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### Course: AP®︎/College Calculus BC>Unit 7

Lesson 6: Finding general solutions using separation of variables

# Identifying separable equations

To solve a differential equation using separation of variables, we must be able to bring it to the form $f\left(y\right)\phantom{\rule{0.167em}{0ex}}dy=g\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$ where $f\left(y\right)$ is an expression that doesn't contain $x$ and $g\left(x\right)$ is an expression that doesn't contain $y$.
Not all differential equations are like that. For example, $\frac{dy}{dx}=x+y$ cannot be brought to the form $f\left(y\right)\phantom{\rule{0.167em}{0ex}}dy=g\left(x\right)\phantom{\rule{0.167em}{0ex}}dx$ no matter how much we try.
In fact, a major challenge with using separation of variables is to identify where this method is applicable. Differential equations that can be solved using separation of variables are called separable equations.
So how can we tell whether an equation is separable? The most common type are equations where $\frac{dy}{dx}$ is equal to a product or a quotient of $f\left(y\right)$ and $g\left(x\right)$.
For example, $\frac{dy}{dx}=\frac{g\left(x\right)}{f\left(y\right)}$ can turn into $f\left(y\right)dy=g\left(x\right)dx$ when multiplied by $f\left(y\right)$ and $dx$.
Also, $\frac{dy}{dx}=f\left(y\right)g\left(x\right)$ can turn into $\frac{1}{f\left(y\right)}\phantom{\rule{0.167em}{0ex}}dy=g\left(x\right)dx$ when divided by $f\left(y\right)$ and multiplied by $dx$.
Here are a few concrete examples:
$\begin{array}{rl}\frac{dy}{dx}& =\stackrel{f\left(y\right)}{\stackrel{⏞}{\mathrm{sin}\left(y\right)}}\stackrel{g\left(x\right)}{\stackrel{⏞}{\mathrm{ln}\left(x\right)}}\\ \\ \frac{1}{\mathrm{sin}\left(y\right)}dy& =\mathrm{ln}\left(x\right)\phantom{\rule{0.167em}{0ex}}dx\\ \end{array}$
$\begin{array}{rl}\frac{dy}{dx}& =\frac{\stackrel{g\left(x\right)}{\stackrel{⏞}{{x}^{3}-5x}}}{\underset{f\left(y\right)}{\underset{⏟}{{e}^{y}}}}\\ \\ {e}^{y}\phantom{\rule{0.167em}{0ex}}dy& =\left({x}^{3}-5x\right)\phantom{\rule{0.167em}{0ex}}dx\\ \end{array}$
$\begin{array}{rl}\frac{dy}{dx}& =\frac{\stackrel{f\left(y\right)}{\stackrel{⏞}{\sqrt{y}}}}{\underset{g\left(x\right)}{\underset{⏟}{\mathrm{cos}\left(x\right)}}}\\ \\ \frac{1}{\sqrt{y}}dy& =\frac{1}{\mathrm{cos}\left(x\right)}dx\end{array}$
Other equations must be slightly manipulated before they are in the form $\frac{dy}{dx}=f\left(y\right)g\left(x\right)$. For example, we need to factor the right-hand side of $\frac{dy}{dx}=xy-7x$ to bring it to the desired form:
$\frac{dy}{dx}=xy-7x=\stackrel{g\left(x\right)}{\stackrel{⏞}{x}}\stackrel{f\left(y\right)}{\stackrel{⏞}{\left(y-7\right)}}$
Problem 1
Can this differential equation be solved using separation of variables?
$\frac{dy}{dx}=3y-{x}^{2}y$

Problem 2
Can this differential equation be solved using separation of variables?
$\frac{dy}{dx}=4x+5y+4$

Problem 3
Can this differential equation be solved using separation of variables?
$\frac{dy}{dx}={2}^{y-x}$

Want more practice? Try this exercise.

## Want to join the conversation?

• In the last exercise, It was not immediately clear to me wether the expression on the right side of the equality was either "2^(y minus x)" or "2^(y times -x)". How can you tell them apart? Are both of them separable?
(1 vote)
• I believe it is a convention that a negative sign before a number not enclosed by anything such as brackets, are meant to be interpreted as a minus sign. If they had wanted you to think that the -x was multiplied, then they would explicitly have written it in the form (y · -x) or (y * (-x)).

TLDR; because there is no multiplication sign present, we may confidently assume that the (y-x) is a subtraction and not multiplication.