Solve Non-Linear First-Order ODE for Parabola Focus point

Question

I'd like to know how to solve this Non-Linear First-Order ODE, step-by-step: $$ \frac{2y'}{{y'}^2-1} = \frac{x}{y-D} $$ where $y=f(x)$ and $D$ is a constant.

Wolfram Alpha gives solution, but hides step-by-step guide: https://www.wolframalpha.com/input/?i=2*f%27%28x%29%2F%281-%28f%27%28x%29%29%5E2%29%3Dx%2F%28D-f%28x%29%29

It also states that it's a D'Alembert's equation, but when I tried to research how to solve it, Google was overwhelmed with the D'Alembert's wave equation instead.

Long story:

I was trying to solve, which function reflects incoming parallel rays of light to a single point. I know it is a parabola, but wanted to prove it myself. The train of thoughts I had:

Out of geometric reasoning and from the reflection of rays one can deduce the followings (see image):

$\frac{x}{y-D}=\tan(\pi-2\alpha)=-\tan(2\alpha)=-\frac{2\tan(\alpha)}{1-\tan^2(\alpha)}=\frac{2y'}{{y'}^2-1}$

And this is where I'm stuck. If anyone has a better method to prove this property, all suggestion is welcome.

Answer 1

$$\frac{2y'}{{y'}^2-1} = \frac{x}{y-D}$$ Substitute $u=x^2$: $$\frac{4xy'}{4x^2{y'}^2-1} = \frac{x}{y-D}$$ $$\frac{4y'}{4u{y'}^2-1} = \frac{1}{y-D}$$ $$y-D=\frac{4u{y'}^2-1}{4y'}$$ $$y=u{y'}-\dfrac 1{4y'}+D$$ This is Clairaut's differential equation $$y=uy'+f(y')$$ To find the general solution just set $y'=C$ in the differential equation and substitute back $u=x^2$: $$y=u{y'}-\dfrac 1{4y'}+D$$ $$y(u)=u{C}-\dfrac 1{4C}+D$$ $$\boxed{y(x)=x^2C-\dfrac 1{4C}+D}$$

Answer 2

Replacing $y = f(x)$ and cross multiplying, we get $$\begin{aligned} 2 f'(x) (f(x) - D) &= x(f''(x)-1) \\ 0 &= xf''(x) - 2f'(x)(f(x)-D)) -x \end{aligned}$$ Let $f'(x) = a$, then this is quadratic in $a$, and solving for the roots of $a$ gives us, $$\begin{aligned} f'(x) &= \frac{2 (f(x)-D) \pm \sqrt{4(f(x)-D)^2 + 4x^2}}{2x} \\ f'(x) &= \frac{f(x) - D}{x} \pm \sqrt{\frac{f(x)-D}{x}+1} \end{aligned}$$ Let $\frac{f(x)-D}{x} = u$. Then $f(x)-D = xu, \partial f(x) = u\partial{x} + x\partial{u}, \ \frac{\partial f(x)}{\partial x} = u + x\frac{\partial u}{\partial x}$

$$\begin{aligned} u + x\frac{\partial u}{\partial x} &= u + \sqrt{u^2+1} \\ x\frac{\partial u}{\partial x} &= \sqrt{u^2+1} \\ \frac{\partial u}{\sqrt{u^2+1}} &= \frac{\partial x}{x} \end{aligned}$$ Integrating on both sides gives us, $$\begin{aligned} \sinh^{-1}u &= \ln{x} + c \\ \sinh^{-1}\left(\frac{f(x)-D}{x}\right) &= \ln{x} + c \\ e^{\sinh^{-1}\left(\frac{f(x)-D}{x}\right)} + c &= x \\ x\sinh{(\ln x+c)}+D &= f(x) \end{aligned}$$