Discussion Forum

If $(2+sin x)\frac{\text{d}y}{\text{d}x}+(y+1)\cos x=0$ and y(0)=1, then $y(\frac{\pi}{2})$ is equal to 

Answer:

Explanation:

We have 

   $(2+sin x)\frac{\text{d}y}{\text{d}x}+(y+1)\cos x=0$

$\Rightarrow  \frac{\text{d}y}{\text{d}x}+\frac{cosx}{2+sinx}y=\frac{-cosx}{2+sinx}$

which is a linear differential equation

$\therefore$     IF= $e^{\int_{}^{} \frac{cosx}{2+sin x}dx}=e^{\log(2+sin x)}$

                      = 2+ sin x

$\therefore$      Required solution is given by

               y+(2+sinx)  = $\int_{}^{} \frac{-cos x}{2+sin x}.(2+sin x)dx +C$

  $\Rightarrow    y(2+sin x)=-\sin x+C$

                            Also y(0)  =1

$\therefore$     $1(2+\sin 0)=-\sin 0 +C$

     $\Rightarrow$  C=2

$\therefore$                             $y= \frac{2-\sin x}{2+\sin x}$

$\Rightarrow$          $y(\frac{\pi}{2})= \frac{2-\sin \frac{\pi}{2}}{2+\sin \frac{\pi}{2}}=\frac{1}{3}$