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Math Help - diff eq's

  1. #1
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    diff eq's

    The original problem: x(dy/dx)=y+xe^(5y/x)

    so far i got...
    (dy/dx)=y/x+e^(5y/x)

    u=y/x, xu=y therefore x(du/dx)=dy/dx

    x(du/dx)=u + e^(5u)

    I'm stuck here.

    By the way, I'm new and I'm assuming you guys don't just give out solutions so a push in the right direction will do. Thanks
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  2. #2
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    Quote Originally Posted by aaronb View Post
    The original problem: x(dy/dx)=y+xe^(5y/x)

    so far i got...
    (dy/dx)=y/x+e^(5y/x)

    u=y/x, xu=y therefore x(du/dx)=dy/dx

    x(du/dx)=u + e^(5u)

    I'm stuck here.

    By the way, I'm new and I'm assuming you guys don't just give out solutions so a push in the right direction will do. Thanks
    u = \frac{y}{x} \Rightarrow y = x u. To get \frac{dy}{dx} you have to use the product rule: \frac{dy}{dx} = u + x \frac{du}{dx}.

    So your DE becomes u + x \frac{du}{dx} = u + e^{5u} \, ....
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