Given differential equation

$x\,dy=\left(\sqrt{x^2+y^2}+y\right)dx, \qquad x>0$

So, $\frac{dy}{dx}=\frac{\sqrt{x^2+y^2}+y}{x}$

Use the substitution

Since the expression contains $\sqrt{x^2+y^2}$, divide by $x$ and set $y=vx$ where $v=v(x)$.

Then $\frac{dy}{dx}=v+x\frac{dv}{dx}$ and $\sqrt{x^2+y^2}=\sqrt{x^2+v^2x^2}=x\sqrt{1+v^2}$ because $x>0$.

Substitute into the differential equation

$v+x\frac{dv}{dx}=\frac{x\sqrt{1+v^2}+vx}{x}=\sqrt{1+v^2}+v$

Hence, $x\frac{dv}{dx}=\sqrt{1+v^2}$

Separate variables

$\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}$

Integrating both sides, $\sinh^{-1}(v)=\ln x+C$

Use the initial condition

The curve passes through $(1,0)$.

At $x=1$, $y=0$, so $v=\frac{y}{x}=0$

Therefore, $\sinh^{-1}(0)=\ln 1 + C \implies 0=0+C \implies C=0$

Thus, $\sinh^{-1}(v)=\ln x$

So, $v=\sinh(\ln x)$

Find $y$ in terms of $x$

Since $v=\dfrac{y}{x}$, $\frac{y}{x}=\sinh(\ln x)$ $y=x\sinh(\ln x)$

Using $\sinh t=\frac{e^t-e^{-t}}{2}$ with $t=\ln x$, $\sinh(\ln x)=\frac{x-1/x}{2}$

Therefore, $y=x\cdot \frac{x-1/x}{2}=\frac{x^2-1}{2}$

Find $\alpha$ at $x=2$

$\alpha=y(2)=\frac{2^2-1}{2}=\frac{4-1}{2}=\frac{3}{2}$

Check options

$\alpha=\frac{3}{2}$

Hence the correct option is B.