Question

For the differential equations, find the general solution:$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

Answer 1

The given differential equation is:

$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

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

$⇒\frac{dy}{dx}=(sec^2\frac{x}{2}-1)$

separating the variables,we get:

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

Now,integrating both sides of this equation,we get:

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

$⇒y=2tan\frac{x}{2}-x+C$

This is the required general solution of the given differential equation.