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Thread: 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.

    $\displaystyle

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

    $


    This can be written as:

    $\displaystyle

    \frac{dY}{dX}+Y=Xe^{-X}

    $

    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
    Behold, the power of SARDINES!
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    $\displaystyle e^{x}+(ye^{x}-x)\frac{dy}{dx}=0$

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

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

    Divide by $\displaystyle e^{x}$ to get

    $\displaystyle \frac{dx}{dy}+y-xe^{-x}=0$
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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.

    $\displaystyle

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

    $


    This can be written as:

    $\displaystyle

    \frac{dY}{dX}+Y=Xe^{-X}

    $

    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

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

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

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

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

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

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

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

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

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

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