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Math Help - Separating Variables? (Sorry everyone! Now I'm totally done :))

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

    Is my answer for this: Separating Variables? (Sorry everyone! Now I'm totally done :))-capture15.jpg 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!
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  2. #2
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    Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

    No, your answer is not correct. How did you attempt it?
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    Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

    Quote Originally Posted by HallsofIvy View Post
    No, your answer is not correct. How did you attempt it?
    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 :/
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    Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

    post some work
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    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
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    Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

    Quote Originally Posted by hollywood View Post
    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
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    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
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    Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

    Quote Originally Posted by hollywood View Post
    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?
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    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
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    Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

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

    - Hollywood
    Hm, weird. Thanks!
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    Re: Separating Variables? (Sorry everyone! Now I'm totally done :))

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    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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