Question

If $z=z\left(x\right)$ and $(2+\cos x)\frac{dz}{dx}+\left(\sin x\right)\cdot z=\sin x$, $z\left(x\right)>0$ and $z\left(\frac{\pi }{2}\right)=3$, then $z\left(\frac{\pi }{3}\right)$ is

• A. $\frac{7}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{7}{2}$

$(2+\cos x)\frac{dz}{dx}+\left(\sin x\right)\cdot z=\sin x\Rightarrow (2+\cos x)\frac{dz}{dx}=-(z-1)\cdot \sin x\Rightarrow \frac{dz}{z-1}=\frac{-\sin x}{2+\cos x}dx$
Integrating
$\Rightarrow \ln |z-1|=\ln |2+\cos x|+C$
It is given $z\left(\frac{\pi }{2}\right)=3$
$\therefore \ln 2=\ln 2+C\Rightarrow C=0$
When $x=\frac{\pi }{3}$
$\ln |z-1|=\ln \frac{5}{2}\Rightarrow z=\frac{7}{2}$