How Do You Solve the Series Solutions for y + xy = 0 and y'' + xy = 0?

[BobMarly]

Series solution for y"+x*y=0

Working on recurance realtion.
Get to (sum(n=2))n*(N-1)*a(n)*X^(n-2)+(sum(n=0))a(n)*x^(n)
Try several things but not sure if their correct.

 

[Marioeden]

Try y = Ʃ a(n)*x^n
where n goes from 0 to infinity

Now y'' = Ʃ n(n-1)*a(n)*x^(n-2)
where n goes from 2 to infinity

and x*y = Ʃ a(n)*x^(n+1)
where n goes from 0 to infinity

We want these to have the same power of x i.e. n-2, so let's write them as follows:

y'' = 2*a(2) + Ʃ n(n-1)*a(n)*x^(n-2)
where n goes from 3 to infinity
(we just pulled the first term out of the summation)

x*y = Ʃ a(n-3)*x^(n-2)
where n goes from 3 to infinity
(index shift)

Thus we have y'' + x*y = 2*a(2) + Ʃ [ n(n-1)*a(n) + a(n-3) ]*x^(n-2)
where n goes from 3 to infinity.

We want this to equal zero, so we have that:
y = Ʃ a(n)*x^n, where
a(0) and a(1) are arbitrary,
a(2) = 0,
and a(n) = -a(n-3) / [n(n-1)]Hope that helps :)