General solution of nonlinear first order differential equation I would like to solve:
$y - y' x - y'^2 = 0$.
In order to do so, we let $y' = t$, and we assume $x$ as a function of $t$. Now, we take derivative with respect to $t$ from the differential equation, and obtain
$\frac{dy}{dt} - x - t \frac{dx}{dt} - 2 t = 0$.
By the chain rule, we have: $dy / dt = t dx / dt$. So, the above simplifies to
$x = - 2 t$.
That is, we have: $x = -2 dy / dx$. Thus, we obtain
$y = - \frac{x^2}{4} + C$.
Now, if we want to verify the solution, it turns out that $C$ must be zero, in other words, $y = - x^2 /4$ satisfies the original differential equation.
I have two questions:
1) What happens to the integration constant? That is, what is the general solution of the differential equation?
2) If we try to solve this differential equation with Mathematica, we obtain
$y = C_1 x + C_1^2$,
which has a different form from the analytical approach. How can we also produce this result analytically?
 A: The differential equation given is a type of Clairaut's Equation :
https://en.wikipedia.org/wiki/Clairaut%27s_equation

You have approached the problem correctly. The reason why the arbitrary constant vanishes is because it represents a singular solution of the differential equation. The singular solution is basically like an envelope of all the solutions of the differential equation. To find the general solution, we do the following,
let $p = \frac{dy}{dx}$. The differential equation is 
$$y-px-p^2=0$$
Differentiating just like you did,
$$p-p-xp'-2pp'=0 \implies p'=0$$
This is the part you forgot to consider.
$$\therefore p=C_1$$ where $C_1$ is an arbitrary constant. When we substitute this in the given differential equation, we get
$$y-C_1x-C_1^2=0 \implies y=C_1x+C_1^2$$
The other possible solution is the one you obtained, i.e the singular solution $$y=-\frac{x^2}{4}$$
A: As an alternative error analysis using your approach of a re-parametrization, your omission is in the first step where you take $t$ as new independent parameter. That is only possible on those solution segments where $t$ is not constant and moreover bijective relative to the old independent $x$, that is, strongly monotonously increasing or falling. 
For that situation you got a solution. Now also consider the solutions where $t=C$ is constant, resulting in the linear solutions $$y=Cx+C^2.$$
