Finding initial conditions for which $\lim_{t \rightarrow \infty}y = 4$ for a differential equation

Question

I am looking for help on a problem that goes as follows:

Find the initial conditions for which $\lim_{t \rightarrow \infty}y = 4$ for the following differential equation: $$y' = y(y-4)(y+4)(y-6)$$

I tried to solve the differential equation and ended with a very large mess but found that it was not possible. Also, by looking at the fact that through a separation of variables, we would end up with $t + c$ on the left hand side so I believe that it is not possible for $\lim_{t \rightarrow \infty}y(t) = 4$ to be true. Is that correct or did i make a mistake somewhere.

I would appreciate some help on this. Thank you

Answer 1

Not like that, no need to solve anything. Notice that $0$, $4$, $-4$ and $6$ are (constant) solutions. On the other hand, solutions cannot jump over one of these because it would violate the uniqueness.

Summing up, you only need to look at what happens on the interval $(0,6)$. For $y\in(0,4)$ you have $y'>0$ (the solutions are increasing) while for $y\in(4,6)$ you have $y'<0$ (the solutions are decreasing). So $$\lim_{t\to+\infty}y(t)=4$$ if and only if $y(0)\in(0,6)$.

Answer 2

$$ f(y)=(y-6)(y-4)y(y+4) $$ is positive for $y<-4$, negative for $-4<y<0$, again positive for $0<y<4$, negative again for $4<y<6$ and positive for $y>6$.

Which means that $y=4$ is an attracting stationary point, since all solutions starting close to it move towards it, falling from above and raising from below.

Thus the only solution having value $4$ at infinity is the constant solution $y(t)\equiv 4$.

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More systematically, around the stationary point $y=4$ one has the linearization for $4+\Delta y$ $$ Δ\dot y=f(4+Δy)\simeq f(4)+f'(4)·Δy=(4-6)4(4+4)·Δy=-64·Δy $$ so that in first order $$ y(t)=4+Δy(0)·e^{-64t} $$ for small values of $Δy(0)$ which clearly converges towards $4$