For the schrodinger picture if you take a state $|\psi(t)\rangle$ satisfying the schrodinger equation $i\hbar\frac{\partial }{\partial t}|\psi(t)\rangle=H|\psi(t)\rangle$, and write a time evolution operator U such that $|\psi(t)\rangle=U(t,t_0)|\psi(t_0)\rangle$, then this gives an operator equation that $U$ must satisfy namely:
$$i\hbar\frac{\partial }{\partial t}U(t,t_0)=HU(t,t_0)$$
The initial condition is that $U(t_0,t_0)=1$.
For $H$ time dependent $U$ is given by the integral equation
$$U(t,t_0)=1+\frac{-i}{\hbar}\int_{t_0}^t H(t')U(t',t_0)\,dt'$$
This is formally solved by writing $U$ as a time ordered exponential:
$$U(t,t_0)=\mathcal{T}\exp\left(\frac{-i}{\hbar}\int_{t_0}^t H(t')\,dt'\right)=1+\frac{-i}{\hbar}\int_{t_0}^t H(t')\,dt'+\frac{-i}{\hbar}\int_{t_0}^{t}\int_{t_0}^{t'} H(t') H(t'')\,dt''\,dt'\ldots$$
The above is the formal solution obtained by subbing in $U$ for itself in the integral equation, but I believe it's in the second volume of Reed and Simon, in their section on time dependent operators, they show some cases where this solution is exact, e.g. if $H$ is a bounded operator, and certain cases of the form $H(t)=H_0+V(t)$, the latter case is done in the interacting picture however.
The dyson series is discussed in Messiah's book on QM, as I mentioned Reed and Simon volume $2$ for a bit more rigour, and there's a nice passage in a book I'm reading by Bohm, Mostafazadeh, Koizumi, Niu, Zwanziger called "The Geometric Phase in Quantum Systems" which discusses it quite nicely also.