MHT CET · Maths · Differential Equations
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
- A \(\frac{-2}{3}\)
- B \(\frac{-1}{3}\)
- C \(\frac{4}{3}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\( (2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0 \)
\( \therefore \frac{1}{y+1} \mathrm{~d} y=\frac{-\cos x}{2+\sin x} \mathrm{~d} x \)
\( \therefore \text { Integrating both sides, we get } \)
\( \log (y+1)=-\log (2+\sin x)+\mathrm{c} \ldots \text { (i) } \)
\( \text { when } x=0, y=1 \)
\( \Rightarrow \mathrm{c}=2 \log 2 \)
\( \therefore (\mathrm{i}) \Rightarrow \log (y+1)=-\log (2+\sin x)+\log 4 \)
\( \Rightarrow \log (y+1)=\log \left(\frac{4}{2+\sin x}\right) \)
\( \Rightarrow y+1=\frac{4}{2+\sin x} \)
\(\text { When } x=\frac{\pi}{2}, \text { (ii) } \Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\)
\( \therefore \frac{1}{y+1} \mathrm{~d} y=\frac{-\cos x}{2+\sin x} \mathrm{~d} x \)
\( \therefore \text { Integrating both sides, we get } \)
\( \log (y+1)=-\log (2+\sin x)+\mathrm{c} \ldots \text { (i) } \)
\( \text { when } x=0, y=1 \)
\( \Rightarrow \mathrm{c}=2 \log 2 \)
\( \therefore (\mathrm{i}) \Rightarrow \log (y+1)=-\log (2+\sin x)+\log 4 \)
\( \Rightarrow \log (y+1)=\log \left(\frac{4}{2+\sin x}\right) \)
\( \Rightarrow y+1=\frac{4}{2+\sin x} \)
\(\text { When } x=\frac{\pi}{2}, \text { (ii) } \Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\)
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