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 Quote by Mårten But, hm... In my control theory text book, it says that when using Laplace transforms in order to more easily handle ODEs, you always assume that $y^{(n)}(0)=0$ for all n. This in order to get the manageable form $Y(s) = G(s)U(s)$. All the calculations in my book assumes this.
well, sure. otherwise you get $Y(s) = G(s)U(s)$ + some other stuff.

 As I understood it, if you not assume this, the whole point with using Laplace transforms is lost, because then you cannot easily multiply the different boxes around the feedback loop for instance. But what I haven't understood yet, is whether the state space representation method is limited to this as well or not (that all the initial values, each of them, are set to 0)? Are you instead allowed to use an initial value state vector X(t_0), as for instance, the one I used above in an earlier message?
sure, if you want to represent the (transformed) output as only the transfer function times a single input, then you have to assume that the system is completely "relaxed" at time 0. but you can model a system in terms of only the input and output (and their initial conditions) with the Laplace Transforms of the input and output. but it won't be a simple transfer function.

the main difference between this and the state-space model is that the state-space model is trying to model what is going on inside the box. it is more general than the simple input-output description of the system.