# series solution of an initial value problem

• Jul 27th 2010, 10:40 AM
Chipps
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
• Jul 27th 2010, 10:44 AM
Ackbeet
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.
• Jul 27th 2010, 10:47 AM
Chipps
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.
• Jul 27th 2010, 10:52 AM
Ackbeet
Well, you can do this:

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