ODE: a specific question Determine the  equation of the curve for which the $y$ intercept of the normal drawn to a point on the curve is equal to the distance of that point from the origin.
My attempt:
Consider an arbitrary point, $(x,y)$, whose slope is $\frac{dy}{dx} $.
Thus the slope of the normal is $\frac{-dx}{dy}$.
Using the $y$ intercept form of a line
$$y= \frac{-dx}{dy}x+ \sqrt{x^2 +y^2}$$
Now I have tried solving this using substitutions,($y=ux$, and $x=uy$), but that didn't work.
To use the method of the integrating factor, Unless I'm mistaken, I'll need a subsitution to do that, but I can't seem to find any appropriate substitutions.
Thanks for the help.
This problem is problem 54, in chapter 1, in the second volume of N.piskunov's differential and integral calculus.
There is a solution, but that does absolutely nothing to explain how to solve this
The solution they have given
$y+\frac{x}{y'}= \sqrt{x^2 +y^2}$
Whence
$x^2=C(2y+C)$
Which is what I'm unable to understand
TL;DR
How did
$y+\frac{x}{y'}= \sqrt{x^2 +y^2}$
Result in
$x^2=C(2y+C)$

$$y= \frac{-dx}{dy}x+ \sqrt{x^2 +y^2}$$
$$ \frac{dx}{dy}x=-y+ \sqrt{x^2 +y^2}$$
$$ \frac{dx}{dy}=-\dfrac yx+ \sqrt{1 +\dfrac {y^2}{x^2}}$$
This looks like an homogeneous DE substitute $x=ty$:
$$t'y+t=-\dfrac 1t+ \sqrt{1 +\dfrac {1}{t^2}}$$
$$t'y=\dfrac {-1-t^2+ \sqrt{1 +t^2}}{t}$$
$$\dfrac 12 (t^2+1)'y= {-1-t^2+ \sqrt{1 +t^2}}$$
The DE is separable.You can integrate and substitute $u^2=t^2+1$ and $udu=tdt$.
Another way:
You can also rewrite the original DE as:
$$y= \frac{-dx}{dy}x+ \sqrt{x^2 +y^2}$$
$$1= -\frac{du}{dv}+ \sqrt{1+\dfrac uv}$$
Where $u=x^2$ and $y^2=v$.
$$(u+v)'= \dfrac { \sqrt{ u+v}}{\sqrt v}$$
Separate and integrate.
$$\sqrt {u+v}= \sqrt v+C$$
$$x^2+y^2=(y+C)^2$$
$$\boxed{x^2=C(C+2y)}$$
