Question

If $(2+\sin x)\frac{dy}{dx}+(y+1)\cos x=0$ and $y(0)=1,$ then $y\left(\frac{\pi }{2}\right)$ is equal to :

• A. $-\frac{2}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{4}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (D)

$\left(2+\sin x\right)\frac{dy}{dx}+\left(y+1\right)\cos x=0$
$y(0)=1,y\left(\frac{\pi }{2}\right)$ = ?
$\frac{1}{y+1}dy+\frac{\cos x}{2+\sin x}dx=0$
$In\left|y+1\right|+In\left(2+\sin x\right)=InC$
$(y+1)(2+\sin x)=C$
Put $x=0,y=1$
$(1+1)\cdot 2=C\Rightarrow C=4$
Now, $(y+1)(2+\sin x)=4$
For, $x=\frac{\pi }{2}$
$(y+1)(2+1)=4$
$y+1=\frac{4}{3}$
$y=\frac{4}{3}-1=\frac{1}{3}$