If the solution of the differential equation

$${{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}}$$ satisfies $$y(0) = 0$$, then the value of y(2) is _______________.

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

$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$ such that $$y(e) = {e \over 3}$$, then y(1) is equal to :

Let $$g:(0,\infty ) \to R$$ be a differentiable function such that

$$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $$, for all x > 0, where c is an arbitrary constant. Then :

Let $$y = y(x)$$ be the solution of the differential equation $$(x + 1)y' - y = {e^{3x}}{(x + 1)^2}$$, with $$y(0) = {1 \over 3}$$. Then, the point $$x = - {4 \over 3}$$ for the curve $$y = y(x)$$ is :