Find the orthogonal trajectories of the family of curves $y^2 + 2cx = c^2$

Question

I start by knowing that I need a differential equation satisfying all members of family $y^2 + 2cx = c^2$, which means $c$ needs to be eliminated.
$$2y \frac{dy}{dx}+ 2c=0\\ c = -y \frac{dy}{dx}\\ y^2 - 2yx\frac{dy}{dx} =y^2 \left(\frac{dy}{dx}\right)^2$$ For orthogonal trajectories, $$y^2 + 2yx\frac{dx}{dy} =y^2 \left(\frac{dx}{dy}\right)^2$$

And then i got stuck, none of the methods i tried to use to solve this helped. The answer must be $$y^2 = 2cx + c^2.$$
What am I doing wrong?

Answer 1

If $$y(x)\ne 0$$ we get $$y(x)-2xy'=yy'^2$$ solving this for $y'$ we get $$y'=\frac{1}{y}(-x+\sqrt{x^2+y^2})$$ so the slope of the traectories are given by $$y'_{orth}=-\frac{y}{-x+\sqrt{x^2+y^2}}$$ solving this equation we get $$y(x)=C\sqrt{C+2x}$$ or $$y(x)^2=C'+2xC'$$ as you has stated

Answer 2

First you must express $c$ as an explicit equation of $x$ and $y$ then differentiate it.$$c^2-2cx+x^2=x^2+y^2\to (c-x)^2=x^2+y^2\to c=x\pm\sqrt{x^2+y^2}$$now by differentiating we have$$1\pm\dfrac{x+yy'}{\sqrt{x^2+y^2}}=0$$by replacing $y'\to-\dfrac{1}{y'}$ we have$$1\pm\dfrac{x-\dfrac{y}{y'}}{\sqrt{x^2+y^2}}=0\to\\y'\pm\dfrac{xy'-y}{\sqrt{x^2+y^2}}=0$$by replacing $y=xu$ we have$$1+xu'(\dfrac{1}{u}\pm\dfrac{1}{u\sqrt{1+u^2}})=0$$ and after solving it we obtain:$$(1) \ \ cx^2=\dfrac{y^2}{\sqrt{x^2+y^2}+x}\\(2)\ \ cx^2={\sqrt{x^2+y^2}+x}$$

Answer 3

The initial equation is equivalent to $$x^2+y^2=(x-c)^2$$ or $$\pm\sqrt{x^2+y^2}=x-c.$$

Then differentiating,

$$\pm\frac{x+yy'}{\sqrt{x^2+y^2}}=1.$$

We replace $y'$ by $-\dfrac1{y'}$ and solve

$$y'\left(x\pm\sqrt{x^2+y^2}\right)=y.$$