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We now know the basics of differential equations and their solutions, and here we're gonna dive a bit deeper as we talk about slope fields. Now remember that a derivative shows us the slope of a function at a given point, and slope fields use this idea to help visualize solutions to a differential equation, even if we don't actually know what those solutions are. This can be especially helpful for more tricky differential equations that we may not even know how to solve. Now the idea of slope fields can seem a bit tricky at first, but I'm going to break it all down for you here. So let's go ahead and get started.

Now slope fields, which you may also hear called direction fields, show us the shape of all possible solutions to a first-order differential equation of the form y′=f(x,y). In order to actually draw a slope field, we're just going to sketch short line segments at every single point on our coordinate system. The way that we determine the slopes of these line segments is by simply plugging each point into our derivative y′. How is this actually going to show us the shape of solutions to a differential equation? Remember earlier, we were reminded that a derivative shows us the slope of a function.

So if we know what those slopes are at every single point on our graph, we'll see some patterns start to emerge, which will show us the shape of solutions to a differential equation. Now this can feel really abstract when we're just talking about it, so let's actually come down to our example and sketch a slope field. Here we're asked to sketch a slope field for the differential equation y′=x-y. We need to draw short line segments at every single point on our coordinate system, but let's start small. We're gonna start by focusing in on just one point, the point (1, 0).

So that point is right here on my graph, and I need to sketch a line segment here. The way that I determine the slope of this line segment, again, is by plugging my point into my derivative. Here my derivative is given to me as y′=x-y. So taking x-y here for the point (1, 0) gives me 1-0=1. So this is the slope of my line segment, and I can go ahead and draw that line segment here on my graph.

Now moving on to my next point, (1, 1). Doing the same thing here, plugging these points into my derivative x-y. So here that's 1-1=0. So at the point (1, 1), I can draw a line segment that has a zero slope, so it's just a horizontal line. Moving on to my next point here, (1, 2), x-y, so 1-2=-1.

So this line will have a slope of -1 at the point (1, 2). Then for (2, 0), my next point here, x-y: 2-0=2. So this is going to be a positively sloped line that's a little bit steeper than what we've seen so far. Then my next point, (2, 1), plugging x-y, gives 2-1=1, the same exact slope that we saw at our first point. Finally, the last point we focus on: 2-2=0.

So I have another zero-sloped line segment, just a flat horizontal line. Now only having drawn the line segments for these six points, I can already see some patterns start to emerge. I can see that every time x and y are the same value, I’m going to get a zero slope. That’s going to happen all along that diagonal. We can see a similar pattern start to emerge along this diagonal as well, where all of these will have a slope of 1.

So with these patterns in mind, I’m gonna fill out the rest of my slope field. Now that we have our completed slope field, this shows us the shape of all possible solutions to our differential equation, which here was y′=x-y. Now let’s actually get a better idea for what this means by focusing in on a particular solution. In part (b) of our example, we’re asked to use the slope field to sketch the particular solution that passes through the point (-1, 2). But how do we actually do this?

Well, in order to sketch a particular solution curve, we’re going to go right through that initial condition, and then we’re going to follow along the slopes on our slope field. So here we’re given the initial condition (-1, 2). I’m going to go ahead and plot that point on my graph here, (-1, 2). Then, in order to actually draw my curve, I’m just going to follow along these slopes. As we go up here, we can see that our slopes get really steep.

So I’m going to draw my curve with a steep slope. Then in the other direction, we curve down and then end up going back up along what looks like a slant asymptote. Now you’ll notice here that I didn’t intentionally go on top of any other points on my slope field, and that’s because the only point that we know our solution goes through is that initial condition given to us in the problem, (-1, 2). So now we know what our particular solution looks like that passes through this point, and we can do this for any other point on our graph. Now we’re asked to do this again for the point (1, -2), sketching that particular solution as well.

So I’m going to go ahead and plot that point at (1, -2), and again, we want this to go through that initial condition and follow along the slopes on our slope field. I’m going to go ahead and follow along my slopes going in this direction. I can see that again, we get really steep here. And then in the other direction, we end up following along that slant asymptote again. So now we can see how a slope field helps us to visualize the solutions of a differential equation, even if we don’t actually know what those solutions are.

Now we’re going to continue getting practice with slope fields coming up in the next couple of videos. I’ll see you there.

