Worked example: forming a slope field | AP Calculus AB | Khan Academy

In drawing the slope field for the differential equation, the derivative of y with respect to x is equal to y minus 2x. I would place short line segments at select points on the xy-plane.

At the point (-1, 1), I would draw a short segment of slope blank. Like always, pause this video and see if you can fill out these three blanks.

When you're drawing the short segments to construct this slope field, you figure out their slope based on the differential equation. So, you're saying when x is equal to -1 and y is equal to 1, what is the derivative of y with respect to x? That's what this differential equation tells us.

So, for this first case, the derivative of y with respect to x is going to be equal to y, which is 1, minus 2 times x. x is -1, so this is going to be negative 2, but you're subtracting it, so it's going to be plus 2. Therefore, the derivative of y with respect to x at this point is going to be 3.

I would draw a short segment or a short segment of slope 3. We keep going. At the point (0, 2), let's see. When x is 0 and y is 2, the derivative of y with respect to x is going to be equal to y, which is 2, minus 2 times 0. Well, that's just going to be 2.

Now, last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is 3, minus 2 times x. x here is 2. So, 2 times 2 is 4, and 3 minus 4 is equal to negative 1.

And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like—I'll try to draw it real fast.

So, let's see. Let me make sure I go to make sure I have space for all of these points here. So, that's my coordinate axes, and I want to get the point (0, 2). That's (0, 2). Actually, I want to go all the way to (2, 3). So, let me get some space here. So, 1, 2, 3; and then 1, 2, 3.

Then we have to go to (-1, 1). We might go right over here. For this first one, this exercise isn't asking us to do it, but I'm just making it very clear how we would construct the slope field.

So, the point (-1, 1)—a short segment of slope 3. Slope 3 would look something like that. Then at the point (0, 2)—a slope of 2. (0, 2), the slope is going to be 2, which looks something like that.

Then at the point (2, 3)—at (2, 3), a short segment of slope negative 1. So, (2, 3), a segment of slope negative 1, would look something like that.

You would keep doing this at more and more points. If you had a computer to do it, that's what the computer would do, and you would draw these short line segments to indicate what the derivative is at those points. You get a sense of, I guess you say, the solution space for that differential equation.