Video Transcript
Which of the following differential equations corresponds to the given slope field? a) d𝑦 by d𝑥 equals 𝑦 over 𝑥. b) d𝑦 by d𝑥 equals negative 𝑦 over 𝑥. c) d𝑦 by d𝑥 equals negative 𝑥 over 𝑦. And d) d𝑦 by d𝑥 equals 𝑥 over 𝑦.
For this question, we’ve been given a number of candidate differential equations. Only one of these will match the slope field which we observe in the question. Although we can solve this question mathematically, it is perhaps more efficient and less time consuming to do so by inspection, eliminating the options until we’re left with only one. The first thing we recall is that d𝑦 by d𝑥 represents the slope or the gradient. We’ve been given a number of sample points for this slope field within our grid.
To begin our inspection method, let us look at the top right-hand quadrant of our grid. In this quadrant, both 𝑥 and 𝑦 take positive values. That is to say, 𝑥 and 𝑦 are greater than zero. We can also observe that all of the slopes are positive. That is to say, d𝑦 by d𝑥 is greater than zero. We now have some information which can be applied to eliminate some of our options.
For options a) and d), d𝑦 by d𝑥 will be a positive number divided by a positive number, either 𝑦 over 𝑥 or 𝑥 over 𝑦. We can see that d𝑦 by d𝑥 will then itself be positive. In other words, d𝑦 by d𝑥 is greater than zero. This does indeed match what we observed on our slope field. On the other hand, for options b) and c), d𝑦 by d𝑥 will be negative one times a positive number divided by a positive number, either negative 𝑦 over 𝑥 or negative 𝑥 over 𝑦. For these options, this means that when both 𝑥 and 𝑦 are positive, d𝑦 by d𝑥 will be negative. In other words, d𝑦 by d𝑥 will be less than zero.
This does not match what we observe on our slope field. Since all of the slopes for options b) and c) should be negative in this quadrant, what we observe that all of the slopes are positive in the quadrant. We can eliminate options b) and c) as potential answers to our question since these cannot be the equations of the slope field we’re observing. Okay, we still have two candidates, option a) and option d). Let’s deal with these now.
To move forward, let us narrow the range of values that we’re inspecting. Here we observe four different values for the slope d𝑦 by d𝑥. However, all four of these sample points share the same 𝑥-value. So here, 𝑥 is constant. On the other hand, while keeping 𝑥 constant, we are looking at four different values of 𝑦. If we pay attention, we might notice that as 𝑦 increases, so does the value of our slope, d𝑦 by d𝑥. So if 𝑥 is constant and 𝑦 increases, d𝑦 by d𝑥 should also increase.
Let’s now see how this relates to our two remaining options. For option a), d𝑦 by d𝑥 is equal to 𝑦 over some constant, which in this case is our 𝑥-value. We can easily see that d𝑦 by d𝑥 is proportional to 𝑦. So as 𝑦 increases, so does d𝑦 by d𝑥. This does indeed match the observation that we just made. Now, how about option d)? For option d), d𝑦 by d𝑥 is equal to 𝑐 over 𝑦. And again, 𝑐 is just the 𝑥-value, which we’re holding constant here. We now see that, instead, that d𝑦 by d𝑥 is proportional to one over 𝑦. Since 𝑦 is in the denominator here, if 𝑦 increases, d𝑦 by d𝑥 will decrease.
This does not match the observation that we made. It’s worth noting here that this observation of increasing 𝑦 and increasing gradient is not just confined to the top right quadrant. In fact, for all the values we see, if 𝑥 is held constant, d𝑦 by d𝑥 increases as 𝑦 increases. It is however worth noting that, in this case, we’re looking at a value of 𝑥 which is constant and positive or positive 𝑐.
If, instead, we were to look at a constant 𝑥 but where 𝑥 is negative, we would instead observe that as 𝑦 increases, d𝑦 by d𝑥 would decrease. Or it would become more negative. This is just a side note however. And the conclusion that we draw is still the same. The behaviour of the equation in option d) does not match the observations that we’ve made in our slope field. And again, we can eliminate this as an option.
Since we’ve now eliminated three of the four options and we verified that option a) does seem to match the slope field that we observe, we have now answered our question. The differential equation that corresponds to the given slope field is option a), d𝑦 by d𝑥 equals 𝑦 over 𝑥. Before we go however, a quick aside. At the beginning of this video, we mentioned that this problem can also be solved mathematically. So here’s a bit of guidance on what that would look like.
Observing our options, we notice that we’ve been given first-order separable differential equations. What this allows us to do is to separate our variables and then to integrate both sides of our equation. We perform our integrations. And we then work on simplifying our equation to find 𝑦 in terms of 𝑥. After a few simplifications, we reach our result, which is that 𝑦 equals some constant times 𝑥. And here we’ve got the constant 𝑑. This is the general form for the solutions to the differential equation given in option a), d𝑦 by d𝑥 equals 𝑦 over 𝑥.
All curves or in this case lines with equations of this form will be solutions to this one differential equation. Since 𝑑, the constant, can take any value if we imagine 𝑑 as one, we have the line 𝑦 equals 𝑥. If 𝑑 is a larger positive number, we’ll have a steeper line or a line which is less steep for a smaller positive number. Perhaps 𝑑 is a negative number. For example, if 𝑑 was negative one, we’d have the line 𝑦 equals negative 𝑥.
Notice that, on this 𝑥, 𝑦 coordinate grid that we’ve drawn, all of the lines which are solutions to the differential equation match the slope field that we’ve been given in the question. All the lines intersect at the origin. And their slopes behave in the same way. As an illustration, we can transpose the line of 𝑦 equals negative 𝑥 onto our slope field.
Okay, we already know that option a) is the correct answer. And hence, we’ll obviously match our slope grid. But what about the other options? Since all of the options are first-order separable differential equations, we could follow the same process as we used for option a). Doing so, we would reach these general solutions. Taking option d) as an example, you might recognise this as the equation of a circle centred at the origin. This circle would have a radius of the square root of 𝑑. But of course, since 𝑑 can take any value, so can the radius of the circle.
If we were to compare any circle with our slope field, we would see that it does not line up at all. Since the solutions to this differential equation do not seem to match our slope field, we would again eliminate this as an option. The same would be true for options b) and c). And again, we would observe that only option a) remains. As mentioned earlier, this method is probably more time consuming and difficult than the initial method shown. However, these tools may be useful for more complicated questions.
