Question

Let a solution $y=y\left(x\right)$ of the differential equation $e^{y}dy-(2+\cos x)dx=0$ satisfy $y\left(0\right)=0$ then the value of $f\left(\frac{\pi }{2}\right)$ 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 $ln(2+\pi )$

Seperable ${\int }_0^y{e}^{y}dy={\int }_0^x(2+\cos x)dx$
$\Rightarrow e^y-1=2x+\sin x$
For $x=\frac{\pi }{2}$, we find $y=ln(2+\pi )$.