I'm working through Jacobi's formula, and I'm really stuck with solving an ODE and would appreciate someone helping me out, as some of my basic maths is a bit rusty.
Essentially proving this corollary. Taking
$$\frac{d}{dt} \det e^{tB} =\operatorname{tr}(B) \det e^{tB}$$
to the solution
$$\det e^{tB} = e^{\operatorname{tr} \left(tB\right)}$$
I assume this boils down to a simple first order ODE
$$\dfrac {d}{dt}\left( y\left( t\right) \right) = q\left( t\right) y\left( t\right)$$
but can't seem to make it work.
Let $f: \Bbb{R} \to \Bbb{R}$ be the function defined by $f(t) = \det e^{tB}$. Suppose you already manage to establish that $f$ satisfies the ODE \begin{align} f'(t) &= \text{tr}(B)\cdot f(t) \end{align} Solving this is pretty easy; in fact this is the protoypical example of a separable ODE. The solution to this ODE is just \begin{align} f(t) &= f(0) \cdot e^{\text{tr}(B) \cdot t} \\ &= \det(I) \cdot e^{\text{tr}(tB)} \\ &= e^{\text{tr}(tB)} \end{align}
And just to answer your other question, the solution to \begin{align} y'(t) = q(t)\cdot y(t) \end{align} subject to the initial conditions $y(t_0) = y_0$ is given by \begin{align} y(t) = y_0\exp\left( \int_{t_0}^t q(s) \, ds \right) \end{align}
The "usual" way of justifying this is to divide both sides by $y(t)$ and then integrate: \begin{align} \dfrac{y'(t)}{y(t)} &= q(t) \\\\ \int \dfrac{y'(t)}{y(t)} \, dt &= \int q(t) \, dt \\ \ln \big( y(t) \big) &= \int q(t) \, dt \end{align} Now exponentiate both sides. I've purposely been sloppy with this method regarding constants of integration, because there is already a problem with this approach: it implicitly assumes $y(t)$ is no-where vanishing (because otherwise we cannot divide by $y(t)$). But hopefully you see where the solution "comes from". A more complete way which I like for solving this ODE is to just recognise that the formula for $y(t)$ I gave above is actually a solution, and then use a uniqueness argument to conclude there aren't anymore.