Let the solution curve of the differential equation $x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$, intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$. Then the value of $\alpha$ is :

<p>${{xdy - ydx} \over {\sqrt {{x^2} + {y^2}} }} = dx$</p> <p>$\Rightarrow {{dy} \over {dx}} = {{\sqrt {{x^2} + {y^2}} } \over x} + {y \over x}$</p> <p>$$ \Rightarrow {{dy} \over {dx}} = \sqrt {1 + {{{y^2}} \over {{x^2}}}} + {y \over x}$$</p> <p>Let ${y \over x} = v$</p> <p>$\Rightarrow v + x{{dv} \over {dx}} = \sqrt {1 + {v^2}} + v$</p> <p>$\Rightarrow {{dv} \over {\sqrt {1 + {v^2}} }} = {{dx} \over x}$</p> <p>OR $\ln \left( {v + \sqrt {1 + {v^2}} } \right) = \ln x + C$</p> <p>at $x = 1,\,y = 0$</p> <p>$\Rightarrow C = 0$</p> <p>${y \over x} + \sqrt {1 + {{{y^2}} \over {{x^2}}}} = x$</p> <p>At $x = 2$,</p> <p>${y \over 2} + \sqrt {1 + {{{y^2}} \over 4}} = 2$</p> <p>$\Rightarrow 1 + {{{y^2}} \over 4} = 4 + {{{y^2}} \over 4} - 2y$</p> <p>OR $y = {3 \over 2}$</p>