Given a first order ode \(y^{\prime }=f\left ( x,y\right ) \) (where \(y\) enters the ode as nonlinear, for example \(y^{2}\) or \(\frac {1}{y}\)) and with initial conditions \(y\left ( x_{0}\right ) =y_{0}\) then we say a solution exists somewhere in vicinity of initial point \(\left ( x_{0},y_{0}\right ) \) if \(f\left ( x,y\right ) \) is continuous at \(\left ( x_{0},y_{0}\right ) \). But we do not know yet if there is only one solution or infinite number of solutions. If \(f\left ( x,y\right ) \) is not continuous at \(\left ( x_{0},y_{0}\right ) \) then we say the theory does not apply and we do not do the next check. Solution could still exist and even be unique, but theory does not say anything about this.
If we found that \(f\left ( x,y\right ) \) is continuous at \(\left ( x_{0},y_{0}\right ) \) then now we check if \(f_{y}\left ( x,y\right ) \) is also continuous at \(\left ( x_{0},y_{0}\right ) \). If it is, then we say there is only one solution curve (i.e. a unique solution) that passes through the initial point \(\left ( x_{0},y_{0}\right ) \) and in some region around it.
If \(f_{y}\left ( x,y\right ) \) turns out not to be continuous at \(\left ( x_{0},y_{0}\right ) \) then theory does not guarantee uniqueness. Solution could still be unique but theory does not say anything about this. We have to solve the ode to find out.