Let y = f(x) be the solution of the differential equation $$\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{6}+4x}{\sqrt{1-x^{2}}},-1 < x < 1$$ such that f(0)=0.If $$6\int_{-\frac{1}{2}}^{\frac{1}{2}}f(x)dx=2\pi - \alpha$$ then $$\alpha^{2}$$ is equal to ______.

The differential equation is $$\frac{dy}{dx} + \frac{x}{x^2-1} \cdot y = \frac{x^6+4x}{\sqrt{1-x^2}}$$, with $$f(0) = 0$$, for $$-1 < x < 1$$.

The integrating factor is:

$$\text{IF} = e^{\int \frac{x}{x^2-1}dx} = e^{\frac{1}{2}\ln(1-x^2)} = \sqrt{1-x^2}$$

Multiplying both sides by the IF:

$$\frac{d}{dx}\left[y\sqrt{1-x^2}\right] = \frac{x^6+4x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} = x^6 + 4x$$

Integrating:

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

Using $$f(0) = 0$$: $$0 = 0 + 0 + C$$, so $$C = 0$$.

Therefore $$f(x) = \frac{\frac{x^7}{7} + 2x^2}{\sqrt{1-x^2}}$$.

Now compute $$6\int_{-1/2}^{1/2} f(x)\,dx$$:

The function $$\frac{x^7/7}{\sqrt{1-x^2}}$$ is odd, so its integral over $$\left[-\frac{1}{2}, \frac{1}{2}\right]$$ is zero.

$$6\int_{-1/2}^{1/2} f(x)\,dx = 6 \int_{-1/2}^{1/2}\frac{2x^2}{\sqrt{1-x^2}}\,dx = 24\int_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}\,dx$$

Substituting $$x = \sin\theta$$, $$dx = \cos\theta\,d\theta$$:

$$= 24\int_0^{\pi/6}\frac{\sin^2\theta}{\cos\theta}\cos\theta\,d\theta = 24\int_0^{\pi/6}\sin^2\theta\,d\theta$$

$$= 24 \cdot \frac{1}{2}\int_0^{\pi/6}(1 - \cos 2\theta)\,d\theta = 12\left[\theta - \frac{\sin 2\theta}{2}\right]_0^{\pi/6}$$

$$= 12\left(\frac{\pi}{6} - \frac{\sin(\pi/3)}{2}\right) = 12\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) = 2\pi - 3\sqrt{3}$$

Since $$6\int_{-1/2}^{1/2} f(x)\,dx = 2\pi - \alpha$$, we get $$\alpha = 3\sqrt{3}$$.

Therefore $$\alpha^2 = 9 \times 3 = 27$$.