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Homework Help: Integration by Parts/Differential Equation

  1. Sep 27, 2006 #1
    If you have [tex] \int e^{x^{2}}x^{2} [/tex] would it be reasonable to choose [tex] u = e^{x^{2}}, dv = x^{2}, du = 2xe^{x^{2}}, v = \frac{x^{3}}{3} [/tex]? And then I get [tex] x^{3}e^{x^{2}} - 2\int x^{4}e^{x^{2}} [/tex]. Would this be equivalent to choosing [tex] u = x, dv = 2xe^{x^{2}}, v = e^{x^{2}}, du = dx [/tex].

    This was for solving a differential equation:

    [tex] \frac{dy}{dx} + 2xy = x^{2} [/tex], where [tex] P(x) = 2x, Q(x) = x^{2}, I = e^{\int 2x} = e^{x^{2}} [/tex]. So [tex] (ye^{x^{2}})' = e^{x^{2}}x^{2} [/tex]

    [tex] y = e^{-x^{2}} \int e^{x^{2}}x^{2} dx + C [/tex]

    Thanks
     
    Last edited: Sep 27, 2006
  2. jcsd
  3. Sep 27, 2006 #2
    You should note that
    [tex]
    2xe^{x^2}=\frac{d}{dx}e^{x^2},
    [/tex]
    so that your integrand can be written as
    [tex]
    e^{x^2}x^2=\frac{x}{2}\left(2xe^{x^2}\right)=\frac{x}{2}\left(\frac{d}{dx}e^{x^2}\right)
    [/tex]
    Then you can integrate this by parts.
    By the way in your expression for [itex]y[/itex] you seem to have lost the homogeneous solution.
    A further note is that you might find it useful to have a look at the error function when trying to evaluate your expression:

    http://mathworld.wolfram.com/Erf.html

    (....and you shouldnt have the constant C in your expression for [itex]y[/itex], it should be attached to your homogeneous solution....)
     
    Last edited: Sep 27, 2006
  4. Sep 27, 2006 #3
    thank you for your help
     
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