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Variation of parameters applied to an ODE

  1. Dec 4, 2012 #1

    fluidistic

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    The ODE to solve via variation of parameters is ##(1-x)y''+xy'-y=(1-x)^2##.
    Knowing that ##e^x## and ##x## are solutions to the homogeneous ODE.
    Now if I call ##y_1=x## and ##y_2=e^x##, the Wronskian is ##W(y_1,y_2)=e^{x}(x-1)##.
    According to http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx, the particular solution of the non homogeneous ODE should be of the form ##-y_1 \int \frac{y_2 (1-x)^2}{W(y_1,y_2)} dx+y_2 \int \frac{y_1(1-x)^2}{W(y_1,y_2)}##. This gave me ##y_p(x)=-\frac{x^3}{2}-x-1## I've even checked out with wolfram alpha the evaluation of the integrals, that is the result. However the answer is ##y_p=x^2+1##.
    I'm clueless on what's going on.
     
  2. jcsd
  3. Dec 4, 2012 #2

    Mute

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    If you take a careful look at that page you linked to, you'll notice that although the author writes the ODE as ##p(x)y'' + q(x)y' + r(x)y = g(x)##, he eventually states that he will assume p(x) = 1. The wikipedia page for variation of parameters similarly makes this assumption. Try diving your differential equation by (1-x) and then applying the ##-y_1 \int \frac{y_2 g(x)}{W(y_1,y_2)} dx+y_2 \int \frac{y_1g(x)}{W(y_1,y_2)}## formula, where g(x), the non-homogeneous term, is (1-x) rather than (1-x)2.
     
    Last edited: Dec 4, 2012
  4. Dec 4, 2012 #3

    fluidistic

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    Thank you very much. I totally missed this.
     
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