 Hello, and welcome to this screencast. In this screencast, we are given a slope field representing a differential equation, and we're asked to draw solutions to the differential equation for specific initial values. Our differential equation is dy dt is equal to y squared minus 2y minus 3. And in a previous screencast, we drew a slope field by hand so we could see a graphical representation of solutions to the differential equation. For this example, I created the slope field using GeoGebra software so that our slope field has many more mini tangent lines than the one we drew by hand, and this is going to make our job of drawing solutions easier. We know that a differential equation has infinitely many solutions, so if we want to find a unique solution to a differential equation, we would need just one specific point that is on that solution that we would like to find, and then we can do the rest. We call that information an initial value, and that's what we have here. So we have four initial values, and we will draw a graph of the unique solution to the differential equation that corresponds to each of the four initial values. We're used to starting with a graph of a function, and then we draw tangent lines for that function, but here we're going to reverse this process because right now what we have is a bunch of little tangent lines in our slope field, and then we're actually going to draw the function by following these tangent lines. We can think of these tangent lines as representing the current in a river, and if we drop a leaf in the river, we can watch as the leaf follows the current, and that's what our solution will be. In part A, our initial value is t equals 0 and y equals 2, so we will start by plotting the point 0, 2 on our graph, and that means we're going to drop the leaf into the current at that point and then see what happens to it. As we follow the tangent lines as t increases, we see that the lines slope downward, and then they flatten out as we get closer to y equals negative 1, and that's where we have a horizontal asymptote in the graph at y equals negative 1. We can also follow this current backwards as t decreases, and so when we do that, we see that our y value increases and it gets closer to y equals 3. This red graph that we've just drawn is the unique solution to the differential equation that passes through the point 0, 2. For our second example, our initial value is written slightly differently but contains the same type of information. Our initial value is y of 2 equals 3, which means the input t is equal to 2 and the output y is equal to 3, so we plot a point there in blue at the point 2, 3. When y is equal to 3, dy dt is equal to 0, so our solution is a horizontal line as t increases, and it's still horizontal as t decreases because the derivative along those points is always exactly 0. This blue graph is a solution to the differential equation with the initial value 2, 3, and it shows us an example of an equilibrium value. So we say that y equals 3 is an equilibrium value of our differential equation because as t increases or decreases, our y value doesn't change. In our third example, our initial value is at 1, negative 2 near the bottom of our screen. As t increases, this solution also has a horizontal asymptote at y equals negative 1, and as t decreases, our y values are going to decrease and the solution goes off of the bottom of our slope field. So there's our third solution given an initial value. In our last example, the initial value is at 5, 4 in the top right corner of our graph. As t increases, our y values also increase, and as t decreases, we see that the solution has a horizontal asymptote at y equals 3. Each of these four graphs is a solution to the differential equation, and there are infinitely more solutions to the differential equation. However, for each initial value that we were given, we were able to use the slope field to help us draw the unique solution to the differential equation that passes through that given initial value. Thanks for watching.