Question

If $y\left(x\right)$ is a solution of the differential equation $\left(\frac{1+sinx}{1+y}\right)\frac{dy}{dx}=-\cos x$ and $y\left(0\right)=1$, then find the value of $y\left(\frac{\pi }{2}\right)$.

Answer 1

$\left(\frac{1+sinx}{1+y}\right)\frac{dy}{dx}=-\cos x$
$\frac{dy}{dx}=\frac{-(1+y)\cos x}{(1+sinx)}$
$\Rightarrow \frac{dy}{(1+y)}=\frac{-\cos x}{(1+sinx)}dx$
Integrating both sides,
Put $1+\sin x=t$
$\Rightarrow \cos xdx=dt$
$\log |1+y|=\int \frac{-dt}{t}$
$\Rightarrow \log |1+y|=-\log |1+\sin x|+\log C$
$\Rightarrow \log (1+y)(1+\sin x)=\log C$
$\Rightarrow C=(1+y)(1+\sin x)$
Given $y\left(0\right)=1$
$\Rightarrow C=2$
So, $2=(1+y)(1+\sin x)$
$\Rightarrow y\left(\frac{\pi }{2}\right)=0$