Verifying a function is a solution of $y' = 3 y^{2/3}$ Let $\alpha \in \mathbb{R}$ and consider $y = (x-\alpha)^3 {\bf 1}_{(\alpha, \infty)}$. I want to verify that $y' = 3 y^{2/3}$. Well, clearly, 
$$ y' = 3(x-\alpha)^2 = 3 [(x-\alpha)^3]^{2/3} = 3 y^{2/3} $$
and if $x \leq \alpha $, then $y = 0$ and so $y' = 0$ and the solution is satisfied trivially.
Im confused as to why do we need to verify the case $x=\alpha$ separetely as my notes says: It says that we check left- and right- hand derivates to see that y(x) at $x= \alpha$ satisfies the ODE.
Can someone clarify this to me? My understanding is that my work on the third line covers this case. What am I missing?
 A: When you write $y = (x-\alpha)^3 {\bf 1}_{(c, \infty)}$ it should be $y = (x-\alpha)^3 {\bf 1}_{(\alpha, \infty)}$.  The point of the ${\bf 1}_{(\alpha, \infty)}$ is that it is $0$ for $x \le \alpha$ and $1$ for $x \gt \alpha$.  Because of the joint in the definition at $x=\alpha$ you need to check that the solution is even differentiable at that point.  The product of differentiable functions is differentiable, but ${\bf 1}_{(\alpha, \infty)}$ is not continuous, let alone differentiable at $\alpha$.  The $(x-\alpha)^3$ factor cures this, but you are asked to demonstrate that.  You can do that by showing that the right hand function approaches $0$ and its derivative approaches $0$ at $\alpha$.
A: Perhaps it may help to consider a different example.  Let $y = x {\bf 1}_{(0,\infty)}$.  Does this satisfy $y' - (y')^2 = 0$?  For $x > 0$, yes: $y' = 1$ and $1 - 1^2 = 0$.
For $x < 0$ yes: $y' = 0$ and $0 - 0^2 = 0$.  For $x = 0$, no: $y$ is not differentiable at $x=0$, because the difference quotients $\dfrac{y(h) - y(0)}{h - 0}$ are $1$ for $h > 0$ and $0$ for $h < 0$,  and the limit as $h \to 0$  does not exist.  
The fact that $x=0$ belongs to the interval $(-\infty,0]$ where $y=0$ is not enough to tell you that $y'(0)=0$.  That derivative depends on what happens as $x \to 0$ both from the right and from the left.
