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How to solve this differential equation $dot{x}=|x|$?

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+Cquadcdotsquadtext{(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 $t$.

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 $t in I$