### Video Transcript

Consider the given slope field graph. Which of the following differential equations is represented on the graph? Option (A) d𝑦 by d𝑥 is equal to the sin of 𝑥. Option (B) d𝑦 by d𝑥 is equal to the cos of two 𝑥. Option (C) d𝑦 by d𝑥 is equal to negative the sin of two 𝑥. Option (D) d𝑦 by d𝑥 is the sin of two 𝑥. Or option (E) d𝑦 by d𝑥 is negative the cos of 𝑥.

In this question, we’re given the slope field graph of a differential equation, and we need to determine the differential equation represented by this slope field graph. To do this, we’re going to need to start by recalling what we mean by the slope field graph of a differential equation. In this graph, at each coordinate 𝑥, 𝑦, instead of being given a point on a curve, we’re instead given the slope of our differential equation. And we can see we’re given five possible options. Each of these are given in the form d𝑦 by d𝑥 is equal to some function of 𝑥. And when we substitute values of 𝑥 into these functions, we’ll be given expressions for d𝑦 by d𝑥, which we know is the slope.

So in actual fact, this gives us two methods to solve this equation. We could substitute values of 𝑥 into all five of our answers and see which one matches with our graph. And this would definitely work. Another method we could use is to read the values of the slopes off of our graph and then see which of our solutions match these values. Either method will work. It’s personal preference which one you would prefer. We’re going to use the graph to determine information about our function. To start, let’s consider all of the values on the line 𝑥 is equal to zero. And if we do this, we can see all of the slopes on this line are perfectly horizontal. That means their slope is equal to zero.

To use this information, let’s start by saying that this slope field graph represents the differential equation d𝑦 by d𝑥 is equal to capital 𝐹 of 𝑥, 𝑦. However, there’s something we can notice already. Either by looking at our answers or by looking at our slope field graph, we can see that the slope does not depend on the value of 𝑦. For example, if we look at our slope field graph, we can see moving vertically will not change the values of our slope. So instead of writing our differential equation as a function in both 𝑥 and 𝑦, we can instead describe this as a function of 𝑥.

Now, let’s move on to finding information about this function. First, remember, we already showed when 𝑥 is equal to zero, our slope lines are all equal to zero. So if d𝑦 by d𝑥 is equal to zero when 𝑥 is equal to zero, this means capital 𝐹 evaluated at zero is equal to zero. We can then substitute 𝑥 is equal to zero into all five of our options to see which ones have this property. We can see in option (A), the sin of zero is equal to zero. So option (A) does have this property. We can see in option (B), when we substitute 𝑥 is equal to zero, we get the cos of zero, which we know is equal to one. This is of course not equal to zero, so option (B) can’t be the correct option.

Substituting 𝑥 is equal to zero into option (C), we get negative the sin of two times zero, which is negative the sin of zero, which we know is equal to zero. So option (C) does have this property. We can do the same for option (D) and option (E). Substituting 𝑥 is equal to zero into option (D), we get the sin of zero, which is zero. So option (D) does have this property. When we do this for option (E), however, we get negative the cos of zero, which is negative one. So this can’t possibly be a slope field graph of the differential equation d𝑦 by d𝑥 is equal to negative the cos of 𝑥. Otherwise, on the line 𝑥 is equal to zero, all of our slopes would have had value of negative one.

We can now choose another value of 𝑥. Let’s try 𝑥 is equal to one. Once again, we can look at all of our slope lines. We see that all of the slopes have the same value, which is approximately equal to one. And we say this is approximately equal to one because this is a sketch. Now, we can once again substitute 𝑥 is equal to one into our three differential equations. However, before we do this, there’s one thing we’ll point out. This is a differential equation in terms of trigonometric functions. This means our inputs will be radians. So let’s start with option (A). We need to evaluate the sin of one. If we do this, we get that it’s approximately equal to 0.84. This is close enough that this could be a graph of the solution curve given in option (A).

Next, we’ll substitute 𝑥 is equal to one into option (C). This gives us negative the sin of two times one. And if we evaluate this, we get approximately negative 0.91. This is a negative value, and we can see that our slopes when 𝑥 is equal to one are all positive, so it can’t possibly be option (C). Finally, we’ll substitute 𝑥 is equal to one into option (D). This gives us the sin of two times one, which is approximately 0.91. And this agrees with our sketch, so this could be our differential equation. So, we’ll clear some space and choose one more value of 𝑥. Let’s try 𝑥 is equal to two.

This time, when 𝑥 is equal to two, we can see from our slope field graph that all of our slopes are negative. We could do what we did before and say this is approximately negative one. However, in this case, we only need that it’s negative. Now, we need to substitute 𝑥 is equal to two into our solutions. Let’s start with option (A). When we substitute 𝑥 is equal to two into the sin of 𝑥, we get the sin of two, which is approximately equal to 0.91 which is positive. So this can’t possibly be a sketch of the slope field graph of the differential equation given in option (A).

This means our answer must be option (D). However, for due diligence, let’s substitute 𝑥 is equal to two into this expression just to make sure. Substituting 𝑥 is equal to two, we get the sin of two times two, which is the sin of four, which, if we calculate, is approximately equal to negative 0.76, which means that option (D) must be our correct answer.

Therefore, given five differential equations and the slope field graph of one of these differential equations, we were able to determine that this must be a sketch of the slope field graph of the differential equation d𝑦 by d𝑥 is equal to the sin of two 𝑥 by looking at the slopes at different values of 𝑥. This was option (D).