A first-order non-linear ordinary differential equation containing various squares The Equation:
Find all differentiable functions $f: I \rightarrow \mathbb{R}$ satisfies:
$$\big(\,f(x)-x\,f'(x)\big)^2 = \big(\,f'(x)\big)^2 + 1 \; \; \; \; \; \text{for all}\,\,\, x \in I,$$
where $I$ is an open interval.
This is not an assignment. And, thanks in advance!
 A: I don't know if my approach is correct, but I found that:
$$f(x) = A \, x \pm \sqrt{1+A^2}, \quad A \in \mathbb{R},$$
is a solution of your ODE. I used Lagrange-Charpit method in order to solve a "false" 1st order PDE for $f = f(x,y)$, assuming that there's no dependence on $y$. 
If you want some more details, just ask for it.

Edit
The Lagrange-Charpit equations state that, for a non-linear partial differential equation written in the form:
$$G(f(x_1,x_2,\ldots,x_N),p_i,x_i) = 0,$$
where $f$ is an unknown function (dependent variable), $p_i = \frac{\partial f}{\partial x_i}$, or in the special case of $N = 2$, $x_i = (x,y) $, in order to simplify the notation, the following relations hold:
$$ \frac{dx}{G_p} = \frac{dy}{G_q} = \frac{df}{pG_p + qG_q} = - \frac{dp}{G_x + pG_f} = -\frac{dq}{G_y + q G_f}, $$
Since:
$$\begin{align}
G & = (f-xp)^2 -p^2-1 = 0,  \\
G_x & = 2(f-xp)(-p),  \\
G_y & = 0,  \\
G_f & = 2 (f-xp),  \\
G_p & =  2 (f-xp)(-x)-2p,  \\
G_q & = 0,
\end{align}
$$ you have:
$$\frac{dx}{2(f-xp)(-x)-2p} = \frac{dy}{0} = \frac{df}{2p(f-xp)(-x)-2p^2} =$$ $$= \frac{dp}{2(f-xp)(-p)+2p(f-xp) } = \frac{dp}{0} = \frac{dq}{2q(f-xp)}.$$
This tells you, among other things, that $dp = 0$, so $p = A$ is a constant. Since $p = f_x$ you can integrate in order to achieve $f$, so:
$$f(x,y) = A x + B(y),$$
for $B(y)$ some arbitrary function of $y$. Substitute back into the original equation, $G = 0$ in order to obtain a relation between $A$ and $B$, which results to be:
$$B^2(y) - A^2 - 1 = 0,$$
so $$B(y) = B = \pm \sqrt{1+A^2},$$ and $B$ does not depend on $y$, as it was expected because $f$ did not depend on $y$ originally. 
So in resume, I have treated your original ODE as a first order non-linear PDE for the dependent variable $f = f(x,y)$, but, as we already knew, $f = f(x)$, as it has been shown. 
Cheers!
A: Two sets of solutions are obtained as shown below :

