$C^1(\Bbb R)$ solutions of $y'=\max\{0,y^{1/3}\}$

Question

By standard computation/integration, one can find $y(x)=({2\over 3} |x|+c)^{3\over2}, c\in\Bbb R_{\ge0}$ as a class of solutions to ${dy\over dx}=y^{1\over 3}$

But is the identically zero solution the only $C^1(\Bbb R)$ solution to $y'=\max\{0,y^{1/3}\}$?

Or is the above solution ok for the problem $y'=\max\{0,y^{1/3}\}$? (Since the $y'=\sqrt{{2\over3}|x|+c}\ge0\ \forall x$ we have that $\max\{0,y^{1\over3}\}=y^{1\over3}$)

So does $\{y(x)=({2\over 3} |x|+c)^{3\over2}\ :\ c\in\Bbb R_{\ge0}\}\cup\{y(x)\equiv0\}$ contain all the $C^1(\Bbb R)$ solutions of $y'=\max\{0,y^{1/3}\}$?

Answer 1

The solutions to $y'=y^{1/3}$ is actually given by

$$y=\pm\left(\frac23x+C\right)^{3/2}$$

for $\frac23x+C\ge0$, and $y=0$ otherwise. There is also the trivial solution given simply by $y=0$ everywhere.

Extending to your problem, the condition that $y'\ge0$ removes the negative solution, leaving us with

$$y=\left(\frac23x+C\right)^{3/2}$$

for $\frac23x+C\ge0$, and $y=0$ otherwise. The trivial solution then extends to $y=C$ for any $C\le0$.