Let's consider a one-dimensional physical system ($x$ is the position and $t$ is time) described by a first order differential equation.
I'm aware of the fact that if the equation can be put in normal form: \begin{equation} \dot{x}=f(x,t) \end{equation} the Cauchy problem, provided some mathematical hypotheses are verified, tells us that the equation admits the existence of an unique solution. Therefore the system is deterministic, i.e., given the initial conditions there is only one admissible motion.
I was wondering what happens when the system is described by an equation that cannot be put in normal form: \begin{equation} g(\dot{x},x,t)=0 \end{equation} Is there anything that proves that the Cauchy problem still provides the existence of an unique solution for an equation not in normal form or are there specific conditions?
For the function $g(v,x,t)$ consider the set $D=\{(v,x,t):\partial_vg(v,x,t)\ne 0\}$. On this set the equation is locally solvable as $v=f(x,t)$ which gives an ODE $\dot x=f(x,t)$. In the complement of this set the equation does not depend on $v$, it is just a functional equation in $x,t$ so that the implicit function theorem again applies in telling you if $x$ is locally a function of $t$.
More generally, if $x$ is a vector, if the implicit equation can not be transformed into an explicit ODE, the next best thing you can hope for is that it is a differential-algebraic equation (DAE). This would be the case if one of the closely related index definitions gives a constant index over some open domain.