Question

For the differential equation find the general solution of
$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

Answer 1

Given differential equation is

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

using $1+\cos x$

$=2{\cos }^2\frac{x}{2}\text{and}1-\cos x=2{\sin }^2\frac{x}{2}$

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

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

Putting

${\tan }^2\frac{x}{2}={\sec }^2\frac{x}{2}-1$

then we get,

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

Integrating both sides
We get,

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

$y=\frac{1}{\frac{1}{2}}\tan \frac{x}{2}-x+c$

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

Final Answer:
Hence, the required general solution is

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