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:
\( \dfrac{1}{2} \)
\( \dfrac{3}{2} \)
\( -\dfrac{3}{2} \)
\( \dfrac{5}{2} \)
