Question

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

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

Answer 1

Given, $\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$
On separating the variables and integrating, we get
$\int dy=\int \frac{1-cosx}{1+cosx}dx\Rightarrow y=\int \frac{2si{n}^2\left(x/2\right)}{2co{s}^2\left(x/2\right)}dx[\because 1-cosx=2si{n}^2(x/2)and1+cosx=2co{s}^2(x/2\left)\right]\Rightarrow y=\int ta{n}^2\left(x/2\right)dx(\because 1+ta{n}^2\frac{x}{2}=se{c}^2\frac{x}{2})\Rightarrow y=\int \left(se{c}^2\right(x/2)-1)dx\Rightarrow y=\int se{c}^2\left(x/2\right)dx-\int 1dx\Rightarrow y=2tan\frac{x}{2}-x+C.(\because \int se{c}^{2}xdx=tanx)$
which is the required general solution.