Ordinary Differential Equation Series Solution

[EnzoF61]

Homework Statement

y' = \sqrt{(1-y^2)<br /> }
Initial condition y(0) = 0
a) Show y = sinx is a solution of the initial value problem.
b) Look for a solution of the initial value problem in the form of a power series about x = 0. Find coefficients up to the term in x^3 in this series.

Homework Equations

part a) was Ok.

The Attempt at a Solution

This is for my part b attempt.

(y')^2 + (y)^2 = 1

(cosx)^2 + (sinx)^2 = 1

\sum(-1)^ {2n} * (x^ {4n} )}/(2n!)^2 + \sum (-1)^{2n} * (x^ {4n+2})/((2n+1)!)^2 = 1

I had also tried the general solution series for y and y'. y = a0 + a1x + ... + an*x^n
y' = a1 + ... + n*an x^n
Please note the values such as a0, a1, ...an have the nth term as a subscript.

Maybe I should not be using part a where y=sinx and y'=cosx because I will have even powers when squared always...?

What series will allow me to evaluate at the x^3 term?

Thanks,
-Adam

 

[Mark44]

Double posted.