Question

For the differential equation in given question find the general solution.​​​​

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

Answer 1

Given, $\frac{dy}{dx}=\sqrt{4-y^2}$
On separating the variables, we get $dx=\frac{dy}{\sqrt{4-y^2}}$
On integrating, we get
$\int \frac{dy}{\sqrt{2^2-y^2}}=\int dx\Rightarrow si{n}^{-1}\frac{y}{2}=x+C(\because \int \frac{1}{\sqrt{a^2-x^2}}dx=si{n}^{-1}\frac{x}{a})$
$\Rightarrow \frac{y}{2}=sin(x+C)\Rightarrow y=2sin(x+C)$ which is the required general solution.