Solving a differential equation and understanding behavior of equilibrium points I'm trying to understand how to solve the differential equation
$$ 
\frac{dx}{dt} = - \sqrt{x}, \; x(0) = 1, \; x(t) \geq 0 \text{ for all $t \geq 0$.}
$$
The solution the instructor gave is
$$
x(t) = \begin{cases}
\frac{(t-2)^2}{4} & \text{ if $t \leq 2$}  \\
0 & \text{ if $t \geq 2$.}
\end{cases}
$$
I am stuck with the case that $t \geq 2$. When I separate variables and use the initial condition, I find that $x(t) = \frac{(t-2)^2}{4}$ when $t \leq 2$.  When I plug in $t = 2$, I find $x(2) = 0$. I can also see from the original differential equation that if $x > 0$, $\frac{dx}{dt} < 0$, so $x = 0$ is a "stable" equilibrium point.
Here is my problem: I do not know how to tie $x = 0$ being a stable equilibrium to the case where $t \geq 2$. It seems that there is no connection to specific values of $t$ because I've only plugged in a value of $x$, not a value of $t$, and I have no other information about how $x(t)$ depends on $t$.
Can someone help me to understand this relationship?
 A: This so defined $x(t)$ is continuously differentiable, its derivative is
$$
x'(t)=\begin{cases}
\frac{t-2}2,&t\le 2\\ 0,&t>2.
\end{cases}
$$
The square root of it is
$$
\sqrt{x(t)}=\begin{cases}
\frac{2-t}2,&t\le 2\\ 0,&t>2,
\end{cases}
$$
so that indeed $x'(t)=-\sqrt{x(t)}$. So it is a solution. There is nothing more required.

When solving the equation you first detect that it is separable. Then check where division-by-zero occurs, and if these curves can be solutions. In this case indeed $x\equiv 0$ is a solution. Then check smoothness, the right side is not smooth, so uniqueness is not automatic there. However for $x\ne 0$ the right side is smooth, so one has to care for anything after finding the solution there with any means.
$x\equiv 0$ is the singular solution of the more general DE $(x')^2=x$. The derivative of this equation is $x'(2x''-1)=0$. Any solution can be divided into segments where one or the other factor is zero. At the intersections both factors have to be zero simultaneously (along with continuity etc.).
With the original equation the first equation gives $x=0$, the second $2x'=t+c\implies 4x=(t+c)^2$, intersections are only possible at $t=-c$. With the fixed sign and reduced domain of the given equation, the family of possible solutions reduces accordingly.
