Asymptotic Behavior of the solution of the DE $y'=-2-y+t$ for $t \to \infty$

Question

I'm new to differential equations, so any help will be grateful.

I've been looking at this problem:

Examine the slope field of the following differential equation. Based on the direction field, determine the behavior of $y$ as $t→∞$. $$y' = -2 -y + t$$

After plotting the slope field, I could not seem to deduce anything, so curiously, I looked at the solution and this is what it stated:

y is asymptotic to $t − 3$ as $t→∞$.

Could someone explain to me what they mean by "asymptotic to ..." My textbook didn't give me an example regarding this type of question so I do not understand what it means.

NOTE I've noticed that the solution to this differential equation is:

$$y = c_1e^{-t} + t -3$$

$t-3$ is apparent in the solution.

Could this be in connection with the solution that they provided?

Again help will be appreciated. Thanks in advance.

Answer 1

The asymptotic behaviour of a function $f(x)$ is about how it looks like for very large $x$ arguments. So $f(x)= 1/x$ would go to zero.

In your solution the $e^{-t}$ goes to zero for large $t$, so $$ y(t) = \underbrace{c_1 e^{t}}_{\to 0} + t -3 \to t-3 $$ for $t\to \infty$. So that non-linear solution $y(t)$ approximates the linear function $t-3$ more and more, the larger $t$ gets.

Answer 2

$y = c_1e^{-t} + t -3$

Notice the negative exponent makes the first term vanish as $t\to \infty$ and the solution almost looks same as $y=(\approx 0) + t-3$

slope field gives the same picture :