Let $$y=y(x)$$ be the solution curve of the differential equation $$ \frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}$$, $$x >1$$ passing through the point $$\left(2, \sqrt{\frac{1}{3}}\right)$$. Then $$\sqrt{7}\, y(8)$$ is equal to :

The differential equation of the family of circles passing through the points $$(0,2)$$ and $$(0,-2)$$ is :

Let the solution curve of the differential equation $$x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :

If $$y=y(x), x \in(0, \pi / 2)$$ be the solution curve of the differential equation

$$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$,

with $$y(\pi / 4)=\mathrm{e}^{-\pi}$$, then $$y(\pi / 6)$$ is equal to :