Here we're asked which of the following differential equations is represented by the slope field given. Now looking at our slope field here, remember that all of these little line segments represent an x, y coordinate being plugged into our differential equation. So for example, looking at this first differential equation,

dy / dx = 3 - x

If I were looking to make a slope field based on this differential equation and I started at my point (0, 0), I would plug in 0 for x, meaning that that line segment would have a slope of 3. Now it clearly doesn't have a slope of 3 looking at my slope field here, so I know that I can already rule that out. But let's focus on the slope field given to us and notice any patterns that are happening here so that we can match that up to the correct differential equation.

The first thing that I notice is that when x is equal to 0, there are no slope field lines. That means that our function would be undefined here, or rather our derivative would be undefined here. So looking at our differential equations here, I know that for

dy / dx = 3 - x , dy / dx = sin ( 2 x ) , and dy / dx = 1 2 cos ( x ) ,

all of these functions would still be defined at x = 0. So I'm already getting pointed to this last differential equation here,

dy / dx = 1 x .

But let's look at a few more details. The other thing that I notice about my slope field is that as x gets larger, my little line segments have a less steep slope. It's getting less steep as x is getting bigger, and the same thing happens but in the negative direction over here on this side of my graph. Now this makes sense for this differential equation,

dy / dx = 1 x .

As x gets bigger and bigger, that denominator gets larger, making for a smaller value and a less steep slope.

So that means that this differential equation,

dy / dx = 1 x ,

is what's represented by this slope field. Let us know if you have any questions, and I'll see you in the next one.

If you haven't already, go ahead and try this problem on your own before checking back in with me. Here, we're asked to match the slope field to its differential equation. So what I wanna do here is look at each of these slope fields and try to notice any patterns that will allow me to then match it to the correct differential equation. Let's take a look at our first slope field here. The first thing that I notice is that along the line where x=0, all of my line segments have a zero slope.

So that means every time x=0, we get a slope of zero. So let's look at our differential equations and see any of these that that might work with. Looking at our first one here, y'=x+y, if x=0, y can still be some other value, meaning that our slope wouldn't always be zero even when x=0. So that's not really a good option here. Looking at our second differential equation, y'=2x, because this is only dependent on x, every time x=0, this will have a zero slope.

So this is looking at like a good option. Looking at our last differential equation here, y'=-x/y, every time x=0 for this differential equation, we will get a zero slope as well. So we need to look at more details here. Coming back to our slope field, the other thing that I notice is that as x gets larger, that slope gets steeper in both directions, so in the positive direction and in the negative direction. And it seems that this happens along the entire value of x, so it doesn't really matter what y is.

As x gets larger, our slope gets larger without any dependence on y. So looking back at our differential equations here, this differential equation did depend on y. So that's not really gonna be a good option for us, meaning that this slope field matches this differential equation that y'=2x. Let's move on to our next slope field. Here we have this one and the first thing that I notice here is that I have this diagonal line that all have zero slopes.

So where exactly is this happening? Well I can see that it's happening at (-1, 1), at (-2, 2), and looking back in our positive x direction, it's also happening at (1, -1) and (2, -2). So what does this tell me? Well, this tells me that if I'm taking these values, (2, -2), and adding them together, that gives me a zero slope.

And that's exactly what this differential equation is stating. y'=x+y, that would result in a zero slope just like is happening here. Now I'm gonna go ahead and check some other details here just to be really sure. And the other thing that I notice is that we have this consistent slope along our diagonal. So why is this happening?

Well, when we consider the values where this is happening, let's look at this a little bit deeper. Here I can see that this is happening at (-1, 0), at (-2, 1), and also over here at (0, -1) and (1, -2). Now, if I consider this differential equation and I take x and y and add them together for all of these coordinate points, that's going to give me a slope of -1. So that would result in this diagonal line where they all have slopes of -1. So I can safely say that this differential equation matches with this slope field here.

Let's take a look at our final slope field. Here, looking at this slope field, I can see a sort of circular pattern forming and the first thing that I notice is that when x=0, we have zero slopes. Now that makes sense based on my last differential equation that I have left because when that numerator x is zero, we would have a zero slope. Now the other thing that I notice here is that along my line y=0, there is no slope. These are undefined.

And this would also make sense for this differential equation because y is in that denominator. It can't be zero. That wouldn't give us a defined value. So it makes sense that our last differential equation here, y'=-x/y, matches up with our last slope field. Let us know if you have any questions, and I'll see you in the next one.

If you haven't already, go ahead and try this problem on your own before checking back in with me. Here, we're asked to sketch a slope field for the following differential equation and then sketch the particular solution curve through the given initial condition. So let's start with that slope field. The differential equation that we're given here is:

d dx y=y+