 Hello, and welcome to this screencast. In this screencast, we will draw an example of a slope field for a differential equation. In this example, we have the differential equation dy dt is equal to y squared minus 2y minus 3. A solution to this differential equation is a function y of t such that when we differentiate that function y, the result that we get is our original function squared minus 2 times our function minus 3, so we get y squared minus 2y minus 3. There are infinitely many solutions to this differential equation, and we can represent these solutions graphically using what's called a slope field, and we'll have an example here. On the left of our screen, we see a set of coordinate axes with t as the horizontal axis and y on the vertical axis, so we have y as a function of t. Our job is to choose points in this plane on this grid and determine the value of dy dt at those points, and this tells us how our function y, our solution to the differential equation, is changing. And we're going to represent this graphically by using small tangent lines, and the slope of those tangent lines is the value of dy dt, the rate of change of our function y. So first, when we look at our differential equation, we notice that it's a polynomial on the right-hand side, and that it factors nicely into two factors, y plus 1 and y minus 3. So that's going to make our first steps a little easier. So let's start with the values where y is equal to negative 1 on our grid. If y is equal to negative 1, then the first factor is going to be equal to 0. So that means the value for dy dt is also going to be equal to 0. So we notice that y is equal to negative 1, but this result doesn't depend on the value of t. So all the points on our grid with a y value of negative 1 and any t value are going to have dy dt equal to 0. So to represent this, we're going to draw little mini tangent lines at those points. So we have here y equals negative 1, t is equal to negative 1, y is equal to negative 1, t is equal to 0, and then across in that horizontal line, all of our little mini tangent lines have a slope of 0. So we have another one that's going to be easier to work with, and that's when y is equal to 3. So when y is equal to 3, our second factor will be 0, and that means again dy dt is equal to 0 at all points on our grid where the y value is 3 and t can have any value. So we'll draw those horizontal tangent lines at y equals 3 as well. For our other y values, we're going to need to work a little bit harder to find the value of dy dt. So let's start at the bottom of our slope field with y is equal to negative 2. When y is equal to negative 2, we're going to go back to using that original form y squared minus 2y minus 3 for our differential equation. When we evaluate this at y is equal to negative 2, we see we get negative 2 squared minus 2 times negative 2 minus 3. And when we simplify this, our result is 5. So that means we're going to draw tangent lines where y is equal to negative 2 that have a steep positive slope. Next, we'll go up a little bit to when y is equal to 0. In this case, we see that when we evaluate dy dt at y equals 0, we get negative 3. So this time, our tangent lines will have a negative slope that is fairly steep, but perhaps not quite as steep as the ones we drew at y is equal to negative 2. So we will continue this process for the y values y equals 1, y equals 2, y equals 4, evaluating dy dt, and then drawing little tangent lines. So you might want to pause this screencast for a moment, try these three y values on your own and create your own slope field, and then you can resume the screencast and we can compare our results. So welcome back. Let's compare our results for the other three y values that we still need to evaluate. So in our next one, when y is equal to 1, we evaluate 1 squared minus 2 times 1 minus 3, and our result is negative 4. So our tangent lines will be just a little bit steeper than the ones we drew just previously. So there we have those black tangent lines that are pretty steep. When y is equal to 2, dy dt, the final result is negative 3. And so these tangent lines are going to be the same slope as the green ones we drew at y is equal to 0, so we can draw those as well. And then our last y value, when y is equal to 4, so we'll evaluate dy dt at 4, and we get a result of 5, and then we get some fairly steep tangent lines again, just like we drew when y is equal to negative 2. So we've completed drawing our result in our slope field, and this gives us a qualitative view of the solutions to our differential equation and how they're changing for different values of y and t. In the following screencast, we'll look at how to find specific solutions to differential equations, unique solutions, instead of the infinitely many possible solutions that exist. Thanks for watching.