Which of the following is true for y(x) that satisfies the differential equation

$${{dy} \over {dx}}$$ = xy $$-$$ 1 + x $$-$$ y; y(0) = 0 :

If y = y(x) is the solution of the differential equation

$${{dy} \over {dx}}$$ + (tan x) y = sin x, $$0 \le x \le {\pi \over 3}$$, with y(0) = 0, then $$y\left( {{\pi \over 4}} \right)$$ equal to :

Let C₁ be the curve obtained by the solution of differential equation

$$2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0$$. Let the curve C₂ be the

solution of $${{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}}$$. If both the curves pass through (1, 1), then the area enclosed by the curves C₁ and C₂ is equal to :

If y = y(x) is the solution of the differential equation,

$${{dy} \over {dx}} + 2y\tan x = \sin x,y\left( {{\pi \over 3}} \right) = 0$$, then the maximum value of the function y(x) over R is equal to: