Could use a little advice here, with a 2nd order ODE

  1. xy'' -(2x+1)y' + (x+1)y = (x ex)2

    I know a solution - (x-1)e2x

    Thus, y= ((x-1)e2x u(x))

    Now, i know how to do the whole reduction of order thing, but when i find y' and y'' and substitute, the u(x) term doesn't cancel out so this doesn't work

    (x2-x)u'' + (2x2-x+1)u' + x2u = x2

    So, how do i approach this? Thanks.
  2. jcsd
  3. You know that the solution can be written in the form [tex]y(x)=y_{p}(x)+y_{h}(x)[/tex] where [tex]y_{p}(x)[/tex] is the particular solution and [tex]y_{h}(x)[/tex] is the solution to the homogeneous equation


    You already know that [tex]y_{p}(x)=(x-1)e^{2x}[/tex].

    For the homogeneous equation, there are two linearly independent solutions, one of them is clearly


    and using the reduction of order, the other is


    So the general solution to the ode is

    If you need further justification, use frobenius method for the homogenous equation to calculate [tex]y_{h_{1}}(x)[/tex], reduction of order for [tex]y_{h_{2}}(x)[/tex], and variation of parameters for [tex]y_{p}(x)[/tex]. That way the problem is fully justified.

    ps. i deleted my first post as it was wrong and confusing.
    Last edited: Oct 5, 2004
  4. I understand the first technique, but i'm not familiar with the Frobenius method, i looked it up on Mathworld and it looks really complex

    With your other post, i can't quite follow how to get the second linearly independant solution. Thanks.
  5. Ignore the first post... is nonsence, thats why i deleted it.

    Given [tex]y_{1}(x)=e^{x}[/tex] using reduction of order to get the second solution


    implies that



    substituting in [tex]xy''_{2}(x)-(2x+1)y'_{2}(x)+(x+1)y_{2}(x)=0[/tex]

    we get that


    do i need to go further?

    ps. Frobenius method is used only if you want to *construct* the first solution [tex]y_{1}=c_{1}e^{x}[/tex]. Just so you know that it didnt came from divine inspiration, you dont need to use it though.
    Last edited: Oct 5, 2004
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