The function y-fx is the solution of the differential equation dy dx - xy x 2 -1 - x 4 -2 - 1- x 2 in -1-1 satisfying f0- Then - - 3 2 - 3-2 f x dx is

The correct option is A π 3 √( 3 ) 2 Hence, Option (B) is the correct answer. ...

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Solving Linear Differential Equations of First Order

The function ...

Question

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

\frac{π}{3} - \frac{\sqrt{3}}{2}

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\frac{π}{3} - \frac{\sqrt{3}}{4}

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\frac{π}{6} - \frac{\sqrt{3}}{4}

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\frac{π}{6} - \frac{\sqrt{3}}{2}

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Similar questions

y = f (x)

\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt{1 - x^{2}}}

\int_{\frac{- \sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f (x) d (x)

y = f (x)

\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt{1 - x^{2}}}

(- 1, 1)

f (0) = 0

\int_{- \frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f (x) d x

y = f (x)

(1 - x^{2}) \frac{d y}{d x} - x y = 1 x \in (- 1, 1)

y (0) = 0

y (\frac{1}{2}) =

\frac{d y}{d x} + 2 y tan x = sin x : y = 0

x = \frac{π}{3}

(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x^{2}}; y = 0

x = 1

\frac{d y}{d x} - 3 y cot x = sin 2 x; y = 2

x = \frac{π}{2}

y = y (x)

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

y (2) = \frac{2}{\sqrt{3}}

y (x) = sec (s e c^{- 1} x - \frac{π}{6})

y (x)

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

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