Laplace transform of the integral function

Question

I try to understand why according to my notes from some lecture the following equality is satisfied

$$ \mathcal{L}\left[ {\exp\left(\int_0^t ds \frac{\dot{x}(s)}{x(s)} \right)} \right] =\frac{1}{u-\frac{\mathcal{L}\left[\dot{x}\right](u)}{\mathcal{L}[x](u)}}, $$

where $\mathcal{L}[\cdot]$ denotes Laplace transform. Is this equation fulfilled for all $x(t)$, and how one can derive it?

Answer 1

The key point is to note, that

$\frac{d}{dt} \ln x(t)= \frac{\dot{x}(t)}{x(t)}$,

accordingly

$ \exp(\int\limits_0^t ds \frac{\dot{x}(s)}{x(s)})=\exp(\int\limits_0^t ds \frac{d}{ds} \ln x(s))=\frac{x(t)}{x(0)}$.

The rest can be obtained by using the formula for the Laplace Transform of the derivative.