Question

If y(x) is a solution of the different equation $\left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x$ and y(0) = 1, then find the value of y(π/2). [CBSE 2014, NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \left(\frac{2+\sin x}{1+y}\right)\frac{dy}{dx}=-\cos x \\ \Rightarrow \frac{1}{1+y}dy=\frac{-\cos x}{2+\sin x}dx \\ \Rightarrow \int \frac{1}{1+y}dy=-\int \frac{\cos x}{2+\sin x}dx \\ \Rightarrow \log \left|1+y\right|=-\log \left|2+\sin x\right|+\log C \\ \Rightarrow \log \left|\left(1+y\right)\left(2+\sin x\right)\right|=\log C \\ \Rightarrow \left(1+y\right)\left(2+\sin x\right)=C.....\left(1\right) \end{gathered}$$
Now, y(0) = 1
$$\begin{gathered} \therefore \left(1+1\right)\left(2+0\right)=C \\ \Rightarrow C=4 \end{gathered}$$
Substituting the value of C in (1), we get
(1 + y)(2 + sinx) = 4
$$\begin{gathered} \Rightarrow 1+y=\frac{4}{2+\sin x} \\ \Rightarrow y=\frac{4}{2+\sin x}-1 \\ \Rightarrow y\left(\frac{\mathrm{\pi }}{2}\right)=\frac{4}{2+\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}-1 \\ =\frac{4}{3}-1 \\ =\frac{1}{3} \end{gathered}$$