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Thread: 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

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

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

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

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

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

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

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

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