Series Solution to ODE IVP: How to Find Initial Values?

[GRB 080319B]

Homework Statement

Find the series solution to the initial value problem.
xy\acute{}\acute{} + y\acute{} + 2y = 0
y(1) = 2
y\acute{}(1) = 4

Homework 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}

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:

(2c_{2} + c_{1} + 2c_{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.

 

[Mark44]

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?