$$x^{'} = 3 x^{\frac23}, \qquad x(0) = 0$$
The answers given are $u(t) = 0$ and $v(t) = t^3$.
I should perhaps mention that it's been a while since I did any mathematics and this was given in class today and we were asked as practice to find more solutions to this IVP. Can anyone perhaps explain why $v(t)$ and $u(t)$ are solutions and how they were found?
when $$u(t)=0 \to u'(t)=0 \to x=u(t) \to 0=3*(0)^{\frac{2}{3}}$$ and $$u(t)=t^3 \to u'(t)=3t^2 \to x=u(t) \to 3t^2=3*(t^3)^{\frac{2}{3}}$$