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.