Math Core

Slope fields show what solutions look like, but often you want an exact formula. For a large and important class of differential equations, you can get one with a single idea: move everything involving $y$ to one side, everything involving $x$ to the other, and integrate.

Separable equations

Definition

Separable differential equation

A first-order differential equation is separable if it can be written as

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

a function of $x$ alone times a function of $y$ alone.

For example, $\dfrac{dy}{dx} = 2xy$ , $\dfrac{dy}{dx} = \dfrac{x}{y}$ and $\dfrac{dy}{dx} = e^{x - y} = e^x e^{-y}$ are all separable. The equation $\dfrac{dy}{dx} = x + y$ is not, because a sum can't be split into a product of an $x$ part and a $y$ part.

The method

Separation of variables

To solve $\dfrac{dy}{dx} = g(x)\,h(y)$ :

Separate: $\dfrac{1}{h(y)}\,dy = g(x)\,dx$ .
Integrate both sides: $\displaystyle\int \frac{1}{h(y)}\,dy = \int g(x)\,dx$ , and add one constant $C$ .
Use the initial condition to find $C$ (do this as soon as you can).
Solve for $y$ , choosing signs and a domain consistent with the initial condition.

Why is this legal? If $y$ is a solution, then $\dfrac{1}{h(y)}\dfrac{dy}{dx} = g(x)$ . Integrating both sides with respect to $x$ , and using substitution on the left ( $dy = \dfrac{dy}{dx}\,dx$ ), gives exactly the equation in step 2. Writing $dy$ and $dx$ as if they were separate pieces is shorthand for that substitution.

You only need one constant. If you wrote $C_1$ on the left and $C_2$ on the right, you could subtract to combine them into a single constant anyway.

Worked example: A solution involving a square root

Find the particular solution of $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 3$ .

Solution. Separate and integrate:

$y\,dy = x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = \frac{x^2}{2} + C.$

Use $y(0) = 3$ : $\dfrac{9}{2} = 0 + C$ , so $C = \dfrac92$ . Multiply by 2: $y^2 = x^2 + 9$ .

Now $y = \pm\sqrt{x^2 + 9}$ , but the initial value $y(0) = 3$ is positive, so take the positive root:

$y = \sqrt{x^2 + 9}.$

Worked example: An exponential solution

Find the particular solution of $\dfrac{dy}{dx} = 2xy$ with $y(0) = 5$ .

Solution. Separate and integrate:

$\frac{1}{y}\,dy = 2x\,dx \quad\Longrightarrow\quad \ln|y| = x^2 + C.$

Use $y(0) = 5$ : $\ln 5 = 0 + C$ , so $C = \ln 5$ . Then

$\ln|y| = x^2 + \ln 5 \quad\Longrightarrow\quad |y| = e^{x^2 + \ln 5} = 5e^{x^2}.$

Since $y(0) = 5 > 0$ and the solution is never 0, $y$ stays positive, so $y = 5e^{x^2}$ .

Check: $\dfrac{dy}{dx} = 5e^{x^2}\cdot 2x = 2x\left(5e^{x^2}\right) = 2xy$ . ✓

Common mistake

Add the constant when you integrate, not after you solve for $y$ . In the example above, a common error is to write $\ln|y| = x^2$ , exponentiate to get $y = e^{x^2}$ , and then tack on " $+ C$ " to get $y = e^{x^2} + C$ . That function is not a solution of $\dfrac{dy}{dx} = 2xy$ unless $C = 0$ . On the AP exam, a missing or misplaced constant of integration costs most of the points for the problem.

The domain of a particular solution

A particular solution is only valid on an interval that contains the initial point and on which the formula (and the differential equation) make sense. Sometimes that interval is much smaller than you'd expect.

Worked example: A solution that blows up

Find the particular solution of $\dfrac{dy}{dx} = xy^2$ with $y(0) = 1$ , and state its domain.

Solution. Separate and integrate:

$y^{-2}\,dy = x\,dx \quad\Longrightarrow\quad -\frac{1}{y} = \frac{x^2}{2} + C.$

Use $y(0) = 1$ : $-1 = 0 + C$ , so $C = -1$ . Then

$-\frac{1}{y} = \frac{x^2}{2} - 1 = \frac{x^2 - 2}{2} \quad\Longrightarrow\quad y = \frac{2}{2 - x^2}.$

This formula is undefined at $x = \pm\sqrt2$ . The solution must be a differentiable function on an interval containing $x = 0$ , so its domain is $-\sqrt2 < x < \sqrt2$ . The pieces of the graph outside that interval belong to the formula, but not to this solution.

y = 2/(2 − x²). Only the middle branch, between the dashed asymptotes, is the solution with y(0) = 1.Open in grapher →

Worked example: Solving with exponentials on both sides

Find the particular solution of $\dfrac{dy}{dt} = e^{t - y}$ with $y(0) = \ln 3$ .

Solution. Rewrite $e^{t - y} = e^t e^{-y}$ , then separate by multiplying both sides by $e^y$ :

$e^y\,dy = e^t\,dt \quad\Longrightarrow\quad e^y = e^t + C.$

Use $y(0) = \ln 3$ : $e^{\ln 3} = e^0 + C$ , so $3 = 1 + C$ and $C = 2$ . Then $e^y = e^t + 2$ , and taking the natural log,

$y = \ln\left(e^t + 2\right).$

Since $e^t + 2 > 0$ for every $t$ , this solution is valid for all real $t$ .

Tip

Always check your final answer in two ways: it should satisfy the initial condition, and differentiating it should give back the original differential equation. Both checks take less than a minute.