Question

The solution of the differential equation $\frac{dy}{dx}={\cot }^2(x+y)$ is

• A. $y=x+1/2\sin 2(x+y)+c$
• B. $y=x-1/2\sin 2(x+y)+c$
• C. $y=x+1/2\cos 2(x+y)+c$
• D. $Noneofthese$

Answer 1

The correct option is A $y=x+1/2\sin 2(x+y)+c$

Given,

$\frac{dy}{dx}=co{t}^2(x+y)$

Let x+y=t

$\frac{dy}{dx}+1=\frac{dt}{dx}$

$\frac{dy}{dx}=\frac{dt}{dx}-1$

$\frac{dt}{dx}-1=co{t}^{2}t$

$\frac{dt}{1+co{t}^{2}t}=dx$

Integrating Both the sides

$\int si{n}^{2}tdt=\int dx$

$\to \int \frac{(1-cos2t)}{2}dt=\int dx$

$\frac{t}{2}-\frac{sin2t}{4}=x+c$

$\frac{x}{2}+\frac{y}{2}-\frac{sin2(x+y)}{4}=x+c$

$y=x+\frac{1}{2}sin2(x+y)+c$