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

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.

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

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 :

• the slope field shows that it follows the linear equation $t-2$ for when $y=0$. How does this relate to $t-3$ – Varun Iyer Aug 11 '14 at 21:56
• it can sing and dance as it wishes in the beginning as the exponential term is dominant for small values of t. the asymptotic behavior tells us about how the long term solution looks like. if we wait long enough, all the solution curves approach this long term solution : $y = t-3$ – AgentS Aug 11 '14 at 22:00
• $t-3$ is already nicely seen in that picture. And it forms kind of a separation curve where the solutions with $c_1>0$ and $c_1 <0$ strive to. – mvw Aug 11 '14 at 22:04
• @VarunIyer Perhaps is not evident from the picture, but all the lines should be arrows, pointing towards the line $y=t-3$. This is a key point, because it shows that all the trajectories are approaching that line. – bartgol Mar 11 '16 at 22:01

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.

• How would you solve this only given a slope field? It seems trivial if you are given the solution to the differential equation. – Varun Iyer Aug 11 '14 at 21:57
• I would solve the differential equation. – mvw Aug 11 '14 at 21:58
• I would do the same thing, but would you know how to see this using only the slope field (not just for this problem, but others as well) – Varun Iyer Aug 11 '14 at 21:59
• You need to find curves in the slope field, where the slopes are tangents to the curves. For water currents the slopes would be the velocity vectors of the moving water. Your task would be to find the trajectories of test particles thrown in the water. In fact some numerical solvers would work like this. – mvw Aug 11 '14 at 22:01