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 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

Let y = y(x) be the solution of the differential equation $$x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0$$, $$x > 1$$, with $$y(2) = - 2$$. Then y(3) is equal to

If the solution curve of the differential equation

$$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is