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 :

The minimum value of the twice differentiable function $$f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$$, $$x \in \mathbf{R}$$, is :

Let $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ be two distinct solutions of the differential equation $$\frac{d y}{d x}=x+y$$, with $$y_{1}(0)=0$$ and $$y_{2}(0)=1$$ respectively. Then, the number of points of intersection of $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ is