The differential equation $${{dy} \over {dx}} = {{\sqrt {1 - {y^2}} } \over y}$$ determines a family of circles with

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

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

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