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Solving A DE

  1. Sep 30, 2011 #1
    I've picked up a bit more since my last problem. I need to solve the following DE:
    [tex]x^{2}\frac{dy}{dx}+x(x+2)y=e^{x}[/tex]

    I decided to use variation of parameters, so I re-arranged it like so:
    [tex]\frac{dy}{dx}=\frac{e^{x}}{x^{2}}-(1+\frac{2}{x})y[/tex]

    Then solved the homogenous DE:
    [tex]\frac{dy}{dx}=-(1+\frac{2}{x})y[/tex]
    [tex]y=e^{-x}x^{-2}c[/tex]

    Now for the particular solution:
    [tex]y_{p}=u(x)e^{-x}x^{-2}c
    [tex]\frac{dy}{dx}=u'(x)e^{-x}x^{-2}-u(x)e^{-x}x^{-2}-2u'(x)e^{-x}x^{-2}[/tex]

    When I shoved this back in I wound up with this for u'(t):
    [tex]u'(x)=e^{x}x^{-2}[/tex]

    It seems...a bit strange. Did I mess up somewhere? It's a bit hard to integrate. I've gone over this several times already.
     
    Last edited: Sep 30, 2011
  2. jcsd
  3. Sep 30, 2011 #2

    HallsofIvy

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    Looks to me like you have an algebra error somewhere. Doing exactly what you say, I get [itex]u'e^{-x}= e^x[/itex] or [itex]u'= e^{2x}[/itex].
     
  4. Sep 30, 2011 #3
    what is u(t)?
     
  5. Sep 30, 2011 #4
    u(t) is the unknown function, that when multiplied by the solution to the homogenous equation, gives you a particular solution to the DE. I forgot to put in the step where I set that part up. It should also be u(x). I'll try going over the algebra again.
     
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