Separation of Variables

Some differential equations can be solved by antidifferentiation alone, once the two variables are on opposite sides of the equation. These equations are called separable.

Separable equations

One constant is enough, since constants on both sides combine into one. On a free-response answer, separation is the step every later point depends on, and an answer missing the constant loses the points after the antiderivatives. Each value of

y

where

h(y) = 0

gives an equilibrium (constant) solution, which dividing by

h(y)

leaves out.

Example 1

Determine which of these equations are separable. Separate the variables in each one that is.

\begin{aligned} &\text{(a)}\quad \frac{dy}{dx} = \frac{y}{e^x} \\[6pt] &\text{(b)}\quad \frac{dy}{dx} = \cos(x + y) \\[6pt] &\text{(c)}\quad \frac{dy}{dx} = e^x\left(1 + y^2\right) \\[6pt] &\text{(d)}\quad \frac{dy}{dx} = \frac{2x}{3 - \cos y} \end{aligned}

(a) A quotient is a product:

\frac{dy}{dx} = y \cdot e^{-x}

. It's separable, and for

y \ne 0

\frac{1}{y}\,dy = e^{-x}\,dx

(b) Suppose

\cos(x + y) = g(x)\,h(y)

. Setting

y = 0

gives

g(x)h(0) = \cos x

, and setting

x = 0

gives

g(0)h(y) = \cos y

. Since

g(0)h(0) = \cos 0 = 1

, multiplying those two equations gives

\cos(x + y) = \cos x \cos y

. That fails at

x = y = \frac{\pi}{4}

, where the left side is

0

and the right side is

\frac{1}{2}

. So this equation isn't separable.

(c) It's already a product, so it's separable:

\frac{1}{1 + y^2}\,dy = e^x\,dx

(d) It's

2x \cdot \frac{1}{3 - \cos y}

, so it's separable:

(3 - \cos y)\,dy = 2x\,dx

Finding general solutions

Example 2

Find the general solution of

\frac{dy}{dx} = \frac{y}{e^x}

The right side is

0

when

y = 0

, so

y = 0

is an equilibrium solution. For

y \ne 0

, separate and antidifferentiate:

\begin{aligned} \int \frac{1}{y}\,dy &= \int e^{-x}\,dx \\[4pt] \ln|y| &= -e^{-x} + C. \end{aligned}

Exponentiate both sides:

|y| = e^{-e^{-x} + C} = e^C e^{-e^{-x}}.

y = Ae^{-e^{-x}}

, where

A = \pm e^C

is any nonzero constant. The value

A = 0

gives the equilibrium solution

y = 0

, so the general solution is

y = Ae^{-e^{-x}}, \quad A \text{ any constant}.

The slope field of dy/dx = y/eˣ with y = Ae^(−e^(−x)) for A = ±1 (dark) and A = ±2 (blue). The x-axis is the solution y = 0, and no other solution crosses it.

Example 3

Find the general solution of each equation.

(a)

\frac{dy}{dx} = e^x\left(1 + y^2\right)

(b)

\frac{dy}{dx} = \frac{2x}{3 - \cos y}

(a) Here

1 + y^2

is never

0

, so there's no constant solution to set aside. Separate and antidifferentiate:

\begin{aligned} \int \frac{1}{1 + y^2}\,dy &= \int e^x\,dx \\[4pt] \arctan y &= e^x + C. \end{aligned}

Take the tangent of both sides:

y = \tan\left(e^x + C\right)

. Since

\arctan y

lies between

-\frac{\pi}{2}

and

\frac{\pi}{2}

, each solution exists only where

e^x + C

stays between those values. For

C = 0

, that means

x < \ln\frac{\pi}{2}

The solutions y = tan(eˣ + C) for C = 0 and C = −1. Each rises without bound at its vertical asymptote (dashed), where eˣ + C = π/2.

(b) Separate and antidifferentiate:

\begin{aligned} \int (3 - \cos y)\,dy &= \int 2x\,dx \\[4pt] 3y - \sin y &= x^2 + C. \end{aligned}

No algebra isolates

y

3y - \sin y

, so the general solution stays in this implicit form. The equation still determines

y

as a function of

x

. The derivative of

3y - \sin y

3 - \cos y \ge 2

, so the left side increases and takes every real value. Each

x

then gives exactly one

y