# Solving differential equation (involves power series)

1. Nov 30, 2009

### EvilKermit

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 = $$\sum$$an(x+2)n (start at n=0, n goes to infinity)

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

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

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

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

This further simplifies to:

0 = $$\sum$$(6an-2ann)(x+2)n + (-x^2 + 4x -3)$$\sum$$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.............

2. Dec 1, 2009

### matematikawan

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.

3. Dec 2, 2009

### Coldatoms

Your equation has regular singularities at x={1,3,$$\infty$$}. 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.

4. Dec 2, 2009

### yungman

How about set y = $$\sum$$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.

5. Dec 2, 2009

### yungman

How about set y = $$\sum$$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.