# Solving an ODE with power series method

1. Mar 9, 2013

### SithsNGiggles

1. The problem statement, all variables and given/known data

Solve $(1-x)y''+y=0$ at the point $x_0=0$. Use this solution to find a solution to $xy''+y=0$ around the point $x_0=1$.

2. Relevant equations

3. The attempt at a solution

$(1-x)y''+y=0$
$(x-1)y''=y$
$\displaystyle\sum_{k=2}^\infty a_k k (k-1)\left(x^{k+3}-x^{k+2}\right)=\displaystyle\sum_{k=0}^\infty a_k x^k$

I tried to find a recurrence relation for $a_k$, but I keep getting 0 for all k. I realize that y = 0 is a solution, but I don't know how that would help me in the second part.

2. Mar 10, 2013

### ehild

Compare the coefficients of xk: they have to be the same on both sides.

What do yo get for the constant term? What is the linear term?

ehild

3. Mar 10, 2013

### SithsNGiggles

Right, that's where my problem occurs:

$y''=\sum_{k=2}^\infty a_k k(k-1)\left(x^{k+3}-x^{k+2}\right)=a_2 (2\cdot1)\left(x^5-x^4\right)+a_3 (3\cdot2)\left(x^6-x^5\right)+a_4 (4\cdot3)\left(x^7-x^6\right)+\cdots$

$y=\sum_{k=0}^\infty a_k x^k=a_0 + a_1x + a_2x^2+ a_3x^3+a_4x^4+\cdots$

Matching up coefficients, I have
$a_0=0\\ a_1=0\\ a_2=0 \\ a_3=0\\ a_4=-(2\cdot1)a_2=0 \\ a_5=(2\cdot1)a_2-(3\cdot2)a_3=0\\ \text{and so on...}$

4. Mar 10, 2013

### ehild

The second derivative of y is wrong. The correct equation is

$$\displaystyle\sum_{k=2}^\infty a_k k (k-1)\left(x^{k-2}-x^{k-1}\right)=\displaystyle\sum_{k=0}^\infty a_k x^k$$

ehild

5. Mar 11, 2013

### SithsNGiggles

Oh, right. I had (k + 2) on my mind for some reason. Thanks for pointing that out!