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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. –  Michael Boratko Aug 23 '12 at 20:31
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3 Answers

The general solution to $y' = 3y^\frac{2}{3}$ is $y = \frac{1}{27}(3x+c)^3$. With the initial condition, we get that $0 = \frac{1}{27}c^3$, so $c = 0$. So yes, the only solutions are the trivial one and $y = x^3$.

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-1, incorrect. See the answer by @kiwi. –  user31373 Aug 23 '12 at 22:02
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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.

In connection with initial problems it is reasonable to consider two solutions as one and the same if they coincide in a neighborhood of the initial point (proceeding to the respective equivalence classes one then talks about germs of solutions). In our example there are exactly four different germs.

Note that the phenomenon observed here is not pathological. It turns up whenever we have an envelope to a given family of curves.

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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.

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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
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