Given, differential equation
$x d y-y d x=\sqrt{x^2+y^2} d x$ $\ldots$ (i)
Putting, $y=v x$ and differentiating w.r.t. $x$, we get
$d y=v d x+x d v$ $\ldots$ (ii)
From Eq. (ii), $x(v d x+x d v)-v x d x=\sqrt{x^2+(v x)^2} d x$
$x v d x+x^2 d v-v x d x=x \sqrt{1+v^2} d x$
$x^2 d v=x \sqrt{1+v^2} d x \Rightarrow \frac{d v}{\sqrt{1+v^2}}=\frac{d x}{x}$
Integrating on both sides, we get
$\ln \left(v+\sqrt{1+v^2}\right)=\ln x+C$
$\left\{\because \int \frac{d x}{\sqrt{1+x^2}}=\ln \left(x+\sqrt{1+x^2}\right)+C\right\}$
Putting $v=\frac{y}{x}$, we get
$\ln \left(\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}\right)=\ln x+C$
$\ln \left(\frac{y}{x}+\frac{1}{x} \sqrt{x^2+y^2}\right)=\ln x+C$
$\ln \left\{\frac{1}{x}\left(y+\sqrt{x^2+y^2}\right)\right\}-\ln x=C$
$\ln \left\{\frac{y+\sqrt{x^2+y^2}}{x^2}\right\}=C$ $\ldots$ (iii)
$\left[\because \ln a-\ln b=\ln \frac{a}{b}\right]$
Given that, $y=1, x=\sqrt{3}$
$\therefore \quad \ln \left\{\frac{1+\sqrt{(\sqrt{3})^2+1^2}}{(\sqrt{3})^2}\right\}=C \Rightarrow \ln \left\{\frac{1+2}{3}\right\}=C$
$\operatorname{lnl}=C$ or $C=0 \quad[\because \ln 1=0]$
Putting the value of $C$ in Eq. (iii), we get
$\ln \left\{\frac{y+\sqrt{x^2+y^2}}{x^2}\right\}=0$
or $\quad \frac{y+\sqrt{x^2+y^2}}{x^2}=e^0=1 \quad\left[\because e^0=1\right]$
$y+\sqrt{x^2+y^2}=x^2$ or $x^2-y=\sqrt{x^2+y^2}$
Squaring on both sides, we get
$\left(x^2-y\right)^2=x^2+y^2$