The general solution of the differential equation $$\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$$ is :

Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$$, represents a circle with center ($$\alpha$$, $$\beta$$). Then, $$\alpha$$ + 2$$\beta$$ is equal to :

If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to

Let the solution curve of the differential equation

$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to: