Question

Consider the differential equation
$\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}=-\cos x$
If $y(0)=1$, then evaluate $6 y\left(\frac{\pi}{2}\right)+5$

Answer 1

$\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}=-\cos x $
$\Leftrightarrow \frac{d y}{y+1}=\frac{-\cos x}{2+\sin x} d x$
Integrating, we get
$\log (y+1)=-\log (2+\sin x)-\log k$
$\Rightarrow k(y+1)(2+\sin x)=1$
$y(0)=1 $
$\Rightarrow k=\frac{1}{4}$
$\Rightarrow(y+1)(2+\sin x)=4$
At $x=\frac{\pi}{2}$
$(y+1)(3)=4$
$\Rightarrow y\left(\frac{\pi}{2}\right)=\frac{1}{3}$
$\Rightarrow 6 y\left(\frac{\pi}{2}\right)+5=7$