Let the solution curve of the differential equation $xdy=\left(\sqrt{x^2+y^2}+y\right)dx, x>0,$

intersect the line $x=1 at y=0$ and the line $x=2 at y=\alpha$. Then the value of $\alpha$ is :

$$\begin{gathered} xdy=\left(\sqrt{x^2+y^2}+y\right)dx \\ \Rightarrow xdy-ydx=\sqrt{x^2+y^2}dx \\ \Rightarrow \frac{xdy-ydx}{x^2}=\left(\sqrt{1+\frac{y^2}{x^2}}\right)·\frac{dx}{x} \\ \frac{d\left({\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\right)}{\sqrt{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}}=\frac{dx}{x} \\ \Rightarrow In\left(\frac{y}{x}+\sqrt{{\left(\frac{y}{x}\right)}^2+1}\right)=In x+\log c \\ \Rightarrow \frac{y+\sqrt{y^2+x^2}}{x}=cx \\ y+\sqrt{y^2+x^2}=c{x}^2 \\ If x=1, y=0\Rightarrow 0+1=C\Rightarrow C=1 \\ Equation of the curve is y+\sqrt{x^2+y^2}=x^2 \\ substituting x=2, y=\alpha \\ \Rightarrow \alpha +\sqrt{4+{\alpha }^2}=4 \\ \Rightarrow 4+{\alpha }^2=16+{\alpha }^2-8\alpha \\ \therefore \alpha =\frac{3}{2} \end{gathered}$$