Let $y(x)$ be a solution of $\frac{(2+\sin x}{(1+y)} \frac{d y)}{d x}=\cos x$. If $y(0)=2$, then…
Let $y(x)$ be a solution of $\frac{(2+\sin x}{(1+y)} \frac{d y)}{d x}=\cos x$. If $y(0)=2$, then $y\left(\frac{\pi}{2}\right)$ equals
$\frac{5}{2}$
2
$\frac{7}{2}$
3
Solution
Given differential equation is
$
\frac{(2+\sin x}{(1+y)} \cdot \frac{d y}{d x}=\cos x
$
which can be rewritten as
$
\frac{d y}{1+y}=\frac{\cos x}{2+\sin x} d x
$
Integrate both the sides, we get
$
\begin{aligned}
& \int \frac{d y}{1+y}=\int \frac{\cos x d x}{2+\sin x} \\
& \Rightarrow \log (1+y)=\log (2+\sin x)+\log C \\
& \Rightarrow 1+y=C(2+\sin x) \\
& \text { Given } y(0)=2 \\
& \Rightarrow 1+2=C[2+\sin 0] \Rightarrow C=\frac{3}{2}
\end{aligned}
$
Now, $y\left(\frac{\pi}{2}\right)$ can be found as
$
1+y=\frac{3}{2}\left(2+\sin \frac{\pi}{2}\right) \Rightarrow 1+y=\frac{9}{2}
$
$
\Rightarrow \quad y=\frac{7}{2}
$
Hence, $y\left(\frac{\pi}{2}\right)=\frac{7}{2}$