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Math Help - [SOLVED] integral of perfect derivative

  1. #1
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    [SOLVED] integral of perfect derivative

    The Differential Equation: 3y^2 y''(x) + 2e^y y'(x)+2y y'(x)+6y(y'(x))^2=0 is a perfect derivative. Find the first integral.
    3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0

    not sure which notation is best so i put 2 ways of the same problem.

    This is the product rule split up but i just cannot see how to do this backwards. Any help would be great! Thanks.
    Last edited by snaes; February 15th 2010 at 08:52 PM. Reason: put second notation on
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  2. #2
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    Quote Originally Posted by snaes View Post
    The Differential Equation: 3y^2 y''(x) + 2e^y y'(x)+2y y'(x)+6y(y'(x))^2=0 is a perfect derivative. Find the first integral.
    3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0

    not sure which notation is best so i put 2 ways of the same problem.

    This is the product rule split up but i just cannot see how to do this backwards. Any help would be great! Thanks.
    I suspect there's a typo in the question. I think it should be
    3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y {\color{red} e^y} \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0.
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    That would definately help. I think its a typo as well. Thanks.
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  4. #4
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    There might be a typo but for your original problem, you could have

     <br />
\frac{d}{dx}\left( 3 y^2 y' + 2 e^y + y^2\right) = 0.<br />
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  5. #5
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    Yeah, the problem had a typo, but its fixed now thanks.
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