Solve $ y=({\rm d}y/{\rm d}x)^2 $ It is obvious that we can differentiate both sides of the equation with respect to $x$ and then discuss the result. But can we just make a square root of both sides and integrate?

Sorry the original version is so naive. The following is more detailed.


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*If we take square root
$$\pm \sqrt y=\frac{{\rm d}y}{{\rm d}x}$$
then
$$\frac{{\rm d}y}{\sqrt y}=\pm {\rm d}x$$
$$2\sqrt y=\pm x+c$$
The solutions are
$$y=(\frac{\pm x+c}{2})^2$$ or $y=0$

*If we differentiate both sides with respect to $x$
$$y'=2y'y''$$
$$y'(2y''-1)=0$$
If we choose $y'=0$, then $y=c$. And plug it in to the original equation, we get $c=0$.
If we choose $y''=\frac12$, then $y=\frac14x^2+dx+f$. Plugging in, we get $y=\frac14x^2+dx+d^2$, which is enough.
(How could the non-trivial solutions differ???-this is a solved question now.)

Now, everything is great. From this example, I know that there are something a little bit different with the liner differential equation in a nonlinear one. Sometimes, one needs to plug in the solution with constants under some circumstances to determine some constant. Besides, the form of the solution of a nonlinear differential equation could be more complicated-it could be piecewise-combined. One could choose a set of solution satisfying the equation and match them at every boundary of each segment correspondingly. Thank @Allawonder and @Gae. S. for pointing them out.
 A: The solutions do not differ.
$$(c\pm x)^2=x^2\pm 2cx+c^2$$
does match
$$x^2+4dx+4d^2$$ with $$\pm c=2d.$$
Notice that, as mentioned by @gaes, as $y=0$ is a solution you can switch from a solution to another when $y=0$, and there are in fact five possible modes.

A: Let's try to make it simple.
We have,
$$y = y'^2 \implies \pm y = y' \implies \pm\frac{dy}{\sqrt y}=dx \implies \pm2\sqrt y = x+c $$
$$\implies 4y = (x+c)^2 \implies\color{red}{ y = \frac{(x+c)^2}{4}}$$
If $c = x$, we have $$y = 0 \implies y' = \frac{d}{dx}0 = 0\implies y=y'^2$$
If $c\ne x$, $$y' = 2\frac{(x+c)}{4} = \frac{x+c}{2} \implies y'^2 = \frac{(x+c)^2}{4} = y$$

Trying it in your way, $y = \frac{(\pm x+c)^2}{4} = \frac{(x\pm c)^2}{4} = \frac{(x+k)^2}{4}$ where $k = \pm c$, which is identical as above.
A: Observe this:
$$y_1=A(x-a)^2 \text{ and } y_2=A(x+a)^2$$
are functions whose derivatives are  
$$y_1'=2A(x-a) \text{ and } y_2'=2A(x+a).$$
The squares of the derivatives are then:
$$(y_1')^2=4A^2(x-a)^2 \text{ and } (y_2')^2=4A^2(x+a)^2.$$
If $A=4A^2$ or if $A=\frac14$ then we have (at least) two solutions.
A: In fact the initial condition is determining which branch is right. For $$\sqrt y=y'$$we have$$2\sqrt y=x+C$$and for $$-\sqrt y=y'$$we obtain$$2\sqrt y=-x+C$$starting with $x=0$, each of the cases above show that $y'(0)$ is determining. For $y'(0)<0$ we have $$y={(x-2\sqrt{y(0)})^2\over 4}\implies y'={x\over 2}-\sqrt{y(0)}\le 0\implies x\le 2\sqrt{y(0)}$$and for $y'(0)\ge0$ we obtain$$y={(x+2\sqrt{y(0)})^2\over 4}\implies y'={x\over 2}+\sqrt{y(0)}\ge 0\implies x\ge -2\sqrt{y(0)}$$
A: Can be factored into two differential equations and so we have two equations of the two parabolas due to the $\pm$ sign.
$$ y'= \pm \sqrt {y} ; \; \dfrac{dy}{\sqrt y}=\pm dx;\;\text{Integrating ,  } 2 \sqrt{y}= \pm x +c $$
which can be separately expressed as
$$ y_1= (x/2-c_1)^2,\quad y_2= (x/2-c_2)^2$$

The parabolas can slide along x-axis so as to have tangential double root
$$ x_{min}= 2c_1,\; x_{min}= 2c_2\;$$
To confirm tangency or double contact we have by differentiation
$$ y= y'^2,\; \to y'= 2y' y''$$
Only when we assume
$$y'\ne 0$$
that is, implying that x-axis is excluded as a solution,  are we allowed a division, and consequently so we have a minima situation:
$$y_{min \;tangency\;point}^{''} = \dfrac12 $$
In the zero arbitrary constant situation both the sets come to the central position of the same parabola of unit focal length.
