Critical Points of $dy/dx=0.2x^2left(1-x/3right)$

I am trying to determine the critical points of the ODE


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 $x=3$.

Mathematics Asked on November 21, 2021

2 Answers

2 Answers

Consider three functions plotted. What you meant was

enter image description here

$$y(x)=frac{x^3-x^4/4}{15}+1 $$

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 $(x=0)$.


and next derivative function to detect max/min is in green.


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 enter image description heretouching 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 $x=2.$

Answered by Narasimham on November 21, 2021

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.

Answered by Henry L on November 21, 2021

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