Let y = f(x) be the solution of the differential equation , – 1 < x < 1 such that f(0) = 0. If , then is equal to __________. [2025]

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Solution

Q.
Let y = f(x) be the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{6} + 4 x}{\sqrt{1 - x^{2}}}$ , – 1 < x < 1 such that f(0) = 0. If $6 \int_{- 1 / 2}^{1 / 2} f (x) d x = 2 π - α$ , then $α^{2}$ is equal to __________. [2025]

Ans.
(27)

From given differential equation, we have

$I . F . = e^{- \frac{1}{2} \int \frac{2 x}{1 - x^{2}} d x} = e^{\frac{1}{2} \ln (1 - x^{2})} = \sqrt{1 - x^{2}}$

Hence, solution of given D.E. is

$y \times \sqrt{1 - x^{2}} = \int (x^{6} + 4 x) d x = \frac{x^{7}}{7} + 2 x^{2} + c$

Given, y(0) = 0, then c = 0

$∴ y = \frac{\frac{x^{7}}{7} + 2 x^{2}}{\sqrt{1 - x^{2}}}$

Let $I = 6 \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^{7}}{7} + 2 x^{2}}{\sqrt{1 - x^{2}}} d x$ ... (i)

$\Rightarrow I = 6 \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{\frac{- x^{7}}{7} + 2 x^{2}}{\sqrt{1 - x^{2}}} d x$ ... (ii)

Adding (i) & (ii), we get

$2 I = 24 \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x = 2 \times 24 \int_{0}^{\frac{1}{2}} \frac{x^{2}}{\sqrt{1 - x^{2}}} d x$

Put $x = \sin θ \Rightarrow d x = \cos θ d θ$

$∴ I = 24 \int_{0}^{\frac{π}{6}} \frac{\sin^{2} θ}{\cos θ} \cos θ d θ$

$= 24 \int_{0}^{\frac{π}{6}} (\frac{1 - \cos 2 θ}{2}) d θ = 12 {[θ - \frac{\sin 2 θ}{2}]}_{0}^{\frac{π}{6}}$

$= 12 (\frac{π}{6} - \frac{\sqrt{3}}{4}) = 2 π - 3 \sqrt{3}$

On comparing, we get $α^{2} = {(3 \sqrt{3})}^{2} = 27$ .

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