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Thread: double functioned differential

  1. #1
    MHF Contributor
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    double functioned differential

    g is differentiable .
    find the solution of
    $\displaystyle y' + g'(x)y=g(x)g'(x)e^{-g(x)}$

    g(x) could appear in the solution

    how to approach it?
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  2. #2
    Math Engineering Student
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    i pressume you want to solve for $\displaystyle y,$ so an integrating factor for the ODE is $\displaystyle g(x)$ multiply the equation and proceed, it's quite straightforward.
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  3. #3
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    i was taught that the integrating factor is k=e^{g(x)}
    $\displaystyle y'e^{g(x)} + g'(x)ye^{g(x)}=g(x)g'(x)$
    $\displaystyle (ye^{g(x)})'=g(x)g'(x)$
    $\displaystyle (ye^{g(x)})=\int g(x)g'(x) +c$

    how to proceed
    how i solve the right side
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  4. #4
    A Plied Mathematician
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    For the RHS, substitute u = g(x). Then du = g'(x) dx.
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