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

વિકલ સમીકરણ \(\left(1+y^2\right)\left(1+\log _e x\right) d x+x d y=0, x>0\) નો ઉકેલ વક્ર \(y=y(x)\) છે, જે બિંદુ \((1,1)\) માંથી પસાર થાય છે તથા \(y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}\) છે. તો \(\alpha+2 \beta =\) ...........

  1. A \(4\)
  2. B \(3\)
  3. C \(8\)
  4. D \(10\)
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Answer & Solution

Correct Answer

(B) \(3\)

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Detailed explanation

\(\int\left(\frac{1}{x}+\frac{\ln x}{x}\right) d x+\int \frac{d y}{1+y^2}=0 \) \( \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=C\) Put \(x=y=1\) \( \therefore C=\frac{\pi}{4} \) \( \Rightarrow \ln x+\frac{(\ln x)^2}{2}+\tan ^{-1} y=\frac{\pi}{4}\) Put \(\mathrm{x}=\mathrm{e}\)…
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