Horizontal Asymptote of a Function with no X values

Question

I am working on some simple differential equation work and it seems I've completely forgotten the rules of limits.

I have a differential equation that looks like so:

$\frac{dy}{dx} = \frac{y}{6}(4 -y)$

Plotting the slope field it's clear to see if you start at the point (0, 6) the solution function will approach $y = 4$ as $x$ tends to infinity. This makes sense, because the derivative goes to 0 at that point.

However I am having difficulty justifying this with limits:

$\lim_\limits{x \to \inf} \frac{y}{6}(4 -y) = ...$

It doesn't seem there are any substitution rules I can do here to prove this is a horizontal asymptote even though it clearly is from the graph.

What am I missing here in order to prove this to myself?

Answer 1

While you might not see $x$ explicitly, $y$ is a function of $x$,

$$\frac{dy}{dx}= \frac{y}{6}(4-y)$$

$$\int \frac{6}{y(4-y)}\, dy=\int\,dx$$

$$ \ln(4-y) - \ln (y)=-\frac23x+C$$ As $x$ goves to $\infty$, RHS goes to $-\infty$, hence LHS has to go to $-\infty$ as well. Hence $y \to 4$.

We can also represent it as

$$\frac{4-y}{y}=A\exp\left( -\frac23 x\right)$$

$$\frac4y=A\exp\left( -\frac23 x\right)+1$$

Answer 2

You are not missing anything.

The differential equation that you have $$ \frac{dy}{dx} = \frac{y}{6}(4 -y)$$

has two equilibrium points, namely $y=0$ and $y=4.$ both of which are represented by horizontal line.

The solution $y=0$ is unstable and the solution $y=4$ is stable.

The solution $y=4$ is actually an attractor which means nearby points will approach $y=4$ as you have notices with the point $(0,6).$

The solution $y=0$ is a repeller which means nearby points will scape from it.