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

Given,

$xdy=\left(\sqrt{x^2+y^2}+y\right)dx$

$\Rightarrow xdy-ydx=\sqrt{x^2+y^2}dx$

$\Rightarrow \frac{xdy-ydx}{x^2}=\sqrt{1+\frac{y^2}{x^2}}·\frac{dx}{x}$

$\Rightarrow \frac{d\left({\displaystyle \frac{y}{x}}\right)}{\sqrt{1+{\left(\frac{y}{x}\right)}^2}}=\frac{dx}{x}$

Now integrating both side we get,

$\Rightarrow \int \frac{d\left({\displaystyle \frac{y}{x}}\right)}{\sqrt{1+{\left(\frac{y}{x}\right)}^2}}=\int \frac{dx}{x}$

$\Rightarrow \ln \left(\frac{y}{x}+\sqrt{{\left(\frac{y}{x}\right)}^2+1}\right)=\ln x+\ln c$

$\Rightarrow \frac{y+\sqrt{y^2+x^2}}{x}=cx$

$\Rightarrow y+\sqrt{y^2+x^2}=c{x}^2$

Now given when $x=1,y=0\Rightarrow 0+1=c\Rightarrow c=1$

So equation of curve is $y+\sqrt{x^2+y^2}=x^2$

Now at $x=2,y=\alpha$

So putting the value in curve we get, $2+\sqrt{4+{\alpha }^2}=4$

$\Rightarrow 4+{\alpha }^2=16+{\alpha }^2=8\alpha$

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