Question

Solve:

$\frac{dy}{dx}=\sqrt{4-y^2}(-2<y<2)$.

Answer 1

Given,

$\frac{dy}{dx}=\sqrt{4-y^2}(-2<y<2)$

or, $\frac{dy}{\sqrt{4-y^2}}=dx$

Now integrating we have,

${\sin }^{-1}\frac{y}{2}=x+c$

or, $y=\sin (x+c)$.