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

    [tex]xy_{h}''(x)-(2x+1)y_{h}'(x)+(x+1)y_{h}(x)=0[/tex].

    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

    [tex]y_{h_{1}}(x)=c_{1}e^{x}[/tex]

    and using the reduction of order, the other is

    [tex]y_{h_{2}}(x)=c_{2}\frac{x^2}{2}e^{x}[/tex].

    So the general solution to the ode is
    [tex]y(x)=c_{1}e^{x}+c_{2}\frac{x^2}{2}e^{x}+(x-1)e^{2x}[/tex]

    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

    [tex]y_{2}(x)=v(x)e^{x}[/tex]

    implies that

    [tex]y'_{2}(x)=[v'(x)+v(x)]e^{x}[/tex]

    [tex]y''_{2}(x)=[v''(x)+2v'(x)+v(x)]e^{x}[/tex]

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

    we get that

    [tex]xv''(x)-v'(x)=0[/tex]

    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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