Does this Cauchy problem $x'=tx^{2/3}$, $x(0)=0$ have unique solution?

Question

Consider the following Cauchy problem: $$ \left\{ \begin{array}{lcc} x'=tx^{2/3} \\ \\ x(0)=0 \\ \end{array} \right. $$ I proceed as usual. First we notice that $x(t)=0 \ \forall t \in \mathbb{R}$ is a constant solution of the separable equation. If we suppose that $x\neq 0$ we get that $$ x^{-2/3}x'=t $$ And integrating with respect to $t$ we get: $$ 3x^{1/3}=\frac{t^2}{2}+C_1 \\ x(t)=\left(\frac{t^2}{6}+C_2\right)^3 $$ with $C$ constant. Here is my doubt: in the definition of solution that was given to me we have to specify an open interval $I$ where $x(t)$ is defined. As I want to solve the Cauchy problem, $0$ must be in $I$. However, in that case $x$ would be zero and to obtain the second solution I imposed that $x\neq 0$. How many solutions does this problem have? I would say only one, the constant solution, but I'm not sure if I should include the second one (to get the constant $C_2$ just substitute and we get that $C_2=0$)