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Dirst order ODE's with x & expontials

  1. Jun 1, 2006 #1
    have a ode that looks like this:

    so i find the particular integral which is e^-(x^2)/2

    but then when i try to solve the eqn i have to do a integration by parts
    ye^-(x^2)/2 = integrate[ -xe^-(x^2)/2 dx]
    but when i do this integration by parts i end up with having to do another, and i never ends..
    i know that there is some trick when dealing with simple x*e^x bla bla bla..

    but don't know how to do it.. would someone please show me, or help me out?
  2. jcsd
  3. Jun 1, 2006 #2


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    Why not solve:

    [tex]\int \frac{dy}{y-1}=\int x dx[/tex] ?
  4. Jun 6, 2006 #3


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    I think there was a slip of the mind on that last step:

    xy - x = x(y-1), so dy/(y-1) = xdx, as originally stated by J77.

    I'm sure it was just an accident, but I figured it warranted correction.
  5. Jun 7, 2006 #4


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    As J77 pointed out, this is a separable first order equation:
    dy/dx= xy- x= x(y-1) so you get
    (edited to correct misprint)
    [tex]\frac{dy}{y-1}= xdx[/tex]
    and so [itex]ln(y-1)= \frac{1}{2}x^2+ c[/itex].
    [tex]y(x)=1+ Ce^{\frac{x^2}{2}[/tex]

    But if you really want to do that integration, just use a simple substitution: your integrand is NOT "[itex]xe^x[/itex]" (which could be done easily by parts) but [itex]xe^{x^2}[/itex]. Let u= x2.
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