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

• Mar 28th 2014, 01:22 PM
canyouhelp
Separating Variables? (Sorry everyone! Now I'm totally done :))
Is my answer for this: Attachment 30556 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!
• Mar 29th 2014, 08:12 PM
HallsofIvy
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
• Mar 30th 2014, 07:10 AM
canyouhelp
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
Quote:

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 :/
• Mar 30th 2014, 07:35 AM
romsek
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
post some work
• Mar 31st 2014, 06:37 AM
hollywood
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
• Mar 31st 2014, 08:56 AM
canyouhelp
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
Quote:

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
• Apr 1st 2014, 09:12 AM
hollywood
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
Which integrates to $-\frac{1}{3}e^{-3y}=\frac{1}{2}e^{2x}+C$. Now you use the initial condition y(0)=1 to find C:

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

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

- Hollywood
• Apr 1st 2014, 12:18 PM
canyouhelp
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
Quote:

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

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

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

- Hollywood

Did I separate them incorrectly or something?
• Apr 1st 2014, 06:21 PM
hollywood
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
• Apr 1st 2014, 06:33 PM
canyouhelp
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
Quote:

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!
• Apr 1st 2014, 10:02 PM
DanielPeyton
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
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• Apr 1st 2014, 10:03 PM
DanielPeyton
Re: Separating Variables? (Sorry everyone! Now I'm totally done :))
That does offer you quality of life I work full-time I manage our company and I what I really feel strongly now and feel like I can go back into uh... give at one hundred Raspberry Ultra Drops percent anchorage anyone that's whether they have gone through art the treatments offered in the states or whether they have not if they'd been diagnosed in with cancer I am courage everyone to at least look at their possibilities make that phone call and inquire about the hours that his treatment because it is an alternative that will provide.
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