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

माना अवकल समीकरण \(\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}}, x \in(-1,1)\) का हल वक्र \(y = f ( x )\) मूल बिंदु से होकर जाता है \(\int \limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f ( x ) dx\) बराबर है :

  1. A \(\frac{\pi}{3}-\frac{1}{4}\)
  2. B \(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
  3. C \(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)
  4. D \(\frac{\pi}{6}-\frac{\sqrt{3}}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)

Step-by-step Solution

Detailed explanation

\(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) \(\text { I.F }=e^{\int \frac{x}{x^{2}-1} d x}\) \(\text { I.F }=\sqrt{1-x^{2}}\) Solution of \(D.E.\) \(y \cdot \sqrt{1-x^{2}}=\int \frac{x^{4}+2 x}{\sqrt{1-x^{2}}} \cdot \sqrt{1-x^{2}} d x\)…
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