y=2xy^{\prime}+\sqrt{1+\left(y^{\prime}\right)^2}\\
y^{\prime}=2y^{\prime}+2xy^{\prime\prime}+\frac{y^{\prime}y^{\prime\prime}}{sqrt{1+\left(y^{\prime}\right)^2}}\\
y^{\prime}=p\\
p=2p+2xp^{\prime}+\frac{p \cdot p^{\prime}}{\sqrt{1+p^2}}\\
-p=2xp^{\prime}+\frac{p \cdot p^{\prime}}{\sqrt{1+p^2}}\\
p=-2xp^{\prime}-\frac{p \cdot p^{\prime}}{\sqrt{1+p^2}}\\
p=p^{\prime}\left(-2x-\frac{p}{\sqrt{1+p^2}}\right)\\
px^{\prime}=-2x-\frac{p}{\sqrt{1+p^2}}\\
px^{\prime}+2x=-\frac{p}{\sqrt{1+p^2}}\\
x^{\prime}+\frac{2}{p}x=-\frac{1}{\sqrt{1+p^2}}\\
x^{\prime}+\frac{2}{p}x=0\\
x^{\prime}=-\frac{2}{p}x\\
\frac{x^{\prime}}{x}=-\frac{2}{p}\\
\frac{\mbox{d}x}{x}=-\frac{2\mbox{d}p}{p}\\
\ln{|x|}=-2\ln{|p|}+C\\
x=Cp^{-2}\\
x\left(p\right)=C\left(p\right)p^{-2}\\
C^{\prime}\left(p\right)p^{-2}-2C\left(p\right)p^{-3}+2C\left(p\right)p^{-3}=-\frac{1}{\sqrt{1+p^2}}\\
C^{\prime}\left(p\right)p^{-2}=-\frac{1}{\sqrt{1+p^2}}\\
C^{\prime}\left(p\right)=-\frac{p^2}{\sqrt{1+p^2}}

\int{\frac{-p^{2}}{\sqrt{1+p^2}}\mbox{d}p}\\

\sqrt{1+p^2}=t-p\\
1+p^2=t^2-2pt+p^2\\
1=t^2-2pt\\\
2pt=t^2-1\\
p=\frac{t^2-1}{2t}\\
t-p=\frac{2t^2-t^2+1}{2t}\\
t-p=\frac{t^2+1}{2t}\\
\mbox{d}p=\frac{2t\cdot 2t-2\left(t^2-1\right)}{4t^2}\mbox{d}t\\
\mbox{d}p=\frac{4t^2-2t^2+2}{4t^2}\mbox{d}t\\
\mbox{d}p=\frac{2t^2+2}{4t^2}\mbox{d}t\\
\mbox{d}p=\frac{t^2+1}{2t^2}\mbox{d}t\\
-\int{\frac{\left(t^2-1\right)^2}{4t^2}\cdot \frac{2t}{t^2+1}\cdot \frac{t^2+1}{2t^2}\mbox{d}t}\\
=-\int{\frac{t^4-2t^2+1}{4t^3}\mbox{d}t}\\
=-\int{\frac{1}{4}t-\frac{1}{2}\cdot \frac{1}{t}+\frac{1}{4}\cdot \frac{1}{t^3}\mbox{d}t}\\
=-\frac{1}{8}t^2+\frac{1}{2}\ln{|t|}-\frac{1}{8}\frac{1}{t^2}+C\\
=-\frac{1}{2}\left(\frac{t^4-1}{4t^2}-\ln{|t|}\right)+C\\
=-\frac{1}{2}\left(p\sqrt{1+p^2}-\ln{|p+\sqrt{1+p^2}|}\right)+C\\
C\left(p\right)=-\frac{1}{2}\left(p\sqrt{1+p^2}-\ln{|p+\sqrt{1+p^2}|}\right)+C\\
x=-\frac{1}{2}\frac{1}{p}\sqrt{1+p^2}+\frac{1}{2p^2}\ln{|p+\sqrt{1+p^2}|}+\frac{C}{p^2}\\
y=-\sqrt{1+p^2}+\frac{1}{p}\ln{|p+\sqrt{1+p^2}|}+\frac{2C}{p}\sqrt{1+p^2}\\
\begin{cases}x=-\frac{1}{2p^2}\left(p\sqrt{1+p^2}-\ln{|p+\sqrt{1+p^2}|}+2C\right)\\y=\frac{1}{p}\left(\ln{|p+\sqrt{1+p^2}|}+2C\right)\end{cases}\\