If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x satisfying y(x) = e is :

The differential equation of the family of curves, x² = 4b(y + b), b $$ \in $$ R, is :

Let y = y(x) be a solution of the differential equation,

$$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1.

If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to :

Let y = y(x) be the solution curve of the differential equation,

$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :