# Thread: Power Series Solving Differencial Equation

1. ## Power Series Solving Differencial Equation

Solve the following by Power Series, then find a basic set of solutions:
(x-2)^2 y" + 2(x-2)y' - 6y = 0

There are two different sets of power series that y", y', and y can equal, so that is confusing, but I also do not really know how to work the problem, so your help would be great. Thanks.

2. Originally Posted by haleyGeorge
Solve the following by Power Series, then find a basic set of solutions:
(x-2)^2 y" + 2(x-2)y' - 6y = 0

There are two different sets of power series that y", y', and y can equal, so that is confusing, but I also do not really know how to work the problem, so your help would be great. Thanks.
Given that LaTeX is currently out on this site, I'm not even going to try to help you directly here. You might find this site helpful though.

You SHOULD have two different power series as solutions. This is a second order linear ODE and thus has two linearly independent solutions.

-Dan

3. I will solve this with series centered at x=0.
Note, x=0 is an ordinary point.

4. What you have done makes perfect sense and is helping me remember some of the class examples, but going back to the original problem, what else do I need to do to be able to completely answer the question?

5. If the power series is exactly 0 then the coefficients of all powers of x must also be 0. So:
8*a_2 + 4*a_1 - 6*a_0 - 6*a_1 = 0 <-- Constant terms
and
-8*a_2 + 24*a_3 + 2*a_1 - 8*a_2 = 0 <-- Linear coefficients of x

From the first equation:
a_2 = (1/8)*(6*a_0 + 2*a_1) = (1/4)*(3*a_0 + a_1)

So the second equation is:
24*a_3 -16*a_2 + 2*a_1 = 0

24*a_3 -16*[(1/4)*(3*a_0 + a_1)] + 2*a_1 = 0

a_3 = (1/24)*(12*a_0 + 2*a_1) = (1/12)*(6*a_0 + a_1)

So given values for a0 and a1 we can find a2 and a3.

Now look at the general term:
n(n-1)*a_n - 4(n+1)n*a_{n+1} + 4(n+2)(n+1)*a_n + 2n*a_n - 4(n+1)*a_{n+1} - 6*a_n = 0

Solving we get:
a_{n+1} = [-n(n+1) - 4(n+2)(n+1) - 2n + 6]/[-4(n+1)n - 4(n+1)] * a_n

a_{n+1} = [5n^2 + 15n + 2]/[4n^2 + 8n + 4]

a_n = [5n^2 + 5n - 8]/[4n^2] = (5/4)*(1 + 1/n - 8/n^2)

This will give you your a_n values for n > 3.

-Dan