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Math Help - [SOLVED] First order non linear DE

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] First order non linear DE

    I wonder what method to use to solve the following DE: y'=\frac{1}{e^y -x} with the initial condition y(1)=0. As a tip they suggest to think of x as a dependent variable of y.
    So I'd start with y'(e^y -x(y))=1. I don't think it's possible to reduce this DE into a linear one. I'm disoriented. Could you provide the name of a method to solve it? So that I check out my book/Internet and try to solve it alone.
    Thanks.
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    Quote Originally Posted by arbolis View Post
    I wonder what method to use to solve the following DE: y'=\frac{1}{e^y -x} with the initial condition y(1)=0. As a tip they suggest to think of x as a dependent variable of y.
    So I'd start with y'(e^y -x(y))=1. I don't think it's possible to reduce this DE into a linear one. I'm disoriented. Could you provide the name of a method to solve it? So that I check out my book/Internet and try to solve it alone.
    Thanks.
    They are correct try this

    \frac{dy}{dx}=\frac{1}{e^y-x}\iff \frac{dx}{dy}=e^y-x
    This gives

    x'+x=e^{y} this can be solved with an integrating factor e^{y}
    Where the derivative is now with respect to y
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  3. #3
    MHF Contributor arbolis's Avatar
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    I reached y=\ln |2x-1| which seems to be wrong if I plug it back into the original equation.
    My work: IF =e^y. So xe^y=\int e^{2y}dy=\frac{e^{2y}}{2}+C.
    Thus x=\frac{e^y}{2}+C \Rightarrow e^y=2x+C \Rightarrow y=\ln |2x+C| and I determined C with the initial condition y(1)=0.
    I wonder where I went wrong.
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    Quote Originally Posted by arbolis View Post
    I reached y=\ln |2x-1| which seems to be wrong if I plug it back into the original equation.
    My work: IF =e^y. So xe^y=\int e^{2y}dy=\frac{e^{2y}}{2}+C.
    Thus x=\frac{e^y}{2}+C \Rightarrow e^y=2x+C \Rightarrow y=\ln |2x+C| and I determined C with the initial condition y(1)=0.
    I wonder where I went wrong.
    This step is correct
    e^{y}x=\frac{1}{2}e^{2y}+C

    This is a quadratic in e^{y} so set z=e^{y}

    To solve for xz=\frac{1}{2}z^2+C multiplying by 2 and cleaning up gives the equation

    z^2-2xz=\hat{C} \iff (z-x)^2=\hat{C-x^2} =
     z=x \pm \sqrt{\hat{C}-x^2} \iff y=\ln\left(x\pm\sqrt{\hat{C}-x^2} \right)
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  5. #5
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    This step is correct
    e^{y}x=\frac{1}{2}e^{2y}+C

    This is a quadratic in e^{y} so set z=e^{y}

    To solve for xz=\frac{1}{2}z^2+C multiplying by 2 and cleaning up gives the equation

    z^2-2xz=\hat{C} \iff (z-x)^2=\hat{C-x^2} =
     z=x \pm \sqrt{\hat{C}-x^2} \iff y=\ln\left(x\pm\sqrt{\hat{C}-x^2} \right)
    I see. I couldn't simply divide C by e^y and consider the result as another constant, C. Thanks for the help.
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