JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=f(x)\) be the solution of the differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x y}{x^2-1}=\frac{x^6+4 x}{\sqrt{1-x^2}},-1 \lt x \lt 1\) such that \(f(0)=0\). If \(6 \int_{-1 / 2}^{1 / 2} f(x) \mathrm{d} x=2 \pi-\alpha\) then \(\alpha^2\) is equal to _______ .
- A 25
- B 26
- C 27
- D 28
Answer & Solution
Correct Answer
(C) 27
Step-by-step Solution
Detailed explanation
I.F. \(\mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ln \left(1-\mathrm{x}^2\right)}=\sqrt{1-\mathrm{x}^2}\) \(y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c\) Given \(y(0)=0 \Rightarrow c=0\)…
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