JEE Mains · Maths · STD 12 - 9. differential equations
If \(\left( {2 + \sin x} \right)\frac{{dy}}{{dx}} + \left( {y + 1} \right)\cos x = 0\) and \(y\left( 0 \right) = 1\) then \(y\left( {\frac{\pi }{2}} \right) = \) . . . .
- A \(\frac{4}{3}\)
- B \(\frac{1}{3}\)
- C \( - \frac{2}{3}\)
- D \( - \frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
We have \((2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0\) \(\Rightarrow \frac{d}{d x}(2+\sin x)(y+1)=0\) On integrating, we get \((2+\sin x)(y+1)=C\) At \(x=0, y=1\) we have \((2+\sin 0)(1+1)=\mathrm{C}\) \(\Rightarrow \mathrm{C}=4\)…
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