

A DE is separable if it can be written as:

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

i.e., the right-hand side factors into a product of a function of $x$ alone times a function of $y$ alone.

The Solution Method

1. Separate: put all $y$'s (and $dy$) on one side, all $x$'s (and $dx$) on the other: $$\frac{dy}{g(y)} = f(x), dx$$

2. Integrate both sides: $$\int \frac{dy}{g(y)} = \int f(x), dx$$

3. Solve for $y$ if possible (implicit solution otherwise).

4. Apply initial condition to find $C$ → particular solution.

Example 1: Simple Separation

Solve $\frac{dy}{dx} = xy$, $y(0) = 2$.

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

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

Exponentiate: $|y| = e^{x^2/2+C} = e^C e^{x^2/2}$

General solution: $y = Ae^{x^2/2}$ (where $A = \pm e^C$, any nonzero constant)

Initial condition: $y(0) = 2$ → $2 = Ae^0 = A$ → $A = 2$

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

Example 2: More Complex Separation

Solve $\frac{dy}{dx} = \frac{x}{y}$, $y(0) = 3$.

Separate: $y, dy = x, dx$

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

$y^2 = x^2 + 2C$; let $K = 2C$: $y^2 = x^2 + K$

Initial condition: $y(0) = 3$ → $9 = 0 + K$ → $K = 9$

Particular solution: $y^2 = x^2 + 9$, i.e., $y = \sqrt{x^2+9}$ (taking positive root since $y(0) = 3 > 0$)

Example 3: Exponential Growth

Solve $\frac{dP}{dt} = kP$.

Separate: $\frac{dP}{P} = k, dt$

Integrate: $\ln|P| = kt + C$

$P = Ae^{kt}$

Initial condition $P(0) = P_0$: $A = P_0$

Particular solution: $P(t) = P_0 e^{kt}$

This is the exponential growth/decay model.

Common Mistakes