Separation of Variables

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In differential equations, we are given an equation in terms of the derivative(s) of some function, and we need to solve for the function that makes the equation true.

For example, a simple differential equation is $y’=2x$, and its solution is just the antiderivative $y=x^2+C$.

$\begin{align*} y' &= 2x \\ (x^2+C)' &= 2x \\ 2x &= 2x \end{align*}$


The simplest differential equations can be solved by separation of variables, in which we move the derivative to one side of the equation and take the antiderivative.

$\begin{align*} 3y' + \cos x &= 6x^2 + y' \\ 2y' + \cos x &= 6x^2 \\ 2y' &= 6x^2 - \cos x \\ y' &= 3x^2 - \frac{1}{2} \cos x \\ y &= \int 3x^2 - \frac{1}{2}\cos x \, dx \\ y&= x^3 - \frac{1}{2} \sin x + C \end{align*}$


Equations with a Higher-Order Derivative

This method can be used to solve simple equations with higher-order derivatives, as well.

$\begin{align*} y''' + x &= 1 \\ y''' &= -x + 1 \\ y'' &= \int -x+1 \, dx \\ y'' &= -\frac{1}{2}x^2 + x + C_1 \\ y' &= \int -\frac{1}{2}x^2 + x + C_1 \, dx \\ y' &= -\frac{1}{6}x^3 + \frac{1}{2}x^2 + C_1 x + C_2 \\ y &= \int -\frac{1}{6}x^3 + \frac{1}{2}x^2 + C_1 x + C_2 \, dx \\ y &= -\frac{1}{24}x^4 + \frac{1}{6}x^3 + C_1 x^2 + C_2 x + C_3 \end{align*}$


Note that, although the antiderivative of $C_1$ is $\frac{C_1}{2}x^2$, the term $\frac{C_1}{2}$ is itself just a constant: $\frac{C_1}{2}x^2$ just means any constant multiplied by $x^2$. But $C_1 x^2$ also means any constant multiplied by $x^2$, so writing the fraction in $\frac{C_1}{2}x^2$ is redundant. To keep the notation simple and free of redundancy, we just write $C_1 x^2$.

Equations with Both Function and Derivative

When differential equations contain $y$ terms as well as $y’$ terms, we can still separate variables by using the differential notation for the derivative and treating it as a fraction.

$\begin{align*} y' y &= x \\ \frac{dy}{dx}y &= x \\ dy \, y &= x \, dx \\ y \, dy &= x \, dx \\ \int y \, dy &= \int x\, dx \\ \frac{1}{2}y^2 &= \frac{1}{2} x^2 + C \\ y^2 &= x^2 + C \\ y &= \pm \sqrt{x^2 + C} \end{align*}$


Even differential equations that contain two different variables multiplied together can sometimes be solved by separation of variables.

$\begin{align*} y' e^y \cos^2x &= 1 \\ y' e^y &= \sec^2 x \\ \frac{dy}{dx}e^y &= \sec^2 x \\ dy \, e^y &= \sec^2 x \, dx \\ e^y \, dy &= \sec^2 x \, dx \\ \int e^y \, dy &= \int \sec^2 x \, dx \\ e^y &= \tan x + C \\ y &= \ln ( \tan x + C) \end{align*}$


But other times, there is no way to separate the variables from each other completely. We will learn more advanced methods to solve such non-separable differential equations in the coming chapters.

$\begin{align*} y' + y &= x \\ \frac{dy}{dx} + y &= x \\ dy + y \, dx &= x \, dx \\ \mbox{(unable to} &\mbox{ separate)} \end{align*}$


Exercises

Solve the following differential equations using separation of variables. (You can view the solution by clicking on the problem.)

$\begin{align*} 1) \hspace{.5cm} y'=4 \end{align*}$
Solution:
$\begin{align*} y=4x+C \end{align*}$


$\begin{align*} 2) \hspace{.5cm} 3y'=x^2 \end{align*}$
Solution:
$\begin{align*}y=\frac{1}{9}x^3+C \end{align*}$


$\begin{align*} 3) \hspace{.5cm} y'=x(1+xy') \end{align*}$
Solution:
$\begin{align*} y=-\frac{1}{2} \ln (x^2-1) + C \end{align*}$


$\begin{align*} 4) \hspace{.5cm} \left( \frac{y'}{x} \right)^3 = \sin^3(x^2) \end{align*}$
Solution:
$\begin{align*} y= -\frac{1}{2} \cos x^2 + C \end{align*}$


$\begin{align*} 5) \hspace{.5cm} y'y = \sin x \end{align*}$
Solution:
$\begin{align*} y= \pm \sqrt{C - 2\cos x} \end{align*}$


$\begin{align*} 6) \hspace{.5cm} y'y = \frac{1}{x}-y' \end{align*}$
Solution:
$\begin{align*} y=-1 \pm \sqrt{C + 2 \ln x} \end{align*}$


$\begin{align*} 7) \hspace{.5cm} (y+1)(1-xe^x) = xy' \end{align*}$
Solution:
$\begin{align*} y=xe^{C-e^x}-1 \end{align*}$


$\begin{align*} 8) \hspace{.5cm} y' \cos y = x \sin y \end{align*}$
Solution:
$\begin{align*} y=\sin^{-1} \left( e^{\frac{1}{2}x^2+C} \right)\end{align*}$


$\begin{align*} 9) \hspace{.5cm} y'' - 4e^{2x} = e^x \end{align*}$
Solution:
$\begin{align*} y= e^x+e^{2x}+C_1x+C_2 \end{align*}$


$\begin{align*} 10) \hspace{.5cm} y''' + \cos x = x \end{align*}$
Solution:
$\begin{align*} y=\frac{1}{24}x^4+\sin x + C_1x^2 + C_2 x + C_3 \end{align*}$


This post is part of a series.