Series Solution of Second Order Linear Equation, IVP Consider the initial value problem
$$y' = \sqrt{1-y^2}$$ 
$$y(0) = 0$$
Look for a solution of the IVP in the form of power series about x=0.
I have started with assuming that  $ y = \sum_{n = 0}^{\infty} {a_nx^n}$. Then $ y' =\sum_{n = 0}^{\infty}{n a_n x^{n-1}}$ .
$\sum_{n = 0}^{\infty}{(n+1) a_{n+1} x^n } = \sqrt{1 - (\sum_{n = 0}^{\infty}{a_nx^n})^2 }$
I got stuck there, because I don't understand how to take power of summation.
Need some help, thanks.
 A: The unique solution of this IVP is $\sin t$, $t\in(-\pi/2,\pi/2)$, and I doubt that it can be obtained by series expansion. 
Note though, that left of $-\pi/2$ and right of $\pi/2$ uniqueness is violated!
A: Not sure if this works, I wanted to leave it as a comment but it is too long.

*

*We have
$$ \left(\sum_{n = 0}^{\infty}{(n+1) a_{n+1} x^n }\right)^2 = 1 - (\sum_{n = 0}^{\infty}{a_nx^n})^2 $$


*Pick pencil and paper.


*Treat the sums as "finite".


*Ask yourself: what is the coefficient of $x^0$ in the left-hand side and in the right-had side? They must be equal.


*Ask yourself: what is the coefficient of $x^1$ in the left-hand side and in the right-had side? They must be equal.


*And so on...


*Hopefully you will get some pattern because as we move to higher powers, it becomes too tiresome to collect at the the coefficients of power $x^n$.


*By Yiorgos S. Smyrlis's answer the solution is $\sin (x)$. This give you additional hints. Using the above, you want to show $a_n=0$ for even $n$ and $a_n=\dfrac{(-1)^{(n-1)/2}}{n!}$ for odd $n$.
