Question

For each of the differential equations, find the general solution:
1.$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$
2.$\frac{dy}{dx}=\sqrt{4-y^2}(-2<y<2)$

Answer 1

$\left(1\right)$

$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

$\frac{dy}{dx}=\frac{1-1+2{\cos }^2\frac{x}{2}}{1+2{\sin }^2\frac{x}{2}-1}$

$\frac{dy}{dx}={\cot }^2\frac{x}{2}$

$dy={\cot }^2\frac{x}{2}dx$

$\int dy=\int ({\csc }^2\frac{x}{2}-1)dx$

$y=-2\cot \frac{x}{2}-x+c$

$\left(2\right)$

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

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

${\sin }^{-1}\left(\frac{y}{2}\right)=x+c$