Toronto Math Forum

MAT244--2018F => MAT244--Lectures & Home Assignments => Topic started by: Nan Choi on September 20, 2018, 04:26:38 AM

Title: Describing Direction Fields
Post by: Nan Choi on September 20, 2018, 04:26:38 AM
To what extent are we expected to describe direction fields?

For example, consider the differential equation from the homework question:

Section 1.1 #28

$y' = e^{-t} + y$

I used the online ODE plotter to determine its behaviour.

As $t → ∞$, $y$ diverges from $0$.
$y$ approaches $∞$ when $t > 0$, and $y$ approaches $-∞$ when $t < 0$.
At $t = 0$ and $y = -1$, the value for $y'$ is zero and the slope line is horizontal.

The question asks us to determine the behaviour of $y$ as $t → ∞$, and whether this behaviour depends on the initial value of $y$ at $t = 0$. Is the explanation above enough?
Title: Re: Describing Direction Fields
Post by: Victor Ivrii on September 20, 2018, 04:37:26 AM
In this case the best would be just to solve this equation analytically.
Title: Re: Describing Direction Fields
Post by: Zhihong Yin on September 22, 2018, 01:51:48 PM
For solving this equation analytically, does it mean that we need to analyze the the equation without just based on the calculation? As y approaches to infinite, should we show the limit?
Incomprehensible. V.I.
Title: Re: Describing Direction Fields
Post by: Nick Callow on September 23, 2018, 01:18:13 AM
A possible analytical approach.

1. If we want to describe the behaviour of $y(t)$ as $t \rightarrow \infty$ then we need to actually find out what $y(t)$ is. We can do this through the method of integrating factors.

Given $y'(t) - y(t) = e^{-t}$,  let $\mu(t)$ represent the integrating factor. $\mu(t) = e^{\int a(t)dt}$, where $a(t) = -1$. Therefore, 
$$ \mu(t) = e^{\int -1dt}, \qquad \mu(t)  = e^{-t}, \qquad y'(t) - y(t) = e^{-t}$$
$e^{-t}(y'(t) - y(t)) = e^{-2t}$
$\frac{d}{dt}(e^{-t}y(t)) = e^{-2t}$
$\int \frac{d}{dt}(e^{-t}y(t))dt = \int e^{-2t}dt$
$e^{-t}y(t) + c_1 = -\frac{1}{2}e^{-2t} + c_2$
$e^{-t}y(t) = -\frac{1}{2}e^{-2t} + k$
$$y(t) = \frac{-\frac{1}{2}e^{-2t} + k}{e^{-t}} = -\frac{1}{2}e^{-t} + k e^t$$

Now that we have an equation for $y(t)$, we can describe its behaviour.

2. Zhihong I think this is what you were talking about where we could find the limit:
$$
\lim_{t \rightarrow \infty} \frac{-\frac{1}{2}e^{-2t} + k}{e^{-t}}
$$
which will depend on what we select $k$ to be. You can then see for what values of $k$ how the function $y(t)$ behaves and this should of course agree with the direction field but be more analytical and accurate.
Title: Re: Describing Direction Fields
Post by: Victor Ivrii on September 23, 2018, 03:47:04 AM
One can write formulae with better (larger) vertical spacing and in more compact form. There are inline formulae (in single dollars) and display formulae (in double dollars, which is the simplest, but  deprecated way).

We will find later that the  behaviour as $t\to \pm \infty$ could be often determined without actually solving equations.