4.2.1

Instructor: Example one, create a direction field for the differential equation y prime equals x squared minus y squared and sketch a solution curve passing to the point negative 1, 2. Use x and y values ranging from negative 5 to 5. I'm going to be up front here and say I'm not going to do a complete created direction field from scratch. 

I mean, I've got one created here from technology here already, but I am going to explain to you sort of the process we would go through to complete this well. First thing I would do is notice that x squared-- well, the y prime equals 0 whenever x squared equals y squared or when x equals plus or minus y. That's when our derivative is 0, so the direction field actually shows what the slope is at every point. So at any point where x and y are either the same or opposite, as it is, then we're going to have slope of 0. 

I'll just draw that on here, overlaying that here. Whenever x and y are the same, that would be points like right here at the origin, or at 1, 1, or at 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, et cetera on that line. And also since they are opposites, that would also be-- when they're opposites, it's equal to 0 as well. That's also down this direction, and that actually continues in both directions, if you notice. 

All of those points on either of those two lines we have a slope of 0. So if you notice, we have horizontal tangents or horizontal line segments drawn there. They're very small line segments, and they represent what's happening at the center of that segment. That's when the slope is 0. 

Now, from there, I would take some other points, and we're going to make a table. I want to take some other points. So I've actually thought, OK, what about 1, 2? So that's above the 1, 1 line. That would give me a representative point, and I would do this for a whole bunch of points, but I'm going to take the 0.12, and my y prime value is x squared minus y squared. 

That will be 1 squared minus 2 squared, or that's 1, and that would be minus 4, so negative 3. My slope at that point is negative 3, so I would draw a line segment at the point 1, 2 that has a slope of negative 3. And if you notice, that segment is already drawn there. It has a slope of negative 3. 

Now let's try another point. Just again, because of symmetry, we can actually determine some other values from these, but I'm not going to do 1,000 of them. Just a few. How about 2, 3? That would be 2 squared minus 3 squared. 

That would be 4 minus negative 5, and, again, if we look at our graph moving to the 0.23, you have something that has a slightly steeper slope of negative 5, or so it appears. All right, and we could continue this with many other points, and all we'd find is that this graph that's here that I've already created with technology matches the slope at every one of those points.