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Is it possible to solve?

  1. Sep 28, 2012 #1
    I've tried almost everything but i still get stuck when finding the integral of :

    (2x^2y-e^(-x^2))dx+(x+1)dy =0

    To attempt to solve it i did the following:----> since the DE is not exact :

    1) [M(x,y)/dy - N(x,y)/dx]/N(x,y) = [2x^2-1/x+1]
    2) ∫[2x^2-1/x+1] =
    3) Integrating Factor: e^(x^2 - 2x +ln|x|) to make it smaller I'm gonna call

    {e^(x^2 - 2x +ln|x|)} = P

    4) P*[(2x^2-e^-(x^2))dx + (x+1)dy] =0

    After simplifying I get: [(2x^2)ye^(x^2)-1]dx = [e^(x^2)](x+1)dy

    After that I don't know what to do.
     
  2. jcsd
  3. Sep 28, 2012 #2
    I get the integrating factor:

    [tex]\mu=e^{x^2-2x+\ln(1+x)}[/tex]

    Once you get the integrating factor, multiply both sides of equation by it, then use the techniques of solving an exact equation (taking those partial integrals and all).
     
  4. Sep 29, 2012 #3
    Hi !
    (2x²y-e^(-x²))dx+(x+1)dy = 0
    is a linear ODE :
    (x+1)y' +(2x²)y = e^(-x²)
    You may solve it in using the classical method to solve linear ODEs.
     
  5. Sep 29, 2012 #4

    micromass

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    Here is a cool LaTeX guide: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
     
  6. Sep 30, 2012 #5
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