The IVP, $x'(t)=3x^{3/2};x(0)=0$ in an interval arount $t=0$ has a

The IVP, $ \dot x(t)=x^{3/2};x(0)=0$ in an interval arount $ t=0$ has a

(a)No solution

(b)Unique solution

(c)Finitely many solutions

(d)Infinitely many solutions.

Solution:- Applying Picard’s -Lindelof Uniqueness and existence theorem. $ f(x,t)=3x^{3/2}$ , Will it be continuous at $ (0,0)$ ? When $ x<0$ , $ f(x,t)$ no more real valued. So, Discontinuous at $ (0,0)$ . So, I can not judge from the theorem. I solved using variable separable form.

I got the solution, $ 3x(t)^{1/3}=t+c$ . When I apply initial condition, I get $ 3x(t)^{1/3}=t$ . A unique solution. But in the answer key it is written that equation has infinitely many solutions. How it is possible? Please explain.