Critical Points of $dy/dx=0.2x^2\left(1-x/3\right)$ I am trying to determine the critical points of the ODE
$$\frac{dy}{dx}=0.2x^2\left(1-\frac{x}{3}\right).$$
Setting the right-hand-side to zero gives two solutions, namely $x=0$ and $x=3$. I was wondering if the $x^2$ means that there's two critical points at $x=0$. That is, there are two critical points at $x=0$ and one critical point at $x=3$.
 A: Consider three functions plotted. What you meant was

$$y(x)=\frac{x^3-x^4/4}{15}+1  $$
is the blue integrated curve with BC $ (x=0, y=1)$. The curve function you gave  is in red, vanishing at $ (x=3, x=0, x=0 )$ where I have repeated writing $(x=0)$.
$$\frac{dy}{dx}=0.2x^2\left(1-\frac{x}{3}\right)$$
and next derivative function to detect max/min is in green.
$$\frac{d^2y}{dx^2}=\frac{x(2-x)}{5}$$
Yes, there are three critical points for $y(x).$ It is a repeated root at $x=0$. This is recognized by characteristic graph. Locally it looks like touching the x-axis tangentially going up and down in opposite ways at the ends of a short interval in a hallmark shape.
The repeated roots occur in general
$$ y(0)=a, y'(a)= 0, y''(a)= 0 $$
Which of the two types of local shape (down/up or up/down) is decided whether $ y''(x) $ goes from negative to positive or vice versa. The local functional representation is like:
$$ y=c( x-a)^2$$
where in our particular case $a=0.$ It is a maximum and minimum at this point as well as has an inflection here.
And at $x=3$, it is maximum because first derivative vanishes and second derivative  is negative as usual. Note inflection at $x=2.$
A: Since $\frac{d^2x}{dt^2}|_{x=0}=0$, it means that there is a reflection point at $x=0$, just like $y=x^3$ at $x=0$.
I won't say that there are two critical points at $x=0$, but there are certainly two roots for $\frac{dx}{dt}$ at $x=0$.
Hope this is helpful.
