Question

For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sqrt{4 - y^2}, - 2 < y < 2\]

Answer 1

We have, 

\[\frac{dy}{dx} = \sqrt{4 - y^2}\]

\[ \Rightarrow \frac{1}{\sqrt{4 - y^2}}dy = dx\]

Integrating both sides, we get

\[\int\frac{1}{\sqrt{4 - y^2}}dy = \int dx\]

\[ \Rightarrow \sin^{- 1} \frac{y}{2} = x + C\]

\[ \Rightarrow \frac{y}{2} = \sin \left( x + C \right)\]

\[ \Rightarrow y = 2\sin \left( x + C \right)\]