ODE: a specific question

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

$x^2=C(2y+C)$

Answer 1

$$y= \frac{-dx}{dy}x+ \sqrt{x^2 +y^2}$$ $$ \frac{dx}{dy}x=-y+ \sqrt{x^2 +y^2}$$ $$ \frac{dx}{dy}=-\dfrac yx+ \sqrt{1 +\dfrac {y^2}{x^2}}$$ 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. $$\sqrt {u+v}= \sqrt v+C$$ $$x^2+y^2=(y+C)^2$$ $$\boxed{x^2=C(C+2y)}$$