# Thread: Separating Variables? (Sorry everyone! Now I'm totally done :))

1. ## Separating Variables? (Sorry everyone! Now I'm totally done :))

Is my answer for this: right? The initial conditions thing is really messing with me because I'm not sure if I'm doing anything right now :/ If I'm wrong which option works? Why? Thank you all so much!

3. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

Originally Posted by HallsofIvy
Well I tried to seperate all the y's to one side and the x's to the other. And then find the antiderivatives.. But the initial conditions thing confuses me. I'm not sure what to do with that :/

4. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

post some work

5. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

You should be able to show us the equation with variables separated (functions of y and dy on one side, functions of x and dx on the other).

- Hollywood

6. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

Originally Posted by hollywood
You should be able to show us the equation with variables separated (functions of y and dy on one side, functions of x and dx on the other).

- Hollywood
Okay here's what I had when I tried to separate the variables: dy/e^(3y)=e^(2x)*dx

7. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

Which integrates to $\displaystyle -\frac{1}{3}e^{-3y}=\frac{1}{2}e^{2x}+C$. Now you use the initial condition y(0)=1 to find C:

$\displaystyle -\frac{1}{3}e^{-3}=\frac{1}{2}e^{0}+C$
$\displaystyle C=-\frac{1}{3}e^{-3}-\frac{1}{2}$

So your solution is $\displaystyle -\frac{1}{3}e^{-3y}=\frac{1}{2}e^{2x}-\frac{1}{3}e^{-3}-\frac{1}{2}$ which is not one of the options.

- Hollywood

8. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

Originally Posted by hollywood
Which integrates to $\displaystyle -\frac{1}{3}e^{-3y}=\frac{1}{2}e^{2x}+C$. Now you use the initial condition y(0)=1 to find C:

$\displaystyle -\frac{1}{3}e^{-3}=\frac{1}{2}e^{0}+C$
$\displaystyle C=-\frac{1}{3}e^{-3}-\frac{1}{2}$

So your solution is $\displaystyle -\frac{1}{3}e^{-3y}=\frac{1}{2}e^{2x}-\frac{1}{3}e^{-3}-\frac{1}{2}$ which is not one of the options.

- Hollywood
Did I separate them incorrectly or something?

9. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

No, I think that's the correct solution. It was probably supposed to be option #3 or #4.

- Hollywood

10. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

Originally Posted by hollywood
No, I think that's the correct solution. It was probably supposed to be option #3 or #4.

- Hollywood
Hm, weird. Thanks!

11. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

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12. ## Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

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