MHT CET · Maths · Differential Equations
If \(y=y(x)\) and \(\frac{2+\sin x}{y+1}\left(\frac{d y}{d x}\right)=-\cos x, y(0)=1\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
- A \(\frac{2}{3}\)
- B \(\frac{1}{3}\)
- C \(-\frac{1}{3}\)
- D \(1\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x \\ & \Rightarrow \int \frac{d y}{y+1}=\int \frac{-\cos x d x}{2+\sin x} \\ & \Rightarrow \log _{\mathrm{e}}|y+1|=-\log _{\mathrm{e}}|2+\sin x|+\log _{\mathrm{e}} C \\ & \Rightarrow \log _e|y+1|+\log _e|2+\sin x|=\log _e C \\ & \Rightarrow \log _e|(y+1)(2+\sin x)|=\log _e C \\ & \Rightarrow(y+1)(2+\sin x)=C \\ & \text { Putting } x=0 \text { and } y=1 \text { we get } c=4 \\ & \Rightarrow(y+1)(2+\sin x)=4 \\ & \end{aligned}\)
Now putting \(x=\frac{\pi}{2}\) we get \(y=\frac{1}{3}\)
Now putting \(x=\frac{\pi}{2}\) we get \(y=\frac{1}{3}\)
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