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Math Help - Beginner Differential Equations

  1. #1
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    Beginner Differential Equations

    Hello,

    I've just started on separating differential equations and am just having a little trouble on how to proceed with this sort of question.

    \frac {dy}{dx} = y^2 \,e^(2x)

    I understand that the y needs to be bough to the left, and then the formal separation of \frac {dy}{dx} occurs - as my tutor puts it.

    =\frac{1}{y^2}\,dy = e^{2x} dx

    = \int \frac{1}{y^2}dy = \int e^{2x}

    = ln\,y^2 = \frac {e^{2x}}{2} + C

    This is where it breaks down a little bit for me...I'd assume that we need to get y to its simplest form, but I don't know how to do that...

    =y^2 = e^{\frac {e^2x}{2}} + C

    That isn't right (I'm assuming), so can someone point me in the direction of the correct step.

    Thanks
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  2. #2
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    Re: Beginner Differential Equations

    Your seperation is good but

    \int \frac{1}{y^2}dy = -\frac{1}{y}

    So you should get

    -\frac{1}{y}=\frac{1}{2}e^{2x}+C
    Thanks from astuart
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Beginner Differential Equations

    First, when you separate the variables to get:

    \frac{1}{y^2}\cdot\frac{dy}{dx}=e^{2x}

    you should note that you are losing the trivial solution y\equiv0.

    Then integrating with respect to x, we have:

    \int y^{-2}\,dy=\int e^{2x}\,dx

    As noted above, you have integrated incorrectly.
    Thanks from astuart
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  4. #4
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    Re: Beginner Differential Equations

    Quote Originally Posted by TheEmptySet View Post
    Your seperation is good but

    \int \frac{1}{y^2}dy = -\frac{1}{y}

    So you should get

    -\frac{1}{y}=\frac{1}{2}e^{2x}+C
    Ahh, of course...It's basically y^{-2}.

    I was treating it as y^{-1} or something like that.

    Thank you!!
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