# Series solution to diffeq, stuck on matching the indices

1. Dec 6, 2013

### freshman2013

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

find the series solution to y''+x^2*y'+y=0

2. Relevant equations
y=summation from n=0 to infinity Cn*x^n

3. The attempt at a solution

y=sum from 0 to inf Cnxn
x^2*y'=sum from 1 to inf nC n xn+1 = sum from 2 to inf (n-1) C n-1 xn = sum from 1 to inf (n-1) C n-1 xn
y''=(n+2)(n+1)Cn+2 xn from 0 to infinity

basically, I cant get all three series to match at starting index n=0 and get them to have the terms x^n

Last edited: Dec 6, 2013
2. Dec 6, 2013

### brmath

It's a little hard to read without Latex, but your y'' doesn't seem right. The y' alone will have terms of the form $nC_nx^{n-1}$ starting at n = 1 (since the constant term vanishes) , so the y'' terms should be $n(n-1)C_nx^{n-2}$ starting at n = 2.

You do not have to fool around with the indices on $x^2y'$. Keep things as simple as you can, and I think it will match up.

3. Dec 6, 2013

### freshman2013

@brmath my y'' is what you have w/ the indices shifted from 2 to 0. How do you manage not shift indices? Don't you need to combine everything into a single summation term and factor x^n, setting the rest equal to zero? That's what I been doing on every problem so far for finding series to diffeqs.

4. Dec 7, 2013

### vela

Staff Emeritus
$$y = \sum_{n=0}^\infty c_n x^n$$
$$x^2 y' = \sum_{n=1}^\infty n c_n x^{n+1} = \sum_{n=2}^\infty (n-1)c_{n-1} x^n = \sum_{n=1}^\infty (n-1)c_{n-1}x^n$$ Looks fine, though that last summation seems unnecessary.

$$y'' = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n$$ That's fine.

You don't have to get them to all start at n=0. You have
$$y'' + x^2 y' + y = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n + \sum_{n=2}^\infty (n-1)c_{n-1} x^n + \sum_{n=0}^\infty c_n x^n$$ Just separate out the first two terms of the first and last series, e.g.
$$y = c_0 + c_1x + \sum_{n=2}^\infty c_nx^n$$ and then you'll have three summations that all start from n=2 so you can combine them.

5. Dec 7, 2013

### freshman2013

what do I do with the extra C0 and C1X?

6. Dec 7, 2013

### HallsofIvy

Staff Emeritus
Vela told you that- separate the n= 0 and n= 1 terms from the first and third sums.
$$y'' + x^2 y' + y = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n + \sum_{n=2}^\infty (n-1)c_{n-1} x^n + \sum_{n=0}^\infty c_n x^n= 0$$

In order that a power series be 0 for all x, we must have each coefficient 0.

If n= 0, $(n+2)(n+1)c_{n+2}= 2c_2$ and $c_n= c_0$ so we must have 2c_2+ c_0= 0.

If n= 1, $(n+2)(n+1)c_{n+2}= 6c_3$ and $c_n= c_1$ so we must have $6c_3+ c_1= 0$.

For $n\ge 2$ we have all three sums so
$(n+2)(n+1)c_{n+2}+ (n- 1)c_{n-1}+ c_n= 0$

In order to solve for $c_2$, you will need to know $c_0$ and, in order to solve for $c_3$, you will need to know $c_1$. Those will be the two "constants" for the general solution.