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## AP®︎/College Calculus BC

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

Lesson 6: Finding general solutions using separation of variables

# Separable equations introduction

"Separation of variables" allows us to rewrite differential equations so we obtain an equality between two integrals we can evaluate. Separable equations are the class of differential equations that can be solved using this method.

## Want to join the conversation?

• Is a further explanation of + possible, please? (I can't quite understand what's happening there with the -2 before the x and the 1/2 etc ..and the subtitles ain't helping) Thanks in advance
• You are integrating:
``⌠     -x²⎮ -x e    dx⌡``

To integrate it you use the substitution `u = -x²`, and it's differential `du = -2x dx`, which reduces the integral to
`1/2∫e^u du = 1/2 e^u`
Replacing the u substitution you get
`1/2 e^(-x²)`
• dy/dx is just a notation not a fraction,so how does multiplying by dx and canceling it make sense?
• This is just a mnemonic device, although it does simply only a tiniest bit, here's a way without separating variables:

first you move all y's on one side, then all x's on the other side
then you move dy/dx to the side where you have all y terms:

`y_terms * dy/dx = x_terms`

now integrate both sides, with respect to x:
`∫ (y_terms * dy/dx) dx = ∫ x_terms dx`

`∫ dy/dx dx = y`; but `∫ 1 dy = y` as well, therefore: `∫ (dy/dx) dx = ∫ (1) dy = y`

So you end up with turning `∫ (y_terms dy/dx) dx = ∫ x_terms dx`
into `∫ y_terms dy = ∫ x_terms dx`

Also take a look at this: https://proofwiki.org/wiki/Separation_of_Variables

In general, you are always able to solve the same problem in calculus without separating dy's and dx's, that includes differential equations as well. Although that might require more time, thinking and space on your paper..

Hope that helps.
• At why does it become e^[(-x^2)/2]? Shouldn't it be e^[-x] because i cancel the 2? Properties of powers, isn't it?
• This is a common mistake that I have seen. The correct answer is indeed e^[-x^2/2].

sqrt (e^[-x^2] ) = (e^[-x^2] )^(1/2) <-- in this instance, you multiply the exponents

e^[-x^2 * 1/2] = e^[-x^2/2]

From EEweb.com

a^[n/m] = (a^[1/m] ) ^n = (a^n)^[1/m]

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• around , Sal sets y equal to the principle root of the left side, because the initial condition only gives a positive y value. But why is this necessary? Wouldn't it be just as correct to say plus or minus?
• I agree with Raviv; I'm fairly certain that the y^2=e^-x2 would also work. The question didn't specify that we had to find a function - since we can differentiate a conic section and end up with an expression for dy/dx, then if working backwards yields a curve that is not a function, that curve should still satisfy the conditions set forth in the problem.
• isn't -x2 always x2? So it could be simplified more.
(1 vote)
• Be careful with your order of operations. Note that -x^2 is equivalent to -1*x^2, but not equal to (-x)^2. If we take a number like 3 for x, we get -1*(3)^2=-1*9=-9, however, for (-x)^2, we get (-3)^2=+9.
• At @ how can he integrate both sides with different "d's "? (one is dx and other is Dy ) . u have to integrate both sides with same dx right ?
• you can do that, here's a good way to think of it:
y dy/dx= -x/(e^(x^2))
integrate both sides with respect to x :
∫ y (dy/dx) dx = ∫ -x/(e^(x^2)) dx
∫ y dy = ∫ -x/(e^(x^2)) dx
and that's the same in the video but he multiplied both sides by dx first then put the integral sign.
(1 vote)
• How can you tell if differential equations are separable?
• If we can write our DE in the form dy/dx = f(x) * g(y), it is called separable.
• Is there a reason that there is no video on solving a first order DE with the "Linear" method?

I'm referring particularly to this:
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Is it that this method isn't as useful or is it possible to do that type of problem with another method?

Or maybe it just hasn't been added yet.
I'm just trying to organize the various methods for solving a DE.
Thanks.
• That method is very useful. I guess it just hadn't been added yet.