The solution of primitive integral equation $$\left( {{x^2} + {y^2}} \right)dy = xy$$
$$dx$$ is $$y=y(x),$$ If $$y(1)=1$$ and $$\left( {{x_0}} \right) = e$$, then $${{x_0}}$$ is equal to

For the primitive integral equation $$ydx + {y^2}dy = x\,dy;$$
$$x \in R,\,\,y > 0,y = y\left( x \right),\,y\left( 1 \right) = 1,$$ then $$y(-3)$$ is

$$\int\limits_{ - 2}^0 {\left\{ {{x^3} + 3{x^2} + 3x + 3 + \left( {x + 1} \right)\cos \left( {x + 1} \right)} \right\}\,\,dx} $$ is equal to

$$a,\,b,\,c$$ are integers, not all simultaneously equal and $$\omega $$ is cube root of unity $$\left( {\omega \ne 1} \right),$$ then minimum value of $$\left| {a + b\omega + c{\omega ^2}} \right|$$ is