Solution for $\frac{dx}{dt} = 1+x^{1/3}; \, x(0)=c$ where $1+x^{1/3}$ is not Lipschitz's

Question

I want to ask that if $f$ is not Lipschitz's and also $f(x)$ is not positive for all $x$ then how can we prove or disprove the uniqueness of solution?

Particularly I am interested to check the uniqueness of the solution of below IVP

$$\dfrac{dx}{dt} = 1+x^{1/3}; \quad x(0)=c$$ and here $1+x^{1/3}$ is not Lipschitz's.

Answer 1

Regarding the differential equation $$\dfrac{dx}{dt} = 1+x^{1/3}; \quad x(0)=c$$ We do not have the Lipschitz's condition to guarantee the uniqueness.

Unfortunately there is no theorem to determine the existence of multiple solutions either.

Note that the differential equation is separable so we can solve it and get the general solution.

We also have the constant solution of $$x(t) =-1$$ if $$x(0)= c=-1.$$

Thus the uniqueness depends on the initial condition.