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

અહી વક્ર \(y=f(x)\) એ વિકલ સમીકરણ \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x \in(-1,1)\) નો ઉકેલ છે કે જે ઉગમબિંદુમાંથી પસાર થાય છે. તો \(\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f ( x ) dx\) ની કિમંત મેળવો.

  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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