$$ \begin{aligned} & \text { I.F. } e^{-\frac{1}{2} \int \frac{2 x}{1-x^2} d x}=e^{-\frac{1}{2} \ln \left(1-x^2\right)}=\sqrt{1-x^2} \\ & y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c \end{aligned} $$ Given $y(0)=0 \Rightarrow c=0$ $$ \begin{aligned} & y=\frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}} \\ & \text { Now, } 6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x^7}{\frac{7}{\sqrt{1-x^2}}+2 x^2} d x=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 x^2}{\sqrt{1-x^2}} d x \\ & =24 \int_0^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} \mathrm{dx} \end{aligned} $$ Put $\mathrm{x}=\sin \theta$ $$ \begin{aligned} & \mathrm{dx}=\cos \theta \mathrm{d} \theta \\ & =24 \int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta}{\cos \theta} \cos \theta \mathrm{~d} \theta \\ & =24 \int_0^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) \mathrm{d} \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{6}} \\ & =12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\ & =2 \pi-3 \sqrt{3} \\ & \alpha^2=(3 \sqrt{3})^2=27 \end{aligned} $$