JEE Mains · Maths · STD 12 - 9. differential equations
Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be the solution of the differential equation, \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1,\) If \(y(\pi)=a\) and \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at \(\mathrm{x}=\pi\) is \(b\), then the ordered pair \((a, b)\) is equal to :
- A \((2,1)\)
- B \(\left(2, \frac{3}{2}\right)\)
- C \((1,-1)\)
- D \((1,1)\)
Answer & Solution
Correct Answer
(D) \((1,1)\)
Step-by-step Solution
Detailed explanation
\(\frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x, y>0\) \(\Rightarrow \frac{d y}{y+1}=\frac{-\cos x}{2+\sin x} d x\) By integrating both sides : \(\ell \mathrm{n}|\mathrm{y}+1|=-\ell \mathrm{n}|2+\sin \mathrm{x}|+\ell \mathrm{n} \mathrm{K}\)…
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