Question

If $y=y\left(x\right)$ is the solution of differential equation $\frac{2+\sin x}{y+1}\left(\frac{dy}{dx}\right)=-\cos x,y\left(0\right)=1,$ then $y\left(\frac{\pi }{2}\right)$ equals

• A. $\frac{-7}{3}$
• B. $\frac{7}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A), (C)

$\frac{1}{3}$

The given differential equation can be rewritten as:
$\frac{1}{y+1}dy=-\frac{\cos x}{2+\sin x}dx$
Integrating both sides, we get
$\int \frac{1}{y+1}dy=-\int \frac{\cos x}{2+\sin x}dx$
$\Rightarrow \ln |y+1|+\ln |c|+\ln (2+\sin x)=0$
$($Here, $c$ is the constant of integration$)$
$\Rightarrow |c(y+1)|(2+\sin x)=1\cdots \left(i\right)$
when $x=0,y=1$
$4c=\pm 1\Rightarrow c=\pm \frac{1}{4}$
From $\left(i\right)$
$|y+1|(2+\sin x)=4$
Now put $x=\frac{\pi }{2}$
$\Rightarrow |y+1|3=4$
$\therefore y=\frac{1}{3}$ or $y=\frac{-7}{3}$