Question

The solution of $\frac{dy}{dx}=\sin (x+y)+\cos (x+y)$ is :

• A. $log[1+\tan \left(\frac{x+y}{2}\right)]+c=0$
• B. $log[1+\tan \left(\frac{x+y}{2}\right)]=x+c$
• C. $log[1-\tan \left(\frac{x+y}{2}\right)]=x+c$
• D. None of these

Answer 1

Answer: (C)

$log[1-\tan \left(\frac{x+y}{2}\right)]=x+c$

$Putx+y=v1+dy/dx=dv/dxdy/dx=dv/dx–1\frac{dy}{dx}=sin(x+y)+cos(x+y)\frac{dv}{dx}-1=sin\left(v\right)+cos\left(v\right)Usingvariableandseparablemethod\frac{dv}{1+cosv+sinv}=dxIntegratebothsides\int \frac{dv}{1+cosv+sinv}=\int dx\int \frac{dv}{1+\frac{1-ta{n}^2\frac{v}{2}}{1+ta{n}^2\frac{v}{2}}+\frac{2tan\frac{v}{2}}{1+ta{n}^2\frac{v}{2}}}=\int dx\int \frac{se{c}^2\frac{v}{2}dv}{2(1+tan\frac{v}{2})}=\int dxln(1+tan\frac{v}{2})=x+cln(1+tan\frac{x+y}{2})=x+c$

Hence, the option $C$ is the correct answer.