Let $n\in\mathbb{N}$, let $I\subset \mathbb{R}$ be an interval, let $U\subset \mathbb{R}^n$ be open, let $F:I\times U\to \mathbb{R}^n$ be continuous, and let $(t_0,x_0)\in I\times U$. Let $J\subset I$ be an interval with $t_0\in J$, and let $x:J\to U$ be continuous.

Call $x$ a solution of the initial value problem if $x(t_0)=x_0$, if $x$ is differentiable at every interior point $t\in J$, if $\dot{x}(t)=F(t,x(t))$ at every such interior point, and if at each endpoint of $J$ that belongs to $J$ and is an accumulation point of $J$ from within, the appropriate one-sided derivative exists and satisfies the same equation.

Then $x$ is a solution of the initial value problem if and only if, for every $t\in J$,

\begin{align*} x(t)=x_0+\int_{t_0}^{t}F(s,x(s))\,d\mathcal{L}^1(s), \end{align*}

with the integral understood in the oriented sense.