Question

For the differential equation find the general solution of
$\frac{dy}{dx}=\sqrt{4-y^2}(-2<y<2)$

Answer 1

Given differential equation is

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

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

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

Integrating both sides
We get,

$\int \frac{dy}{\sqrt{2^2-y^2}}=\int dx$

As we know that,

$\int \frac{dx}{\sqrt{a^2-x^2}}$

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

$\frac{y}{2}=\sin (x+c)$

$y=2\sin (x+c)$

Final Answer:
Hence, the required general solution is

$y=2\sin (x+c)$