Question

The general solution of the differential equation $\frac{dy}{dx}+\frac{1+\cos 2y}{1-\cos 2x}=0$ is given by:

• A. $\tan y+\cot x=c$
• B. $\tan y-\cot x=c$
• C. $\tan x-\cot y=c$
• D. $\tan x+\cot x=c$

Answer 1

Answer: (B)

$\tan y-\cot x=c$

$\frac{dy}{dx}+\frac{1+\cos 2y}{1-\cos 2x}=0$

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

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

$\int {\sec }^{2}ydy=-\int {\csc }^{2}xdx$

$\tan y+c=\cot x+C$

$\tan y-\cot x=c$