$y'= y(1 − y)$, with $y(0) \in [0, 1]$. Then $\lim_{t\to\infty} y(t) = 1$

Question

Let $y(t)$ be a real valued function defined on the real line such that $y'= y(1 − y)$, with $y(0) \in [0, 1]$. Then $\lim_{t\to\infty} y(t) = 1$.

The solution is given as false .But i have no idea about that. I try some counterexample but it won't work.

How can i find solution with 3 minutes?

Answer 1

Take $y(t)=0$. Then $y(0)\in[0,1]$ and $y'=0=y(1-y)$. But $\lim_{t\to\infty}y(t)=0\neq1$.

Answer 2

$$\frac{dy}{dx} = y(1-y) \implies dx = \frac{dy}{y(1-y)}$$

so $$x+C = \ln\frac{y}{1-y} \implies Ce^x = \frac{y}{1-y}$$

and therefore $$y = \frac{Ce^x}{1+Ce^x}$$

The initial condition is

$$y(0) = \frac{C}{1+C} \implies C = \frac{y(0)}{1-y(0)}$$

so $$y = \frac{\left(\frac{y(0)}{1-y(0)}\right)e^x}{1+\left(\frac{y(0)}{1-y(0)}\right)e^x}$$

If we pick $y(0) = 0$ then $y = 0$ so $\lim_{t\to\infty} y(t) = 0 \ne 1$.

Answer 3

One way is to solve this logistic equation and find the limit, but it might take more than $3$ minutes.

The other way is qualitative analysis.

Find the equilibrium states which are $y=0$ and $y=1$

Note that if $ y_0 \in (0,1) $ then your $y'=y(1-y)>0$ and it stays positive, so your function increases and approaches the equilibrium state of $y=1$.

Note that this argument fails if $y_0 =0$ because then you stay at $y=0$ for ever and you do not approach $1$