$y'=y(y-1)(y-2)$

Question

Consider the differential equation $$y'=y(y-1)(y-2)$$ Which of the following statements is true ?

1. If $y(0)=0.5$ then y is decreasing.
2. If $y(0)=1.2$ then y is increasing.
3. If $y(0)=2.5$ then y is unbounded.
4. If $y(0)<0$ then y is bounded below.

My Attempt: I have solved the given differential equation by variable separable method and got it $$\frac{y(y-2)}{(y-1)^2}=ce^{2x}$$ where c is arbitrary constant. Furthermore, please guide me! Or suggest other way to tackle this question. Thanks ☺

Answer 1

You can study the behaviour of the solutions. First of all, there are three constant solutions, which correspond to the roots of the polynomial $$y(y-1)(y-2),$$ and these are the horizontal lines $y=0$, $y=1$ and $y=2$. By uniqueness of the solutions you know that other solutions do not intersect these three lines. Hence, for $0<y<1$ you have $y>0$, $y-1 < 0$ and $y-2 < 0$, thus $y'=y(y-1)(y-2)$ is positive, so solutions between the lines $y=0$ and $y=1$ are increasing. This implies the first statement is false. Can you see the others using a similar reasoning?

Answer 2

I had trouble getting the formulas correct, so I am showing solutions for the related $$ y' = y^3 - y = (y-1)(y+1)y $$ When an ODE has constant solutions, this is what happens: the inner curves are asymptotic to the constants. Then, when the degree of the polynomial is at least two, the curves in the two outer regions blow up to infinity in finite time.

Same thing for $$ y' = y^2 - 1 = (y-1)(y+1) $$