Series solution to diffeq, stuck on matching the indices

[freshman2013]

Homework Statement

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

Homework Equations

y=summation from n=0 to infinity Cn*x^n

The Attempt at a Solution

y=sum from 0 to inf C_nxⁿ
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)C_n+2 xⁿ from 0 to infinity

basically, I can't get all three series to match at starting index n=0 and get them to have the terms x^n

 

[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.

 

[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.

 

[vela]

Homework Statement

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

Homework Equations

y=summation from n=0 to infinity Cn*x^n

The Attempt at a Solution

y=sum from 0 to inf C_nxⁿ

$$y = \sum_{n=0}^\infty c_n x^n$$

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

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

y''=(n+2)(n+1)C_n+2 xⁿ from 0 to infinity

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

basically, I can't get all three series to match at starting index n=0 and get them to have the terms x^n

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.

 

[freshman2013]

what do I do with the extra C0 and C1X?

 

[HallsofIvy]

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.