If $y=y(x)$ is the solution of the differential equation $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d…
If $y=y(x)$ is the solution of the differential equation $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0 \quad$ with $y(0)=1$, then $y\left(\frac{\pi}{2}\right)$ is equal to
$\frac{1}{3}$
$\frac{2}{3}$
1
$\frac{4}{3}$
Solution
Given differential equation can be rewritten as
$\frac{d y}{1+y}+\frac{\cos x}{2+\sin x} d x=0$
Put $2+\sin x=t \Rightarrow(\cos x) d x=d t$
$\therefore \quad \frac{d y}{1+y}+\frac{1}{t} d t=0$
On integrating, we get
$\begin{array}{rlrl}
\log (1+y)+\log t =\log C \\
\Rightarrow (1+y) t =C \\
\Rightarrow \quad (1+y)(2+\sin x) =C
\end{array}$
when $y(0)=1$
$\begin{array}{ll} & (1+1)(2+0)=C \Rightarrow C=4 \\ \therefore \quad & (1+y)(2+\sin x)=4 \\ \text { At } x=\frac{\pi}{2}\end{array}$
$\begin{array}{rlrl} {\left[1+y\left(\frac{\pi}{2}\right)\right]\left[2+\sin \frac{\pi}{2}\right]=4} \\ \Rightarrow 1+y\left(\frac{\pi}{2}\right) =\frac{4}{3} \\ \Rightarrow y\left(\frac{\pi}{2}\right) =\frac{1}{3}\end{array}$