JEE Mains · Maths · STD 12 - 9. differential equations
Let the solution curve \(y=f(x)\) of the differential equation \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x \in(-1,1)\) pass through the origin. Then \(\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f ( x ) dx\) is equal to
- A \(\frac{\pi}{3}-\frac{1}{4}\)
- B \(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
- C \(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)
- D \(\frac{\pi}{6}-\frac{\sqrt{3}}{2}\)
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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