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

Lesson 6: Finding general solutions using separation of variables- Separable equations introduction
- Addressing treating differentials algebraically
- Separable differential equations
- Separable differential equations: find the error
- Worked example: separable differential equations
- Separable differential equations
- Worked example: identifying separable equations
- Identifying separable equations
- Identify separable equations

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# Identifying separable equations

To solve a differential equation using separation of variables, we must be able to bring it to the form $f(y){\textstyle \phantom{\rule{0.167em}{0ex}}}dy=g(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx$ where $f(y)$ is an expression that doesn't contain $x$ and $g(x)$ 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 $f(y){\textstyle \phantom{\rule{0.167em}{0ex}}}dy=g(x){\textstyle \phantom{\rule{0.167em}{0ex}}}dx$ 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 $f(y)$ and $g(x)$ .

For example, $\frac{dy}{dx}}={\displaystyle \frac{{g(x)}}{{f(y)}}$ can turn into ${f(y)}dy={g(x)}dx$ when multiplied by ${f(y)}$ and $dx$ .

Also, $\frac{dy}{dx}}={f(y)}{g(x)$ can turn into $\frac{1}{{f(y)}}}{\textstyle \phantom{\rule{0.167em}{0ex}}}dy={g(x)}dx$ when divided by ${f(y)}$ and multiplied by $dx$ .

Here are a few concrete examples:

Other equations must be slightly manipulated before they are in the form $\frac{dy}{dx}}=f(y)g(x)$ . For example, we need to factor the right-hand side of $\frac{dy}{dx}}=xy-7x$ to bring it to the desired form:

*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.(17 votes)

- what about a partial differential equation where can I find it?(3 votes)