Question

For the differential equation, find the general solution:

`dy/dx = sqrt(4-y^2)      (-2 < y < 2)`

Answer 1

We have,

`dy/dx = sqrt(4 - y^2)  (-2 < y < 2)`

⇒ `intdy/ sqrt(4 - y^2) = intdx`

On integrating

`int dy/ sqrt(2^2 - y^2) = int dx`

`sin^-1 (y/2)` = x + C

`y/2` = sin (x + C)

y = 2 sin (x + C)