Posted by Jen on Saturday, April 21, 2012 at 5:18pm.
Seek power series solution of the given differential equation about the given point x0; find the recurrence relation.
(1x)y'' + y = 0; x0 = 0

math  MathMate, Saturday, April 21, 2012 at 9:16pm
I'll give it a try.
(1x)y"+y=0 at x0=0
A. Check for singular points.
Write as y"+(1/1x)y=0
and find Lim x>0 (1/1x) = 1
Therefore x0=0 is not a singular point.
B. Assume y=∑an x^n for n=0 to ∞
then
(1x)y"=∑n=0 to ∞ (n)(n1)an x^(n2)
=∑n=2 to ∞ (n)(n1)an x^(n2) ∑n=1 to ∞ (n)(n1)an x^(n1)
Now switch indices
=∑n=0 to ∞ (n+2)(n+1)an+2 x^(n) ∑n=0 to ∞ (n+1)(n)an+1 x^(n)
Add to y gives
y(x)
∑n=0 to ∞ (n+2)(n+1)an+2 x^(n) ∑n=0 to ∞ (n+1)(n)an+1 x^(n)+∑ n=0 to ∞ an x^n =0
This gives
an+1 = [(n)(n+1)an+1  an]/[(n+2)(n+1)]
The coefficients an are dependent on values of a0 and a1, which are arbitrary constants depending on the given initial conditions.
Check my arithmetic.
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