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Math Help - Having Trouble With ODE Step

  1. #1
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    Having Trouble With ODE Step

    Hey. Looking through my notes here and I'm having trouble with a jump they've made.

    <br /> <br />
e^X+(Ye^X-X)\frac{dX}{dY}=0<br /> <br />


    This can be written as:

    <br /> <br />
\frac{dY}{dX}+Y=Xe^{-X}<br /> <br />

    I've been sitting trying to see how the top line can be written as the bottom line but I seem to be stumped.

    Any help would be greatly appreciated .
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  2. #2
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    e^{x}+(ye^{x}-x)\frac{dy}{dx}=0

    Mult by \frac{dx}{dy} to get

    e^{x}\frac{dx}{dy}+(ye^{x}-x)=0

    Divide by e^{x} to get

    \frac{dx}{dy}+y-xe^{-x}=0
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  3. #3
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    Ahh, It's seems very simple now. Thanks man .
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  4. #4
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    Quote Originally Posted by Macca567 View Post
    Hey. Looking through my notes here and I'm having trouble with a jump they've made.

    <br /> <br />
e^X+(Ye^X-X)\frac{dX}{dY}=0<br /> <br />


    This can be written as:

    <br /> <br />
\frac{dY}{dX}+Y=Xe^{-X}<br /> <br />

    I've been sitting trying to see how the top line can be written as the bottom line but I seem to be stumped.

    Any help would be greatly appreciated .
    It is easy

    e^x + \left(ye^x - x\right)\frac{dx}{dy} = 0

    e^x\frac{dy}{dx} + e^xy = x

    \left(e^xy\right)^{'} = x

    e^xy = \frac{x^2}{2} + C

    y = \left(\frac{x^2}{2} + C\right)e^{-x}
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  5. #5
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    Quote Originally Posted by Ment View Post
    It is easy

    e^x + \left(ye^x - x\right)\frac{dx}{dy} = 0

    e^x\frac{dy}{dx} + e^xy = x
    The question was how to get to this line! You can't say "It is easy" and them just write it down!

    \left(e^xy\right)^{'} = x

    e^xy = \frac{x^2}{2} + C

    y = \left(\frac{x^2}{2} + C\right)e^{-x}
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