### Video Transcript

Which of the following differential equations corresponds to the slope field shown? a) dπ¦ by dπ₯ is equal to negative π₯ over π¦. b) dπ¦ by dπ₯ is equal to π₯ over π¦. c) dπ¦ by dπ₯ is equal to π¦ over π₯. Or d) dπ¦ by dπ₯ is equal to negative π¦ over π₯.

A slope field is a graph showing the slope of a function at various points in the π₯π¦-plane. If we look, for example, at a point in the top-right quadrant where both π₯ and π¦ are positive, in our slope field dπ¦ by dπ₯, that is the slope at that point, is negative. If we look at any point on the vertical axis, say, where π₯ is equal to zero, because the line representing slope is flat, the slope at that point is equal to zero. Weβve been given four differential equations where the slope dπ¦ by dπ₯ is a function of π₯ and π¦. And to determine which of the given differential equations corresponds to the slope field shown. We can choose points on the slope field and compare the direction of slope at those points with our differential equations. We can then eliminate any differential equation where the signs donβt much.

Letβs use a table to compare with the differential equations down the left-hand side and the slope field points across the top. If we choose as our first point a point on the π¦-axis where π₯ is equal to zero and π¦ is positive. We know that for all such points, dπ¦ by dπ₯ is equal to zero. Now, letβs see how such a point would look in our first differential equation. Here we have dπ¦ by dπ₯ is equal to negative π₯ over π¦. If π₯ is zero and π¦ is positive, dπ¦ by dπ₯ is negative zero over a positive number. And thatβs equal to zero. This matches with our slope field dπ¦ by dπ₯ is equal to zero.

Now, letβs try this with our second differential equation. dπ¦ by dπ₯ is equal to π₯ over π¦. In this case, if π₯ is zero and π¦ is positive, dπ¦ by dπ₯ is zero over a positive number. And thatβs equal to zero. So for this type of point, our second differential equation also matches the slope field. If we try this with our third differential equation dπ¦ by dπ₯ is π¦ over π₯, we have a positive number divided by zero. But any number divided by zero is undefined, so cannot equal zero. This does not match the direction of the derivative in our slope field. So we can eliminate the third differential equation.

Now, looking at our fourth differential equation where dπ¦ by dπ₯ is equal to negative π¦ over π₯, we have a negative number divided by zero. We again have a division by zero which is undefined. And so, the slope does not equal zero. So now we can eliminate the fourth differential equation. Weβre left with two possible differential equations to match with our slope field. So letβs choose another point on the slope field and see if we can eliminate one of these. Taking any point in the first quadrant where both π₯ and π¦ are positive, the slope of any such point is negative. So dπ¦ by dπ₯ is less than zero. In our first differential equation where dπ¦ by dπ₯ is negative π₯ over π¦, we have the negative of a positive number over a positive number, which is negative. So dπ¦ by dπ₯ is less than zero. And this matches with our slope field.

We now have one remaining differential equation to compare with the slope field. Thatβs equation b where dπ¦ by dπ₯ is equal to π₯ over π¦. In this case, for a point in the first quadrant, we have a positive number over a positive number. So dπ¦ by dπ₯ is greater than zero. This does not match our slope field where points of this type have a negative slope. So we can eliminate differential equation b. Weβre now left with a single differential equation, dπ¦ by dπ₯ is equal to negative π₯ over π¦. We can check that this differential equation does actually match the slope field by comparing the slopes for points in each of the different quadrants. We know that the differential equation matches the slope field in the first quadrant and on the π¦-axis. So letβs compare the slopes for points in the second, third, and fourth quadrants.

In the second quadrant, points have a negative π₯ and a positive π¦. In the slope field, we can see that the slope is positive. In our differential equation then, we have dπ¦ by dπ₯ is a negative of a negative over a positive. That gives us a positive number over a positive number which is greater than zero. This matches with our slope field. Points in the third quadrant have negative π₯ and negative π¦. The slope of such a point in the slope field is negative. In our differential equation, such a point has dπ¦ by dπ₯ with negative of a negative over a negative. This is a positive number over a negative number, which is negative. So dπ¦ by dπ₯ in our differential equation is negative and matches the slope field.

Points in the final quadrant have positive π₯ and negative π¦ and a positive slope. For such a point in our differential equation, minus π₯ over π¦ gives us a negative over a negative, which is greater than zero. And this matches our slope field. The slopes of points in all four quadrants match our differential equation. So in fact, the differential equation dπ¦ by dπ₯ is negative π₯ over π¦ corresponds to the slope field shown, and that is equation a.