Definite Integral of Solution to Linear First Order Differential Equation

To find the value of the definite integral, we first solve the given first-order linear differential equation:

$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}, \quad x \in (-1, 1)$

We can rewrite the equation in standard form $\frac{dy}{dx} + P(x)y = Q(x)$, where: $P(x) = \frac{x}{x^2 - 1} = -\frac{x}{1 - x^2}$ $Q(x) = \frac{x^4 + 2x}{\sqrt{1 - x^2}}$

The integrating factor ($\text{I.F.}$) is given by: $\text{I.F.} = e^{\int P(x) \, dx} = e^{-\int \frac{x}{1 - x^2} \, dx} = e^{\frac{1}{2}\ln(1 - x^2)} = \sqrt{1 - x^2}$

Multiplying the differential equation by the integrating factor: $\frac{d}{dx}\left(y \sqrt{1 - x^2}\right) = \sqrt{1 - x^2} \cdot \frac{x^4 + 2x}{\sqrt{1 - x^2}} = x^4 + 2x$

Integrating both sides with respect to $x$: $y \sqrt{1 - x^2} = \int (x^4 + 2x) \, dx = \frac{x^5}{5} + x^2 + C$

Using the initial condition $f(0) = 0$: $0 \cdot \sqrt{1 - 0} = \frac{0^5}{5} + 0^2 + C \implies C = 0$

Thus, the function $f(x)$ is: $f(x) = \frac{x^5}{5\sqrt{1 - x^2}} + \frac{x^2}{\sqrt{1 - x^2}}$

Now, we evaluate the definite integral: $I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \, dx = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \left( \frac{x^5}{5\sqrt{1 - x^2}} + \frac{x^2}{\sqrt{1 - x^2}} \right) dx$

Since $\frac{x^5}{5\sqrt{1 - x^2}}$ is an odd function, its integral over the symmetric interval $\left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right]$ is zero: $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^5}{5\sqrt{1 - x^2}} \, dx = 0$

Since $\frac{x^2}{\sqrt{1 - x^2}}$ is an even function: $I = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1 - x^2}} \, dx$

Using the trigonometric substitution $x = \sin\theta$, with $dx = \cos\theta \, d\theta$:

• When $x = 0$, $\theta = 0$
• When $x = \frac{\sqrt{3}}{2}$, $\theta = \frac{\pi}{3}$

Substituting these into the integral: $I = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2\theta}{\cos\theta} \cos\theta \, d\theta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2\theta \, d\theta$ $I = \int_{0}^{\frac{\pi}{3}} (1 - \cos 2\theta) \, d\theta$ $I = \left[ \theta - \frac{\sin 2\theta}{2} \right]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\sin\left(\frac{2\pi}{3}\right)}{2} = \frac{\pi}{3} - \frac{\frac{\sqrt{3}}{2}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$

Therefore, the correct option is (B).