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

माना अवकल समीकरण \(\frac{d y}{d x}+\frac{1}{x^2-1} y =\left(\frac{ x -1}{ x +1}\right)^{\frac{1}{2}}, \quad x > 1\) का हल वक्र \(y = y ( x )\) बिंदु \(\left(2, \sqrt{\frac{1}{3}}\right)\) से होकर जाता है। तब \(\sqrt{7} y (8)\) बराबर है

  1. A \(11+6 \log _{ e } 3\)
  2. B \(19\)
  3. C \(12-2 \log _{ e } 3\)
  4. D \(19-6 \log _{ e } 3\)
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Answer & Solution

Correct Answer

(D) \(19-6 \log _{ e } 3\)

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Detailed explanation

\(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}\) \(\frac{d y}{d x}+P y=Q\) \(I \cdot F .= e ^{\int Pdx }=\left(\frac{ x -1}{ x +1}\right)^{\frac{1}{2}}\) \(y\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}=\int\left(\frac{x-1}{x+1}\right)^{1} d x\)…
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