# How to solve the differential equation $y'=3y^{\frac{2}{3}}$, $y(0)=0$

I found the ODE $\displaystyle y'=3y^{\frac{2}{3}}$ under the assumption of $y(0)=0$ somewhere and tried to solve it.

I think there are infinitely many solutions to the problem but couldn't find more than $y=x^3$ and $y=0$.

Can you verify that there are infinitely many solutions or that there aren't?

Thanks for the help!

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I can't agree with that – Tomer Galanti Aug 23 '12 at 20:18
Sry, you are right – Simon Markett Aug 23 '12 at 20:19
This is known as a Bernoulli differential equation. – process91 Aug 23 '12 at 20:31

This is a standard example in connection with uniqueness of solutions to initial value problems. The initial value problem $$y'=3y^{2/3}\ ,\quad y(0)=0$$ does not satisfy the essential technical assumption of the existence and uniqueness theorem, because $$\lim_{y\to0}{|y|^{2/3}\over |y|}=\infty\ .$$ Therefore we cannot expect a unique solution. As other contributors have noted the functions $x\to0 \ (x\in{\mathbb R})$ and $x\to x^3 \ (x\in{\mathbb R})$ are solutions; and as the differential equation is "$x$-free" an infinity of further solutions can be "spliced" using these and their translates.
You can obtain infinitely many solutions by splicing the two already given $\dots$ $$y=\cases{0,& x\le a\\(x-a)^3,& x>a}$$ for any $a\ge 0$. There are lots of other combinations you can make also.
Lots of others? Well, you could have $$y = \cases{(x-b)^3 & x \le b \cr 0 & x > b\cr}$$ for $b \le 0$ or $$y = \cases{(x - b)^3 & x \le b\cr 0 & b < x \le a\cr (x - a)^3 & x > a\cr}$$ where $b \le 0 \le a$, but I think that's it. – Robert Israel Aug 24 '12 at 6:36