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

If $$y=y(x)$$ and it follows the relation $$x\cos \,y + y\,cos\,x = \pi $$ then $$y''(0)=$$

The area bounded by the parabola $$y = {\left( {x + 1} \right)^2}$$ and
$$y = {\left( {x - 1} \right)^2}$$ and the line $$y=1/4$$ is

$$a,\,b,\,c$$ 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