Question

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

Answer 1

Answer:

Step-by-step explanation:

Concept:

I have applied variable separable method to find solution the differential quation

Formula used:

$cosA=1-2sin^2\frac{A}{2}\\\\cosA=2cos^2\frac{A}{2}-1$

Now,

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

Integrating on both sides

$\int{dy}=\int{tan^2\frac{x}{2}}\:dx\\\\\int{dy}=\int{(sec^2\frac{x}{2}-1)}\:dx\\\\y=\frac{tan\frac{x}{2}}{\frac{1}{2}}-x+c\\\\y=2tan\frac{x}{2}-x+c$