Question

What is the horizontal asymptote for y(t) for the differential equation dy dt equals the product of 2 times y and the quantity 1 minus y over 8 ? (3 points)
y equals one fourth
y = 2
y = 4
y = 8

Answer 1

First, we need to solve the differential equation.
$\frac{d}{dt}\left(y\right)=2y\left(1-\frac{y}{8}\right)$
This a separable ODE. We can rewrite it like this:
$-\frac{4}{y^2-8y}{dy}=dt$
Now we integrate both sides.
$\int \:-\frac{4}{y^2-8y}dy=\int \:dt$
We get:
$\frac{1}{2}\ln \left|\frac{y-4}{4}+1\right|-\frac{1}{2}\ln \left|\frac{y-4}{4}-1\right|=t+c_1$
When we solve for y we get our solution:
$y=\frac{8e^{c_1+2t}}{e^{c_1+2t}-1}$
To find out if we have any horizontal asymptotes we must find the limits as x goes to infinity and minus infinity. 
It is easy to see that when x goes to minus infinity our function goes to zero.
When x goes to plus infinity we have the following:
$\lim_{x\to\infty} f(x) =y=\frac{8e^{c_1+\infty}}{e^{c_1+\infty}-1} = 8$
When you are calculating limits like this you always look at the fastest growing function in denominator and numerator and then act like they are constants. 
So our asymptote is at y=8.