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

I am trying to determine the critical points of the ODE

$$frac{dy}{dx}=0.2x^2left(1-frac{x}{3}right).$$

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 $$x=3$$.

Mathematics Asked on November 21, 2021

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 $$(x=0)$$.

$$frac{dy}{dx}=0.2x^2left(1-frac{x}{3}right)$$

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 $$x=2.$$

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 $$x=0$$.

Answered by Henry L on November 21, 2021

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