# ODE IVP Series Solution

1. Oct 26, 2009

### GRB 080319B

1. The problem statement, all variables and given/known data
Find the series solution to the initial value problem.
xy$$\acute{}$$$$\acute{}$$ + y$$\acute{}$$ + 2y = 0
y(1) = 2
y$$\acute{}$$(1) = 4

2. Relevant equations

y=$$\sum^{\infty}_{n=0}c_{n}(x-1)^{n}$$

t = (x-1), x = (t+1)

y = $$\sum^{\infty}_{n=0}c_{n}t^{n}$$

y$$\acute{}$$= $$\sum^{\infty}_{n=1}c_{n}(n)t^{n-1}$$

y$$\acute{}$$$$\acute{}$$= $$\sum^{\infty}_{n=2}c_{n}(n)(n-1)t^{n-2}$$

3. The attempt at a solution
I substituted the above series into the DE, adjusted the series so they all had t$$^{n}$$, and took out terms so that they all had the same starting index. By grouping the terms and the series, I got:

(2$$c_{2}$$ + $$c_{1}$$ + 2$$c_{0}$$) + $$\sum^{\infty}_{n=1}t^{n}[ (n+2)(n+1)c_{n+2} + (n+1)^{2}c_{n+1} + 2c_{n}]$$ = 0

Setting the terms and series equal to zero and finding several constants:

$$2c_{2}$$ + $$c_{1}$$ + $$2c_{0}$$ = 0

$$(n+2)(n+1)c_{n+2}$$ + $$c_{n+1}$$(n+1)$$^{2}$$ + $$2c_{n}$$] = 0

$$c_{0}$$ = -(2c_{2}[/tex]+$$c_{1}$$)/2

$$c_{1}$$ = -2($$c_{0}$$+$$c_{2}$$)

$$c_{2}$$ = -($$c_{1}$$+$$2c_{0}$$)

$$c_{3}$$ = $$(2/3)c_{0}$$

$$c_{4}$$ = $$(c1-4c0)/(3\cdot4)$$

$$c_{5}$$ = $$-(c1+5c0)/(3\cdot5)$$

$$c_{6}$$ = $$(9c1+46c0)/(2\cdot3\cdot5\cdot6)$$

I don't understand how find the solution for the initial values. I can't determine a pattern for the constants for $$c_{n}$$. Are you supposed to group these constants: y = $$c_{0}$$[ 1 + x + ...] + $$c_{1}$$[ 1 + x +...] to get the solution, and if so how do we find the constants for the initial values? Any help will be greatly appreciated.

2. Oct 27, 2009

### Staff: Mentor

Your initial conditions y(1) = 2 and y'(1) = 4 are all in terms of x. Your power series is in terms of t = x - 1, so x = 1 corresponds to t = 0. Does that help?