# 4.3 Separable equations

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• Use separation of variables to solve a differential equation.
• Solve applications using separation of variables.

We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.

## Definition

A separable differential equation    is any equation that can be written in the form

$y\prime =f\left(x\right)g\left(y\right).$

The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of $x$ times a function of $y.$ Examples of separable differential equations include

$\begin{array}{}\\ \\ y\prime =\left({x}^{2}-4\right)\left(3y+2\right)\hfill \\ y\prime =6{x}^{2}+4x\hfill \\ y\prime =\text{sec}\phantom{\rule{0.1em}{0ex}}y+\text{tan}\phantom{\rule{0.1em}{0ex}}y\hfill \\ y\prime =xy+3x-2y-6.\hfill \end{array}$

The second equation is separable with $f\left(x\right)=6{x}^{2}+4x$ and $g\left(y\right)=1,$ the third equation is separable with $f\left(x\right)=1$ and $g\left(y\right)=\text{sec}\phantom{\rule{0.1em}{0ex}}y+\text{tan}\phantom{\rule{0.1em}{0ex}}y,$ and the right-hand side of the fourth equation can be factored as $\left(x+3\right)\left(y-2\right),$ so it is separable as well. The third equation is also called an autonomous differential equation    because the right-hand side of the equation is a function of $y$ alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables    .

## Problem-solving strategy: separation of variables

1. Check for any values of $y$ that make $g\left(y\right)=0.$ These correspond to constant solutions.
2. Rewrite the differential equation in the form $\frac{dy}{g\left(y\right)}=f\left(x\right)dx.$
3. Integrate both sides of the equation.
4. Solve the resulting equation for $y$ if possible.
5. If an initial condition exists, substitute the appropriate values for $x$ and $y$ into the equation and solve for the constant.

Note that Step 4. states “Solve the resulting equation for $y$ if possible.” It is not always possible to obtain $y$ as an explicit function of $x.$ Quite often we have to be satisfied with finding $y$ as an implicit function of $x.$

## Using separation of variables

Find a general solution to the differential equation $y\prime =\left({x}^{2}-4\right)\left(3y+2\right)$ using the method of separation of variables.

Follow the five-step method of separation of variables.

1. In this example, $f\left(x\right)={x}^{2}-4$ and $g\left(y\right)=3y+2.$ Setting $g\left(y\right)=0$ gives $y=-\frac{2}{3}$ as a constant solution.
2. Rewrite the differential equation in the form
$\frac{dy}{3y+2}=\left({x}^{2}-4\right)dx.$
3. Integrate both sides of the equation:
$\int \frac{dy}{3y+2}=\int \left({x}^{2}-4\right)\phantom{\rule{0.1em}{0ex}}dx.$

Let $u=3y+2.$ Then $du=3\frac{dy}{dx}dx,$ so the equation becomes
$\begin{array}{ccc}\hfill \frac{1}{3}\int \frac{1}{u}du& =\hfill & \frac{1}{3}{x}^{3}-4x+C\hfill \\ \hfill \frac{1}{3}\phantom{\rule{0.1em}{0ex}}\text{ln}|u|& =\hfill & \frac{1}{3}{x}^{3}-4x+C\hfill \\ \hfill \frac{1}{3}\phantom{\rule{0.1em}{0ex}}\text{ln}|3y+2|& =\hfill & \frac{1}{3}{x}^{3}-4x+C.\hfill \end{array}$
4. To solve this equation for $y,$ first multiply both sides of the equation by $3.$
$\text{ln}|3y+2|={x}^{3}-12x+3C$

Now we use some logic in dealing with the constant $C.$ Since $C$ represents an arbitrary constant, $3C$ also represents an arbitrary constant. If we call the second arbitrary constant ${C}_{1},$ the equation becomes
$\text{ln}|3y+2|={x}^{3}-12x+{C}_{1}.$

Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base $e\right).$
$\begin{array}{ccc}\hfill {e}^{\text{ln}|3y+2|}& =\hfill & {e}^{{x}^{3}-12x+{C}_{1}}\hfill \\ \hfill |3y+2|& =\hfill & {e}^{{C}_{1}}{e}^{{x}^{3}-12x}\hfill \end{array}$

Again define a new constant ${C}_{2}={e}^{{c}_{1}}$ (note that ${C}_{2}>0\right)\text{:}$
$|3y+2|={C}_{2}{e}^{{x}^{3}-12x}.$

This corresponds to two separate equations: $3y+2={C}_{2}{e}^{{x}^{3}-12x}$ and $3y+2=\text{−}{C}_{2}{e}^{{x}^{3}-12x}.$
The solution to either equation can be written in the form $y=\frac{-2±{C}_{2}{e}^{{x}^{3}-12x}}{3}.$
Since ${C}_{2}>0,$ it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant $C$ is entirely arbitrary, and can be dropped. Therefore the solution can be written as
$y=\frac{-2+C{e}^{{x}^{3}-12x}}{3}.$
5. No initial condition is imposed, so we are finished.

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