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

Question

Aicharya · Accepted Answer

$\begin{array}{ll} & (2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0 \ 	herefore \quad & \frac{1}{y+1} \mathrm{~d} y=\frac{-\cos x}{2+\sin x} \mathrm{~d} x \ & 	ext { Integrating on both sides, we get } \ & \log (y+1)=-\log (2+\sin x)+\mathrm{c} \ & 	ext { When } x=0, y=1 \ & \Rightarrow \mathrm{c}=2 \log 2=\log 4 \ 	herefore \quad & 	ext { (i) } \Rightarrow \log (y+1)=-\log (2+\sin x)+\log 4 \ & \Rightarrow \log (y+1)=\log \left(\frac{4}{2+\sin x}ight) \end{array}$ $\begin{aligned} & \Rightarrow y+1=\frac{4}{2+\sin x} \ & \Rightarrow y=\frac{4}{2+\sin x}-1 \ 	herefore \quad & y\left(\frac{\pi}{2}ight)=\frac{4}{3}-1=\frac{1}{3}\end{aligned}$

If $(2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0$ and $y(0)=1$, then…

Solution