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Differential Equation

  1. Oct 20, 2009 #1
    [tex]x^2[/tex][tex]d^2[/tex]y/[tex]dx^2[/tex]+[tex]3x[/tex][tex]dy[/tex]/[tex]dx[/tex]+[tex]5y[/tex]=[tex]3x[/tex]


    I don't know where to start with the question ,can anyone here help me plzz.
     
  2. jcsd
  3. Oct 20, 2009 #2
    This equation looks to be solvable using the variation of parameters method or a green's function. To rearrange:

    [tex]y''+3x^{-1}y'+5x^{-2}y=3x^{-1}[/tex]

    The method states that if you have

    [tex]y''+a_1(x)y'+a_2(x)y=F[/tex]

    then let y1 and y2 be solutions to the associated homogeneous equation.

    Then the particular solution is

    [tex]y_p=u_1y_1+u_2y_2[/tex]

    where u1 and u2 satisfy both:

    [tex]y_1u_1'+y_2u_2'=0[/tex]

    [tex]y_1'u_1'+y_2'u_2'=F[/tex]

    and the general solution:

    [tex]y(x)=c_1y_1+c_2y_2+y_p[/tex]

    To use a green's function to find yp, then

    [tex]y_p(x)=\displaystyle\int_{x_0}^xK(x,t)F(t)dt[/tex]

    where the green's function, K(x,t), is defined as

    [tex]K(x,t)=\frac{y_1(t)y_2(x)-y_2(t)y_1(x)}{W[y_1,y_2](t)}[/tex]

    and the Wronskian, W[y1,y2](t), is defined as

    [tex]W[y_1,y_2](t)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=y_1y_2'-y_2y_1'[/tex]

    unless I made a typo somewhere.
     
    Last edited: Oct 20, 2009
  4. Oct 20, 2009 #3
    Wow U r legend!Thanks a lot matey ;)
     
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