Given the ODE: $$ \left\{ \begin{array} ttx'(t)=x(t)-1 \\ x(0)=1 \end{array} \right. $$
I have found that there are infinitely many solutions to the equation: $x(t)=Ct+1, C \in \Bbb R$.
I am wondering how does it agree with Picard's existence theorem. If I write $x'(t)=f(t,x(t)) \;$ where $\; f(t,x)=\frac{x-1}{t}$, then $f$ is not continuous in $0$. Can I conclude the theorem doesn't work in the other direction?