The Frenet-Serret frame has the following property:
$$\begin{bmatrix}T'\\N'\\ B'\end{bmatrix}=\begin{bmatrix}0&\kappa&0\\-\kappa&0&\tau\\0&-\tau&0\end{bmatrix}\begin{bmatrix}T\\N\\B\end{bmatrix}$$
I have learnt that we can usually solve O.D.Es of this form as follows $v'(s)=Mv(s)\implies v=\exp(Ms)\times v_0$ if $M$ is a matrix. But I hesitate to say that we can do that here, because my second thought, after this being my first, was that both $\kappa$ and $\tau$ are functions of $s$, the arc length, so they also change as $s$ changes so the matrix $M$ would be in flux...
Would it still be correct to write $v=\exp(Ms)\times v_0$, where $v$ is $\begin{pmatrix}T\\N\\B\end{pmatrix}$?
Or is there some other trick to solve this system for all future $s$, once the base position $v_0$ is known?
The more general "trick" is the following. Consider an ODE of the form: $$ x'(s) = M(s)x(s), \ \ \text{given }x(0) $$ where $M(s)$ is a matrix. You can solve this problem by the integrating factor technique. Just multiply the whole equation by $\exp\left(\int_0^s -M(r)dr\right)$: $$ \exp\left(\int_0^s -M(r)dr\right)x'(s)-\exp\left(\int_0^s -M(r)dr\right)M(s)x(s) = 0 $$ And notice that this resembles the product rule of the derivative: $$ \frac{d}{ds}\left(\exp\left(\int_0^s -M(r)dr\right)x(s)\right) = 0 $$ which can be solved by integrating from $0$ to any $s'$: $$ \int_0^{s'} \left(\frac{d}{ds}\left(\exp\left(\int_0^{s} -M(r)dr\right)x(s)\right)\right)ds = \exp\left(\int_0^{s'} -M(r)dr\right)x(s') - \exp\left(\int_0^0 -M(r)dr\right)x(0) = 0 $$ or equivalently: $$ x(s) = \exp\left(\int_0^s M(r)dr\right)x(0) $$ Note that if $M(r)=M$ a constant matrix, then $\exp\left(\int_0^s M(r)dr\right) = \exp(Ms)$ which is what you tried before. In general you need to compute the integral of $M(s)$.
(See the comments for some technical care one need to have when using this reasoning)