Let $$y=y(x)$$ be a solution curve of the differential equation.

$$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$$.

If the line $$x=1$$ intersects the curve $$y=y(x)$$ at $$y=2$$ and the line $$x=2$$ intersects the curve $$y=y(x)$$ at $$y=\alpha$$, then a value of $$\alpha$$ is :

Let $$f$$ be a differentiable function such that $${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt} $$, $$f(1) = {2 \over 3}$$. Then $$18f(3)$$ is equal to :

If the solution curve $$f(x, y)=0$$ of the differential equation

$$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$$,

passes through the points $$(1,0)$$ and $$(\alpha, 2)$$, then $$\alpha^{\alpha}$$ is equal to :

Let $$\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$$ be the solution of the differential equation $$2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2}$$. Then $$\alpha+\beta-\gamma$$ equals :