# Thread: series solution of an initial value problem

1. ## series solution of an initial value problem

Please help me get the power series solution of this problem. I have a problem with getting rid of the square(^2) in the resulting equation.

(dy/dx)^2 = 1 - y^2 y(0) = 0

2. Clarification: on the LHS, is that the second derivative, or is it, as it's written, the square of the first derivative?

If it is the square of the first derivative, does it have to be a series solution? That equation is separable.

3. it's actually the square of the first derivative. I would like to get the series solution but it's the square that is a problem.

4. Well, you can do this:

$\displaystyle{\frac{dy}{dx}=\pm\sqrt{1-y^{2}},}$ and solve each equation separately.