the shortest methot of solving $y' = 3y^{2/3} \quad y(0) = 0$?

Question

I have this equation from my pre-exam:

$$y' = 3y^{2/3} \quad y(0) = 0$$

For the first look it reminds Cauchy equation wit all the following steps, but my tutor told me that this is actually a joke, it can be solved without any complex calculus, but I still can't find the way, any ideas?

Answer 1

Look at the answer of Holo. Then you should obtain the solution $y(x)=x^3$. But there are a lot of other solutions:

$y(x)=0$ is a solution;

if $a>0$ , then a further solution is given by

$y(x)=0$ if $x \le a$ and $y(x)=(x-a)^3$ if $x \ge a$.

Answer 2

$$y' = 3y^{2/3}, y(0) = 0\\\dfrac{dy}{dx}=3y^{2/3}\\\dfrac1{dx}=3y^{2/3}\dfrac1{dy}\\dx=\frac1{3y^{2/3}}dy\\\int dx=\int\frac1{3y^{2/3}}dy$$ now after integration isolate $y$ and find the constant using $y(0)=0$

Answer 3

The question is if you have to find all solution or just one. You should be capable to see without any calculation that $y\equiv0$ is a solution of this problem as $0$ is a root of the right side and thus a stationary point.