Let us consider a curve, y = f(x) passing through the point ($$-$$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x². Then :

Let y(x) be the solution of the differential equation 2x² dy + (e^y $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :

Let y = y(x) be a solution curve of the differential equation $$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$, $$x \in \left( {0,{\pi \over 2}} \right)$$. If $$\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1$$, then the value of $$y\left( {{\pi \over 4}} \right)$$ is :

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

equation (x $$-$$ x³)dy = (y + yx² $$-$$ 3x⁴)dx, x > 2. If y(3) = 3, then y(4) is equal to :