How to solve $\dot{x}=|x|$ How to solve this differential equation $\dot{x}=|x|$? 
 A: One observes that $x=0$ is the trivial solution of the given ODE.
Now find nontrivial solutions of the equation.
Using separation of variables, one gets $$\frac{\dot{x}}{|x|}=1.$$
Integrating both sides gives
\begin{align}\text{sgn}(x)\log|x|=t+C\quad\cdots\quad\text{(a)}\end{align} for some constant $C$, where $\text{sgn}(x)=\begin{cases}1,&x> 0\\-1,&x<0.\end{cases}$
Therefore, $x(t)=\begin{cases}C_1e^t,&x>0\\C_2e^{-t},&x<0.\end{cases}$
Now find the solution of explicit form.
From the equation, $\dot{x}$ is always nonnegative, so $x$ is a continuously differentiable and nondecreasing function of $t$. This shows that $C_1$ must be positive and $C_2$ must be negative.
Since $x$ is a continuous function of $t$ and a nontrivial solution of $x$ cannot take 0 by the above observation, it follows that $x(t)=C_1e^t,C_1>0$ for all $t$ or $x(t)=C_2e^{-t},C_2<0$ for all $t$.
A: bellcircle has provided a solution to the problem. However, I would like to express the solution in a more compact form and also provide an answer using a proposition-proof style.
$(Dx)(t) = |x(t)|$ (1)
Proposition. If $I$ is an open interval in $\mathbb{R}$ and $C \in \mathbb{R}$, then $x(t)=Ce^{\mathsf{sgn}(C) t}$ is a solution of the differential equation (1) on $I$. If $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $\mathbb{R}$, then there exists $C \in \mathbb{R}$ such that $x(t)=Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$. 
Suppose $I$ is an open interval in $\mathbb{R}$. Suppose $C \in \mathbb{R}$ and $x$ is such that $x(t)=Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$. Then $(Dx)(t)=\mathsf{sgn}(C)Ce^{\mathsf{sgn}(C) t}$. Also, $|x(t)|=\mathsf{sgn}(C)Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$. Therefore, indeed $x(t)=Ce^{\mathsf{sgn}(C) t}$ is a solution of (1) on $I$.
Suppose there is a solution $x_2(t)$ of (1) on $I$ such that there is no $C \in \mathbb{R}$ such that $x_2(t) = Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$. Then take $T \in I$. Define $A=\frac{x_2(T)}{e^{\mathsf{sgn}(x_2(T)) T}}$. Then $x(t) = A e^{\mathsf{sgn}(A)t}$ is a solution of (1) on $I$ with $x(T)=x_2(T)$. The uniqueness of the solution of (1) on $I$ with $x(T)=x_2(T)$ follows from the Lipschitz continuity of $|\cdot|$ by the Picard–Lindelöf theorem (link) and the global uniqueness theorem (e.g. link). Therefore $x_2(t) = A e^{\mathsf{sgn}(A)t}$ for all $t \in I$ and, thus, by contradiction, if $x(t)$ is a solution of the differential equation (1) on an open interval $I$ in $\mathbb{R}$, then there exists $C \in \mathbb{R}$ such that $x(t)=Ce^{\mathsf{sgn}(C) t}$ for all $t \in I$
