Maximal solution of $x'+f\sin x - g\cos x + g =0$ Given an open interval $I$ and two smooth functions $f,g \colon I \to \mathbb{R}$, I would like to study the non-linear differential equation in $x \colon I \to \mathbb{R}$ defined by
\begin{equation}
x'(t)+f(t) \sin(x(t)) - g(t)\cos(x(t)) + g(t) =0.
\end{equation}
What can one say about the maximal domain of existence of the corresponding initial value problem? In particular, is there a simple condition on $f$ and $g$ guaranteeing global solvability – i.e., that the solution is defined on the entire $I$?
 A: A simple condition guaranteeing global solvability of our ODE is that $f$ and $g$ are bounded.
Let $F \colon I \times \mathbb{R} \to \mathbb{R}$ be defined by:
\begin{equation*}
F(t,x) = g(t)\cos(x) -f(t)\sin(x) - g(t).
\end{equation*}
We show that, if $f$ and $g$ are bounded, then $F$ is globally Lipschitz, i.e., for all $t\in I$ and all $x,y \in \mathbb{R}$, there exists $K>0$ such that:
\begin{equation*}
\left\lvert F(t,x)-F(t,y)\right\rvert \leq K \left\lvert x-y \right\rvert.
\end{equation*}
Computing
\begin{align*}
\left\lvert F(t,x)-F(t,y)\right\rvert &= \left\lvert g(t)\left(\cos(x)-\cos(y)\right) +f(t)\left(\sin(y)-\sin(x)\right) \right\rvert\\
&\leq \left\lvert g(t)\left(\cos(x)-\cos(y)\right)\right\rvert + \left\lvert f(t)\left(\sin(y)-\sin(x)\right) \right\rvert,
\end{align*}
we observe that, if $f$ and $g$ are bounded by $M$ and $N$, respectively, then
\begin{equation*}
\left\lvert F(t,x)-F(t,y)\right\rvert \leq 2(M+N).
\end{equation*}
Hence $F$ is globally Lipschitz, as desired.
