Question

The solution of the differential equation$\frac{dy}{dx}=\sin (x+y)\tan (x+y)-1$

• A. $\csc (x+y)+\tan (x+y)=x+c$
• B. $x+\csc (x+y)=c$
• C. $x+\tan (x+y)=c$
• D. $x+\sec (x+y)=c$

Answer 1

The correct option is B $x+\csc (x+y)=c$

Given, $\frac{dy}{dx}=sin(x+y)tan(x+y)-1$

Put $(x+y=z)$ $\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}$

$\therefore \frac{dz}{dx}-1=sinztanz-1$

$\Rightarrow \frac{cosz}{si{n}^{2}z}dz=dx$

Put $\sin z=t\Rightarrow \cos zdz=dt$

$\therefore \frac{1}{t^2}dt=x-c\Rightarrow -\frac{1}{t}=x-c$

$\Rightarrow -\csc z=x-c$

$\Rightarrow x+\csc x+y=c$