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JEE Mains · Maths · STD 12 - 9. differential equations

माना अवकल समीकरण \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1\) का हल \(y=y(x)\) है। यदि \(y(\pi)=a\) और \(x=\pi\) पर \(\frac{d y}{d x}\) का मान \(b\) है तो क्रमित युग्म \((a, b)\) बराबर है

  1. A \((2,1)\)
  2. B \(\left(2, \frac{3}{2}\right)\)
  3. C \((1,-1)\)
  4. D \((1,1)\)
Verified Solution

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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