MAT244-2014F > MAT244 Math--Lectures

Problem 1.1.28

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Joseph Mandozzi:
Hello all!

Problem: For y'= e-t + y . Draw the direction field. And determine the behavior of y, as t approaches + infinity.

Here is the directional field:

If we notice as t-> infinity, if y>0 then y-> infinity, if y<0 y-> negative infinity.

If we notice the x-axis, y=0
Thus, all values (t,0), have positive slopes. Thus as t-> infinity y-> positive infinity if at some point y=0

However the solution says there exists an interval curve in which as t-> infinity y->0 . However, as I look at the directional fields as t-> infinity. If y>0 it goes to positive infinity eventually, if y=0 it goes to positive infinity eventually, and if y<0 it goes to negative infinity.

So what integral curve would exhibit such property that as t-> infinity y-> 0 ?

Also for this problem is there a way to analytically find which values of y(0), will give infinity, 0, or negative infinity as t-> infinity. Because graphically, we can see it is when y(0)>- 0.5 (approximate).  But is there a way to do this algebraically?

Thanks.

Victor Ivrii:
Observe that this is 1st order linear equation which could be solved analytically (not algebraically):
\begin{equation}
y'=e^{-t}+y.
\end{equation}
Solving it you will easily find a solution $y(t)$ s.t. $y(t)\to 0$ as $t\to +\infty$ (this solution separate solutions tending to $+\infty$ and to $-\infty$ as $t\to +\infty$.

Joseph Mandozzi:
I understand the solution is y(t)=cet -0.5e-t

And if c=0... then y(t)-> 0 as t->infinity

Is there a way to tell this from the directional field? Because when y=0, or is near 0 (approaching 0 from the left or right). The directional field points away from y=0. So I don't understand how it can approach 0, if the direction of the field is sending it in a different direction?


Thanks for the reply!

Victor Ivrii:

--- Quote from: Joseph Mandozzi on September 14, 2014, 09:47:07 PM ---I understand the solution is y(t)=cet -0.5e-t

And if c=0... then y(t)-> 0 as t->infinity

Is there a way to tell this from the directional field? Because when y=0, or is near 0 (approaching 0 from the left or right). The directional field points away from y=0. So I don't understand how it can approach 0, if the direction of the field is sending it in a different direction?


Thanks for the reply!

--- End quote ---

In some sense you are right: this solution is unstable. What does it mean we will learn later in Chapter 9.

Xinyi Li:
Hi, I think you have mixed up something here.
For the directional field, when y = 0, if you let t goes from -infinity to +infinity,
the slope y' basically follow the function e^(-t) which finally approach 0 at some point.

That means for function y, although it keep increasing but the rate of increasing is keep getting smaller. Just like the -e^(-t) function, it keep increasing for t increases, but finally it approaches 0.You can definitely see it approaches 0 though the directional filed graph without any calculation of the solution. I don't see any confusion here.

Thanks

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