Clarification on the definition of General Solution Given the differential equation $\frac{dy}{dx}=3{y^\frac{2}{3}}$, the general solution is $\sqrt[3]{y}=x+C$. But, there are two solutions curves that pass through (2,0) namely, $\sqrt[3]{y}=x-2$ and $y=0$.
Why do we call $\sqrt[3]{y}=x+C$ a general solution?  I thought "general solution" meant that it describes all solution curves, one for each value of $C$.  But it clearly doesn't, as $y=0$ is a solution curve and it's not included.
I believe I need help defining a "general solution". This seems to be glossed over in the textbook I am reading.
 A: The  differential equation $\frac{dy}{dx}=3{y^\frac{2}{3}} \tag{i}$ is separable, that means you can seperate the variables by dividing $(i)$ by $\frac{y^\frac{2}{3}}{dx}$ but while doing so, you made a tacit assumption that $y^\frac{2}{3}\neq0$.
Now regarding $y$ as the dependent variable we consider the situation that occurs if $y^\frac{2}{3}=0$ i.e. $y=0$ and we notice that $y=0$ is indeed a solution of $(i)$. But this $y=0$ is not a member of one parameter family of solution you obtained (with that assumption) for $(i)$. Therefore, we conclude that it is a solution which was lost in the separation process.
Always remember while separating the variables to check if any solutions are lost in the process due to the assumption that any factor by which we divide is not zero.
As such your general solution would be $\sqrt[3]{y}=x+C$ or $y=0$ where C is an arbitrary constant.
$\mathbf{Note:}$
In elementary texts, this lost solution $y=0$ is often ignored.
A: The answer of Aman is incomplete, but I cannot explain myself reasonably in just a few comments so I will post an answer.
There are many more solutions that flip from the unstable $y=0$ to $y= (x-x_0)^3$ at an arbitrary point $x_0\in\mathbb R$, which parameterises the family of solutions $(y^{x_0})_{x_0\in\mathbb R}$:
$$y^{x_0}(x) := \begin{cases} 0 &x\le x_0\\
(x-x_0)^3 &x>x_0\end{cases}$$
In fact you could also allow it to have the cubic behavior at $x=-\infty$ and then just flatten out once it reaches $y=0$:
$$y_{x_0}(x) := \begin{cases} 
(x-x_0)^3 &x\le x_0 \\ 0 &x> x_0\end{cases}$$
Note that $y_{2}$ also solves $y(2)=0$. There's also $y_{x_1}^{x_2}$ for $x_1<x_2$... I'll leave the enumeration of all possible solutions as an exercise in book keeping.
I say that the $y=0$ solution is "unstable" because if you perturb it to be positive at some point, then it must follow the cubic solution from that point on (interpreting $x$ as time marching forward.) Intuitively the $y^{2/3}$ term amplifies small errors near $0$, as for $t\in(0,1)$, $t^{2/3}>t$.
questions you might have from reading comments (not all were directed at me)

*

*Is there some initial condition given with the differential equation in the question?

Yes, but besides the point. It is given that $(2,0)$ is on the graph of the solution, i.e. $y(2)=0$. Note that giving an initial condition does not give uniqueness, and it shouldn't be expected anyway because this ODE fails to satisfy the assumptions of Picard's theorem (or even the Osgood lemma.)


*Is this a "solution" that is not even differentiable?

Weak solutions are a very interesting and useful idea, but it turns out the above solution is differentiable. It is IMO a good exercise to show by direct verification  that in fact they are $C^{2,1}$, i.e. twice differentiable with Lipschitz second derivative. (check directly from the definition at the transition point $x=x_0$.) In fact any solution to $y'=y^{2/3}$, which I meant precisely that any differentiable function $y$ satisfying the ODE at every $x$, must also be continuous. Note that this is different from $y'+y=f$ where $f$ is not continuous (as in the linked post.) In fact, if $f$ has a jump discontinuity, then $y$ cannot be differentiable as derivatives satisfy the intermediate value property.
tl;dr the conclusion of Aman is correct but misses some technical details. I agree that usually, textbooks at this level say "general solution" to mean "the family of solutions you can find via separation of variables".
