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).

Answer 1

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