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Math Help - Differential Equation Help

  1. #1
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    Differential Equation Help

    i need help on how to solve this differential equation

    y'' = y' * e^y
    initial conditions
    y(0) = 0
    y'(0) = 2

    first i use u = y' and y'' = u*du/dy substitution to get
    u*du/dy = u * e^y ---> du/dy = e^y

    separating variables and integrating both sides give
    u = e^y + c1

    substituting y' back into u and separating variables give
    dy / (e^y + c1) = dx

    integrating gives
    [ln(e^y) - ln(e^y + c1)] / c1 = x + c2
    y - ln(e^y + c1) = c1*(x + c2)

    from here i am stuck. I can't get it to be in the form of y =
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  2. #2
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    Quote Originally Posted by TshingY View Post
    i need help on how to solve this differential equation

    y'' = y' * e^y
    initial conditions
    y(0) = 0
    y'(0) = 2

    first i use u = y' and y'' = u*du/dy substitution to get
    u*du/dy = u * e^y ---> du/dy = e^y

    separating variables and integrating both sides give
    u = e^y + c1 (*)

    substituting y' back into u and separating variables give
    dy / (e^y + c1) = dx

    integrating gives
    [ln(e^y) - ln(e^y + c1)] / c1 = x + c2
    y - ln(e^y + c1) = c1*(x + c2) (**)

    from here i am stuck. I can't get it to be in the form of y =
    Why not use your IC here (*) to find the constant c1 (=1). Second, leave the ln alone

    \ln e^y - \ln (e^y + 1) = \ln \frac{e^y}{e^y+1} = \cdots so you can solve for e^y then take a ln to get y.
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  3. #3
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    If y''=\mathrm e^yy' then integration gives y'=\mathrm e^y+c.

    Since y(0)=0 and y'(0)=2 we see that c=1.

    Thus \frac{\mathrm d y}{\mathrm e^y+1}=\mathrm d x.

    Write this as \frac{\mathrm e^{-y}\mathrm d y}{1+\mathrm e^{-y}}=\mathrm d x.

    Now you may integrate to get -\ln(1+\mathrm e^{-y})=x+k.

    Since y=0 when x=0 this means k=-\ln 2.

    Hence x=\ln\left(\frac 2{1+\mathrm e^{-y}}\right) and so \mathrm e^x=\frac 2{1+\mathrm e^{-y}}.

    Rearrange to get y=-\ln(2\mathrm e^{-x}-1).
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