The function y = f (x) is the solution of differential equation \frac{dy}{dx} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } \frac{xy}{x^{2} - 1} \textrm{ }\textrm{ } = \textrm{ }\textrm{ } \frac{x^{4} + 2 x}{\sqrt{1 - x^{2}}} in (–1, 1) satisfying f (0) = 0. Then \int_{- \frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f \left(\right. x \left.\right) dx is

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Mathematics

Linear Differential Equation

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The function y = f (x) is the solution of differential equation $\frac{dy}{dx} + \frac{xy}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt{1 - x^{2}}}$ in (–1, 1) satisfying f (0) = 0. Then $\int_{- \frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f (x) dx$ is

$\frac{π}{3} - \frac{\sqrt{3}}{4}$

$\frac{π}{6} - \frac{\sqrt{3}}{2}$

$\frac{π}{3} - \frac{\sqrt{3}}{2}$

$\frac{π}{6} - \frac{\sqrt{3}}{4}$

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$\frac{dy}{dx} - \frac{x}{1 - x^{2}} y = \frac{x^{4} + 2 x}{\sqrt{1 - x^{2}}}$

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

$y \sqrt{1 - x^{2}} = \int (x^{4} + 2 x) dx = \frac{x^{5}}{5} + x^{2} + C$

x = 0, y = 0

$∴$ C = 0

$\frac{\frac{x^{5}}{5} + x^{2}}{\sqrt{1 - x^{2}}}$

$∴ \int_{\frac{- \sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{\frac{x^{5}}{5} + x^{2}}{\sqrt{1 - x^{2}}} dx = 0 + 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1 - x^{2}}} dx; put x = sinθ$

$∴ I = \frac{π}{3} - \frac{\sqrt{3}}{4}$