Let  $y=y\left(x\right)$ be a solution curve of the differential equation $(1-{\mathrm{x}}^2{\mathrm{y}}^2)\mathrm{dx}=\mathrm{ydx}+\mathrm{xdy}$. If the line x = 1 intersects the curve $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ at $\mathrm{y}=2$ and the line $\mathrm{x}=2$ intersects the curve $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ at $\mathrm{y}=\mathrm{\alpha }$, then a value of $\mathrm{\alpha }$ is 
 

$$\begin{gathered} (1-{\left(\mathrm{xy}\right)}^2)\mathrm{dx}=\mathrm{ydx}+\mathrm{xdy} \\ \Rightarrow \int \mathrm{dx}=\int \frac{\mathrm{d}\left(\mathrm{xy}\right)}{1-{\left(\mathrm{xy}\right)}^2} \\ \Rightarrow \mathrm{x}=\frac{1}{2}.\ln \left|\frac{1+\mathrm{xy}}{1-\mathrm{xy}}\right|+\mathrm{c} \end{gathered}$$ i 

is passing through (1,2) .....(1) 

$1=\frac{1}{2}.\ln 3+\mathrm{c}$$\Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\boxed{\hspace{0.17em}\hspace{0.17em}\mathrm{c}=1-\frac{1}{2}\ln 3}$

substitute $\mathrm{x}=2,\mathrm{y}=\mathrm{\alpha }$ in (1) 
$2=\frac{1}{2}\ln \left|\frac{1+2\mathrm{\alpha }}{2\mathrm{\alpha }-1}\right|+1-\frac{1}{2}\ln 3$
$$\begin{gathered} \Rightarrow 1=\frac{1}{2}\ln \frac{(2\mathrm{\alpha }+1)}{3(2\mathrm{\alpha }-1)} \\ \Rightarrow 3{\mathrm{e}}^2=\frac{2\mathrm{\alpha }+1}{2\mathrm{\alpha }-1} \\ \Rightarrow \frac{3{\mathrm{e}}^2+1}{3{\mathrm{e}}^2-1}=\frac{4\mathrm{\alpha }}{2} \\ \Rightarrow \frac{1}{2}\left(\frac{3{\mathrm{e}}^2+1}{3{\mathrm{e}}^2-1}\right)=\mathrm{\alpha } \end{gathered}$$