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Suppose for $x\in\mathbb{R}^n$, $$\frac{dx}{dt}=A(t,x(t))x(t)+\frac{dg}{dt}, \quad x(0)=0$$ has a solution that exists for all time where $A(t,x(t)),$ $\frac{dg}{dt}$ are continuous and bounded functions. I'd like to show the solution is unique. To this end, I integrate and have $$x(t)=g(t)+\int_0^tA(\tau,x(\tau))x(\tau)d\tau. $$ Define $$Tx=g(t)+\int_0^tA(\tau,x(\tau))x(\tau)d\tau $$ and attempt to show $T$ is a contraction mapping. $$\left|Tx-Ty\right|\leq\int\left|A(\tau,x(\tau))x(\tau)-A(\tau,y(\tau))y(\tau)\right|d\tau.$$ I'm stuck trying to estimate this quantity since $A(t,x(t))$ is not Lipschitz continuous. I've tried manipulating so I could use Gronwall's inequality among other basic estimates, but to no avail. Is there another strategy?

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It need not hold. It suffices to find a counter-example for a neighborhood of a point (easiest when different from the origin). Consider e.g. for $x$ in a neighborhood of $1$ the initial value problem: $$ \dot{x} = 2\sqrt{|\ln (x(t)) |}\; x(t) , \; \; \; x(0)=1$$

Then $x(t)\equiv 1$ is a solution, but you may also take $x(t)=1, t<0$ and for $0\leq t \leq \delta$: $$ x(t) = e^{t^2} ,$$ since for $t\geq 0$: $$ \dot{x} = 2 t \;e^{t^2} = 2 \sqrt{\ln x(t)} \; \; x(t) $$ You may modify this example to a global one in ${\Bbb R}$ verifying the stated conditions. You need a suitable $g$ that takes you from an initial position at the origin (at the origin the solution is unique) to a non-zero $x$ value, say $x=1$, and then employ my counter-example in the vicinity of that value.

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