Question

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

Answer 1

The given differential equation is:

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

Separating the variables, we get:

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

Now, integrating both sides of both sides, we get:

$∫\frac {dy}{\sqrt {4-y^2}} =∫dx$

$⇒sin^{-1}\frac y2=x+C$

$⇒\frac y2=sin(x+C)$

$⇒y=2sin(x+C)$

This is the required general solution of the given differential equation.