Initial value problem!

Question

richyw · Answer

Prove that the IVP $x'=x^{2/3},\quad x(0)=0$ has infinitely many solutions. 

So $x(t)=0$ is a solution. Now solving I get
$$\frac{dx}{dt}=x^{2/3}$$$$x^{-2/3}dx=dt$$$$x=\left(\frac{t+c}{3}\right)^3$$

richyw · Answer

my solution tells me that for any $a>0$, the function 
$x(t)=0$ if $t\leq a$
$$x(t)=\left(\frac{t-a}{3}\right)^3\text{  if  }t\geq a $$
is also a solution of the IVP. Could someone please explain this to me?

anonymous · Answer

I think it should be "if t > a," and not "t ≥ a." The "or equal to" part would contradict the first solution (x(t) = 0 if t ≤ a). I'd say it's a typo in your textbook or instructor's mistake.

richyw · Answer

ok so even with that (and it is that in the solution, I double checked). I still don't see how this is proving anything.

anonymous · Answer

Hmm, maybe someone else can shed some light on this? "Proving" in the context of diff.eq.'s isn't my forte.