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Math Help - differential equation

  1. #1
    Junior Member
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    differential equation

    solve dy/dx = (y^2)(e^2y), y(0) = 1.

    e^2y = x
    lnx = 2y.
    lnx/2 = y

    dy/y^2 = (0.5)(lnx)dx

    -y^-1 = (.5)(xlnx-x) + C

    C = -1

    y = -2/(xlnx -x -1)

    Is this correct?

    Im stuck here since lnx is not determinable.
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  2. #2
    Member
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    Quote Originally Posted by lord12 View Post
    solve dy/dx = (y^2)(e^2y), y(0) = 1.

    e^2y = x
    lnx = 2y.
    lnx/2 = y

    dy/y^2 = (0.5)(lnx)dx



    -y^-1 = (.5)(xlnx-x) + C

    C = -1

    y = -2/(xlnx -x -1)

    Is this correct?

    Im stuck here since lnx is not determinable.
     \frac{dy}{dx} = y^{2}e^{2y}, \ y(0) = 1

     \ln x = 2y \implies y = \frac{\ln x}{2} .

    How did you get  \frac{dy}{y^2} = \frac{1}{2} \ln x \ dx ?

    According to your substitution,  \frac{dy}{y^2} = x \ dx .
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  3. #3
    MHF Contributor kalagota's Avatar
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    first, where does this come from?

    x = e^{2y} or is it included in the given?
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