Let the solution curve of the differential equation

$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to:

Let x = x(y) be the solution of the differential equation

$$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$$ such that x(1) = 0. Then, x(e) is equal to :

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $$\tan x(\cos x - y)$$. If the curve passes through the point $$\left( {{\pi \over 4},0} \right)$$, then the value of $$\int\limits_0^{\pi /2} {y\,dx} $$ is equal to :

Let the solution curve $$y = y(x)$$ of the differential equation

$$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$$

pass through the points (1, 0) and (2$$\alpha$$, $$\alpha$$), $$\alpha$$ > 0. Then $$\alpha$$ is equal to