# Slope Fields with Mathematica

## Making Slope Fields by Yourself

### dy/dx = y, on the region -3 ≤ x ≤ 3, and -3 ≤ y ≤ 3

Don't cheat! If you didn't do the exercise in Mathematica before you came here to see the discussion, go back and do it now!

Assuming that you did the exercise correctly, you should have produced a picture that was simply a larger version of the following:

Slope Field for dy/dx = y

Hmmm! This is the first example we have seen since we started studying slope fields where the isoclines are not vertical lines. In fact it looks like this time the isoclines are horizontal lines.

Now, what about looking for any kind of recognizable general trend in the overall flow of the picture? It's pretty hard to say this time. Taken as a whole the picture doesn't really remind us of any of the "memorized" function graphs we're familiar with. Here's a hint: consider just the portion of the field lying above the x-axis by itself. Does it look a bit more familiar now? Kind of an exponential growth feel to it now, right?

Now consider just the lower portion of the picture taken by itself. Isn't this just the top half flipped upside down? So we're now talking about negative exponential functions. Do you see it?

Well, we fudged our way through the "what does it look like" analysis. Let's see if we can come up with something a bit more concrete to back up our analysis.

We can solve this problem by using integration, but not in quite as direct a manner as we did before. Let's go through the steps.

Starting with:

dy/dx = y

We first divide both sides by y, to get:

(1/y) (dy/dx) = 1

Then we integrate both sides individually to get:

ln|y| = x + C

A little algebra, (exponentiate both sides, rename constants, etc.) allows us to isolate y giving us:

y = A ex

At last, an explicit solution in the form of a family of curves, where we get a different curve for each different value of A. But what about that "top-half, bottom-half" business?

Well, if A takes on a positive value, we'll get a "right side up" exponential function, whereas if A takes on a negative value, we'll get an "upside down" exponential function. Can you see the connection now between this analytic solution and the slope field picture?

Well, we're done with the discussion of this problem, so let's go back to the exercises.

 If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.

ODE Laboratories: A Sabbatical Project by Christopher A. Barker

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