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Math Help - Help solving: y" = y' * e^y

  1. #1
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    Help solving: y" = y' * e^y

    Initial conditions: y(0) = 0, y'(0) = 1

    I asked a question this morning which was promptly answered by mr. fantastic (thanks again if he is reading this).

    I have come to the end of my problem set and the only problem remaining is the one in the thread title.

    I have done similar problems involving setting u = y' and applying the chain rule, and then performing integration by means of separable variables with the initial conditions. Basically, I think I could complete it mostly on my own, but I can't quite figure out what to do with the e^y term. If anyone could provide some insight it would be appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by LucasV View Post
    Initial conditions: y(0) = 0, y'(0) = 1

    I asked a question this morning which was promptly answered by mr. fantastic (thanks again if he is reading this).

    I have come to the end of my problem set and the only problem remaining is the one in the thread title.

    I have done similar problems involving setting u = y' and applying the chain rule, and then performing integration by means of separable variables with the initial conditions. Basically, I think I could complete it mostly on my own, but I can't quite figure out what to do with the e^y term. If anyone could provide some insight it would be appreciated. Thanks.
    Substitute v = dy/dx and note that \frac{dv}{dx} = \frac{dv}{dy} \cdot \frac{dy}{dx} = \frac{dv}{dy} \cdot v. Solve for v and use this solution to solve for y.
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    Alright, so I let u = y', which makes y" = u * du/dy....

    which changes the eq to:

    u * du/dy = u * e^y (u's cancel)

    I apply the method of separable variables and this gives:

    u = e^y + c

    when x=0, y(0)=0, y'(0)=1

    therefore 1 = e^0 + c which yields c = 0

    I then change u back to dy/dx and the eq becomes:

    dy/dx = e^y

    by performing separable variables again, this yields

    -1/(e^y) = x, solving for y gives -> y = ln(-1/x) + c

    this seems a bit off to me because solving for c requires taking the natural log of infinity which doesn't help.

    have I made a mistake somewhere?
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  4. #4
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    Quote Originally Posted by LucasV View Post
    Initial conditions: y(0) = 0, y'(0) = 1

    I asked a question this morning which was promptly answered by mr. fantastic (thanks again if he is reading this).

    I have come to the end of my problem set and the only problem remaining is the one in the thread title.

    I have done similar problems involving setting u = y' and applying the chain rule, and then performing integration by means of separable variables with the initial conditions. Basically, I think I could complete it mostly on my own, but I can't quite figure out what to do with the e^y term. If anyone could provide some insight it would be appreciated. Thanks.
    You could simply integrate both sides with respect to \displaystyle x...

    \displaystyle \frac{d^2y}{dx^2} = e^y\,\frac{dy}{dx}

    \displaystyle \int{\frac{d^2y}{dx^2}\,dx} = \int{e^y\,\frac{dy}{dx}\,dx}

    \displaystyle \int{\frac{d^2y}{dx^2}\,dx} = \int{e^y\,dy}

    \displaystyle \frac{dy}{dx} = e^y + C_1

    \displaystyle \frac{dx}{dy} = \frac{1}{e^y + C_1}

    \displaystyle \frac{dx}{dy} = \frac{e^y}{e^{2y} + C_1e^y}

    \displaystyle \int{\frac{dx}{dy}\,dy} = \int{\left(\frac{1}{e^{2y} + C_1e^y}\right)e^y\,dy}

    \displaystyle x = \int{\frac{1}{u^2 + C_1u}\,du} after making the substitution \displaystyle u = e^y. You can now solve using partial fractions.
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  5. #5
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    Thanks to both of you. For some reason I couldn't see that I should have found my c2 PRIOR to taking ln of both sides. If I take my solution now and stuff it into the ode it yields the desired result, with some very light algebraic manipulation. Solved.
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  6. #6
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    Quote Originally Posted by LucasV View Post
    Alright, so I let u = y', which makes y" = u * du/dy....

    which changes the eq to:

    u * du/dy = u * e^y (u's cancel)

    I apply the method of separable variables and this gives:

    u = e^y + c

    when x=0, y(0)=0, y'(0)=1

    therefore 1 = e^0 + c which yields c = 0

    I then change u back to dy/dx and the eq becomes:

    dy/dx = e^y

    by performing separable variables again, this yields

    -1/(e^y) = x, solving for y gives -> y = ln(-1/x) + c

    this seems a bit off to me because solving for c requires taking the natural log of infinity which doesn't help.

    have I made a mistake somewhere?
    Your mistake is here:

    u = e^y + c

    when x=0, y(0)=0, y'(0)=1

    therefore 1 = e^0 + c which yields c = 0
    dy/dx = e^y + C is correct. But there's no given boundary condition that relates dy/dx to y. y'(0) = 1 is a boundary condition between dy/dx and x. C = 0 is not correct.
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