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Math Help - Another general solution

  1. #1
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    Another general solution

    find the general solution of the first order diff. equation.

    e^y + 2xy + (xe^y + x^2)dy/dy = 0

    I divided accross by (xe^y + x^2) to get 2ln(y) - ln(x) + dy/dx = 0
    Not sure where to take it from here.Help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by matty888 View Post
    find the general solution of the first order diff. equation.

    e^y + 2xy + (xe^y + x^2)dy/dy = 0

    I divided accross by (xe^y + x^2) to get 2ln(y) - ln(x) + dy/dx = 0
    Not sure where to take it from here.Help would be greatly appreciated.
    I think what you meant is

     <br />
e^y + 2xy + (xe^y + x^2)\frac{dy}{dx} = 0<br />

    Really not sure what you did but multiplying by dx gives

     <br />
(e^y + 2xy)dx + (xe^y + x^2)dy = 0<br />

    and this ODE is exact meaning that there exist a \phi such that

     <br />
d \phi = \frac{\partial \phi}{\partial x}\; dx + \frac{\partial \phi}{\partial y}\; dy = (e^y + 2xy)dx + (xe^y + x^2)dy<br />
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