The slope of the tangent to a curve $$C: y=y(x)$$ at any point $$(x, y)$$ on it is $$\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}$$. If $$C$$ passes through the points $$\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$$ and $$\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$$, then $$\mathrm{e}^{\alpha}$$ is equal to :

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