Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving xy'(x) - y(x) = x^2 Exp[x] using the power series method

  1. Jan 19, 2007 #1
    I've tried solving the equation xy'(x) - y(x) = x^2 Exp[x] using the power series method. I assume that y has the form:

    [tex] y = \sum_{n=0}^{\infty} a_n x^n [/tex]

    Inserting this in the diff. eq. gives:

    [tex] \sum_{n=0}^{\infty} n a_n x^n - \sum_{n=0}^{\infty} a_n x^n = x^2 e^x [/tex]

    Now, in the other types of diff. eqations I've solved using this method I found a recursive formula for the coefficients, but I don't see how I can do that here? How would I got about solving this problem?

  2. jcsd
  3. Jan 19, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    You could try using the series for ex on the RHS. It's just a suggestion, so I don't know whether it'll work, but give it a go!
  4. Jan 19, 2007 #3


    User Avatar
    Science Advisor

    Yes, if you are required to use the series method, expand ex in Taylor's series and compare corresponding powers of x.

    Of course, that's an "Euler type" equation so it should be easier to solve it directly, perhaps by converting to an equation with constant coefficients.
  5. Jan 20, 2007 #4
    My problem is that I don't see how to obtain a recusion formula for the coefficients. Even when trying to solve just the corresponding homogeneous equation [tex]x y'(x) - y(x) = 0[/tex] I end up with something like [tex]n a_n - a_n = 0 [/tex] and I cant see what to do.
  6. Jan 20, 2007 #5


    User Avatar
    Science Advisor

    Obviously, [itex]na_n- a_n= 0[/itex] is true as long as [itex]a_n= 0[/itex] for any n other than 1. For n= 1, the equation becomes [itex]a_1= a_1[/itex] which is also true- [itex]a_1[/itex] becomes the undetermined constant in your general solution. What that tells you is that the general solution to xy'(x)- y(x)= 0 is y(x)= Cx.
    Since [itex]e^x= \Sigma_{n=0}^\infty\frac{1}{n!}x^n[/itex], [itex]x^2e^x= \Sigma_{n=0}^\infty\frac{1}{n!}x^{n+2}[/itex]. If you let i= n+2, then you get [itex]x^2e^x= \Sigma_{i=2}^\infty\frac{1}{(i-2)!}x^i[/itex] so you can compare "like powers".
    You get [itex](0)a_0- a_0= 0[/itex] so a_0= 0. [itex](1)a_1- a_1= 0[/itex] which is always true- [itex]a_1[/itex] will be an "unknown constant".

    For n> 1, [itex]na_n- a_n= a_n(n-1)= \frac{1}{(n-2)!}[/itex] so
    [tex]a_n= \frac{1}{(n-2)!(n-1)}[/itex].
  7. Jan 21, 2007 #6
    Hmm, of course. Why did I not see that. I guess I was to focused on obtaining a recursion formula for the coefficients, because I thought I always had to do that. I see now that I don't. Thanks alot for the help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook