Question

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

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

Answer 1

The correct option is A $\frac{1}{3}$

$(2+\sin x)\frac{dy}{dx}=-(y+1)\cos x$

$\Rightarrow \frac{-dy}{y+1}=\frac{\cos x}{2+\sin x}dx$

Integrate on both sides. In RHS, take $2+\sin x=t\Rightarrow \cos xdx=dt$

$\Rightarrow \int -\frac{dy}{y+1}=\int \frac{dt}{t}$

$\Rightarrow -\ln (y+1)=\ln \left(t\right)+c=\ln (2+\sin x)+c$

It is given that $y\left(0\right)=1$, i.e., for $x=0,y=1$

$\Rightarrow -\ln (1+1)=\ln (2+0)+c$

$\Rightarrow c=-\ln 4$

Therefore, the solution of the differential equation is:

$-\ln (y+1)=\ln (2+\sin x)-\ln 4$

For $x=\frac{\pi }{2}$,

$-\ln (y+1)=\ln (2+1)-\ln 4$

$\Rightarrow -\ln (y+1)=\ln \frac{3}{4}$

$\Rightarrow y+1=\frac{4}{3}$

$\Rightarrow y\left(\frac{\pi }{2}\right)=\frac{1}{3}$