- #1
GRB 080319B
- 108
- 0
Homework Statement
Find the series solution to the initial value problem.
xy[tex]\acute{}[/tex][tex]\acute{}[/tex] + y[tex]\acute{}[/tex] + 2y = 0
y(1) = 2
y[tex]\acute{}[/tex](1) = 4
Homework Equations
y=[tex]\sum^{\infty}_{n=0}c_{n}(x-1)^{n}[/tex]
t = (x-1), x = (t+1)
y = [tex]\sum^{\infty}_{n=0}c_{n}t^{n}[/tex]
y[tex]\acute{}[/tex]= [tex]\sum^{\infty}_{n=1}c_{n}(n)t^{n-1}[/tex]
y[tex]\acute{}[/tex][tex]\acute{}[/tex]= [tex]\sum^{\infty}_{n=2}c_{n}(n)(n-1)t^{n-2}[/tex]
The Attempt at a Solution
I substituted the above series into the DE, adjusted the series so they all had t[tex]^{n}[/tex], and took out terms so that they all had the same starting index. By grouping the terms and the series, I got:
(2[tex]c_{2}[/tex] + [tex]c_{1}[/tex] + 2[tex]c_{0}[/tex]) + [tex]\sum^{\infty}_{n=1}t^{n}[ (n+2)(n+1)c_{n+2} + (n+1)^{2}c_{n+1} + 2c_{n}][/tex] = 0
Setting the terms and series equal to zero and finding several constants:
[tex]2c_{2}[/tex] + [tex]c_{1}[/tex] + [tex]2c_{0}[/tex] = 0
[tex](n+2)(n+1)c_{n+2}[/tex] + [tex]c_{n+1}[/tex](n+1)[tex]^{2}[/tex] + [tex]2c_{n}[/tex]] = 0
[tex]c_{0}[/tex] = -(2c_{2}[/tex]+[tex]c_{1}[/tex])/2
[tex]c_{1}[/tex] = -2([tex]c_{0}[/tex]+[tex]c_{2}[/tex])
[tex]c_{2}[/tex] = -([tex]c_{1}[/tex]+[tex]2c_{0}[/tex])
[tex]c_{3}[/tex] = [tex](2/3)c_{0}[/tex]
[tex]c_{4}[/tex] = [tex](c1-4c0)/(3\cdot4)[/tex]
[tex]c_{5}[/tex] = [tex]-(c1+5c0)/(3\cdot5)[/tex]
[tex]c_{6}[/tex] = [tex](9c1+46c0)/(2\cdot3\cdot5\cdot6)[/tex]
I don't understand how find the solution for the initial values. I can't determine a pattern for the constants for [tex]c_{n}[/tex]. Are you supposed to group these constants: y = [tex]c_{0}[/tex][ 1 + x + ...] + [tex]c_{1}[/tex][ 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.