General Solutions by Separation of… — AP® Calculus AB Notes

You can now write a differential equation, check a proposed solution and sketch solutions from a slope field. This page finds solutions by formula. The method, separation of variables, works whenever the variables can be pulled apart. On the AP® Calculus AB exam it is the only solving method you need.

When an equation is separable

A differential equation $\frac{dy}{dx} = F(x, y)$ is separable when the right side is a product of a function of $x$ alone and a function of $y$ alone:

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

Equation	Separable?	Why
$\frac{dy}{dx} = 2xy^2$	yes	$(2x)(y^2)$
$\frac{dy}{dx} = e^{x - y}$	yes	$e^{x}e^{-y}$ , by an exponent rule
$\frac{dy}{dx} = \frac{x^2}{y}$	yes	$x^2 \cdot \frac{1}{y}$
$\frac{dy}{dx} = x + y$	no	a sum cannot be written as $g(x)h(y)$

The method in four steps

Separate. Multiply and divide so that every $y$ is on the side with $dy$ and every $x$ is on the side with $dx$ : $\frac{1}{h(y)}\,dy = g(x)\,dx$ . Both $dy$ and $dx$ stay in numerators.
Integrate both sides. $\int \frac{1}{h(y)}\,dy$ $= \int g(x)\,dx$ .
Add the constant now. Each side has its own constant, but $C_2 - C_1$ is just one constant, so write a single $+ C$ , usually on the $x$ side. Do it in this step, not later.
Solve for $y$ if you can. The result, with $C$ still in it, is the general solution: a whole family of curves.

Worked example 1. Find the general solution of $\frac{dy}{dx} = 2xy^2$ .

Separate: divide by $y^2$ and multiply by $dx$ : $y^{-2}\,dy = 2x\,dx$ .

Integrate: $\int y^{-2}\,dy = \int 2x\,dx$ , so $-\frac{1}{y} = x^2 + C$ .

Solve: take reciprocals and change the sign: $y = -\frac{1}{x^2 + C}$ .

Check one case: with $C = 1$ , $y = -(x^2 + 1)^{-1}$ and $\frac{dy}{dx}$ $= 2x(x^2 + 1)^{-2}$ $= 2xy^2$ . It works.

Look at what would happen if you waited to add the constant: $-\frac{1}{y} = x^2$ gives $y = -\frac{1}{x^2}$ , and then " $+ C$ " gives $y = -\frac{1}{x^2} + C$ . That family is different, and for $C \ne 0$ it does not satisfy the equation. The constant must go in when you integrate.

Logarithms and the absolute value

Many separable equations put $\frac{1}{y}$ on the left. Its antiderivative is $\ln|y|$ , with the absolute value. Keep it; it decides which solutions you find.

Worked example 2. Find the general solution of $\frac{dy}{dx} = xy$ .

Separate: $\frac{dy}{y} = x\,dx$ (this assumes $y \ne 0$ ; see below).

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

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

Remove the absolute value: $y = \pm e^{C}e^{x^2/2}$ . The number $\pm e^{C}$ can be any nonzero constant; call it $A$ . So $y = Ae^{x^2/2}$ .

Finally, $y = 0$ also satisfies $\frac{dy}{dx} = xy$ (both sides are 0), but dividing by $y$ hid it. Allowing $A = 0$ brings it back. General solution: $y = Ae^{x^2/2}$ for any constant $A$ .

The slope field of dy/dx = xy on −2.5 ≤ x ≤ 2.5 and −4 ≤ y ≤ 4, with the curves y = Ce^(x²/2) for C = 0.25, 0.5, 1, 2 in crimson above the x-axis, for C = −0.25, −0.5, −1, −2 in blue below it, and C = 0, the x-axis itself, in green. Each curve is flattest at x = 0 and follows the segments. — **Figure 1.** The general solution $y = Ae^{x^2/2}$ of $\frac{dy}{dx} = xy$ drawn on its slope field: $A > 0$ above the axis, $A < 0$ below, $A = 0$ on it. LevlPrep original diagram.

Two algebra facts do most of the work here. First, $e^{a + b} = e^{a}e^{b}$ , which turns $e^{x^2/2 + C}$ into a constant times $e^{x^2/2}$ ; it is not $e^{x^2/2} + C$ . Second, $\ln|y| = u$ means $|y| = e^{u}$ ; you undo a logarithm by exponentiating the whole other side.

Shifted variables

The cooling equation $\frac{dT}{dt} = -0.05(T - 20)$ separates as $\frac{dT}{T - 20} = -0.05\,dt$ . The whole quantity $T - 20$ stays together in the denominator, and its antiderivative is $\ln|T - 20|$ :

$\ln|T - 20| = -0.05t + C, \qquad T = 20 + Ae^{-0.05t}.$

Common slips

Leaving an $x$ on the $dy$ side or a $y$ on the $dx$ side. If the variables will not separate, the method does not apply.
Missing $+ C$ , or adding it only after solving for $y$ .
Writing $\ln y$ instead of $\ln|y|$ , which loses the negative solutions.
Writing $e^{u + C} = e^{u} + C$ . It is $e^{C}e^{u}$ , a constant times $e^{u}$ .