The solution of the differential equation

$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$

where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by :

If   f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x} $$   $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :

If a curve $$y=f(x)$$ passes through the point $$(1,-1)$$ and satisfies the differential equation, $$y(1+xy) dx=x$$ $$dy$$, then $$f\left( { - {1 \over 2}} \right)$$ is equal to :

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

$$\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).$$ Then $$y(e)$$ is equal to :