Question

Let $y=y\left(x\right)$ be a solution of the differential equation $e^{y}dy-(2+\cos x)dx=0$ satisfying $y\left(0\right)=0.$ Then the value of $y\left(\frac{\pi }{2}\right)$ is equal to

• A. $\ln \pi$
• B. $\ln (2+\pi )$
• C. $\ln (1+\pi )$
• D. does not exist

Answer 1

The correct option is B $\ln (2+\pi )$

$e^{y}dy=(2+\cos x)dx$
Integration on both sides,
$\int e^{y}dy=\int (2+\cos x)dx$
$\Rightarrow e^y=2x+\sin x+C$
Since $y\left(0\right)=0,$
$\therefore 1=0+0+C$
$\Rightarrow C=1$
$\therefore e^y=2x+\sin x+1$
For $x=\frac{\pi }{2},$ we have $e^y=\pi +1+1$
$\Rightarrow y\left(\frac{\pi }{2}\right)=\ln (2+\pi )$