### Author Topic: Describing Direction Fields  (Read 1927 times)

#### Nan Choi

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##### Describing Direction Fields
« 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?

#### Victor Ivrii

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##### Re: Describing Direction Fields
« Reply #1 on: September 20, 2018, 04:37:26 AM »
In this case the best would be just to solve this equation analytically.

#### Zhihong Yin

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##### Re: Describing Direction Fields
« Reply #2 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.
« Last Edit: September 22, 2018, 02:43:36 PM by Victor Ivrii »

#### Nick Callow

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##### Re: Describing Direction Fields
« Reply #3 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.
« Last Edit: September 23, 2018, 03:50:28 AM by Victor Ivrii »

#### Victor Ivrii

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