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

Solving differential equation (involves power series)

  1. Nov 30, 2009 #1
    Hey guys. I'm new here. I've been trying to figure out how to solve this problem, and I'm still confused.

    (-x^2 + 4x -3)* d2y/dx2 - 2(x-2) * dy/dx + 6y = 0

    y(-2) = 1
    dy/dx(-2) = 0

    I set y = [tex]\sum[/tex]an(x+2)n (start at n=0, n goes to infinity)

    dy/dx = [tex]\sum[/tex]ann(x+2)n-1 (start at n=0, n goes to infinity)

    d2y/dx2 = [tex]\sum[/tex]ann(n-1)(x+2)n-2 (start at n=0, n goes to infinity


    So:
    0 = 6[tex]\sum[/tex]an(x+2)n - 2(x+2)[tex]\sum[/tex]ann(x+2)n-1 + (-x^2 + 4x -3)[tex]\sum[/tex]ann(n-1)(x+2)n-2

    I can simplify this to:
    0 = [tex]\sum[/tex]6an(x+2)n - [tex]\sum[/tex]2ann(x+2)n + (-x^2 + 4x -3)[tex]\sum[/tex]ann(n-1)*(x+2)n-2

    This further simplifies to:

    0 = [tex]\sum[/tex](6an-2ann)(x+2)n + (-x^2 + 4x -3)[tex]\sum[/tex]ann(n-1)*(x+2)n-2

    However I don't know how to further simplify it, and where to go from here. Am I on the right track? I believe the answer will look like this:

    y = a1+ a2x + a3x2 + a4x3.............

    If somebody could please help me out, I would very much appreciate it. Thank You
     
  2. jcsd
  3. Dec 1, 2009 #2
    Look like you are on the right track but there is still a lot more to do.

    I would prefer you start with let t=x+2 and replace all x in the DE with t-2.
     
  4. Dec 2, 2009 #3
    Your equation has regular singularities at x={1,3,[tex]\infty[/tex]}. Are these points within the range of the desired solution? If you need solutions that are near to these singularities you might have better luck using the Frobenius series technique and find solutions that cover your desired range. Start with the indicial equation.

    Right now it looks to me like you're trying to find a Taylor series solution about the point where your initial conditions are defined (x=-2) but this is so close to the singular point (x=1) that you could have problems. However, it might be good enough if you're only concerned about solutions in the neighborhood of x=-2.
     
  5. Dec 2, 2009 #4
    How about set y = [tex]\sum[/tex]an(x)n (start at n=0, n goes to infinity)

    Just do it the regular way!!! then you absorb the polynomial into the series. It is not as pretty, but you don't end up with the polynomial times a series.
     
  6. Dec 2, 2009 #5
    How about set y = [tex]\sum[/tex]an(x)n (start at n=0, n goes to infinity)

    Just do it the regular way!!! then you absorb the polynomial into the series. It is not as pretty, but you don't end up with the polynomial times a series.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook