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 𝑥 squared plus five. Option (B) d𝑦 by d𝑥 is equal to five minus 𝑥 squared. Option (C) d𝑦 by d𝑥 is equal to 𝑥 squared minus five. Option (D) d𝑦 by d𝑥 is equal to 𝑥 plus five. Or option (E) d𝑦 by d𝑥 is equal to 𝑥 minus five.
In this question, we’re given the slope field graph of a differential equation. And we need to determine which of five possible differential equations this is the slope field graph of. To do this, we need to recall what the slope field graph of a differential equation is. What this means is at the point 𝑥, 𝑦 on our slope field graph, we plot the slope of our differential equation. In other words, we substitute this value of 𝑥 and 𝑦 into our differential equation, and it gives us a value for the slope. And by knowing this, we find the several different methods we could use to solve this problem.
We could substitute values of 𝑥 into all of our answers to determine information which should be present on our graph. Alternatively, we could also look at our graph to determine information about our differential equation. We could then check our answers to see which one has these properties. Both of these methods would work, and in fact, it’s personal preference which one you would like to use. We’re going to look at the sketch to determine information about our differential equation.
So to start this, we’ll say that this is a sketch of the differential equation d𝑦 by d𝑥 is equal to capital 𝐹 of 𝑥, 𝑦. And the first thing we can notice about our sketch is the value of 𝑦 does not change the value of the slopes. In other words, the vertical position of the slopes on our slope field graph does not change the value of these slopes. And if our values of 𝑦 are not changing the values of our slopes d𝑦 by d𝑥, then d𝑦 by d𝑥 must be entirely in terms of 𝑥. In other words, it’s just a function of 𝑥. So we can just write this d𝑦 by d𝑥 is equal to capital 𝐹 of 𝑥.
We now want to look at our sketch to determine information about capital 𝐹 of 𝑥. Let’s start when 𝑥 is equal to zero. We can see from our sketch when 𝑥 is equal to zero, all of our slope lines are negative. And remember, the slope of these lines will be d𝑦 by d𝑥 when 𝑥 is equal to zero. And if this is less than zero, then 𝐹 evaluated at zero is less than zero. So the first piece of information about our differential equation is capital 𝐹 of zero is less than zero. So we’ll substitute values of 𝑥 is equal to zero into our five answers.
Let’s start with option (A). Substituting 𝑥 is equal to zero, we get zero squared plus five, which we know is equal to five. And this is a positive value, and we know this is supposed to be negative. So this can’t possibly be a sketch of the slope field graph of the differential equation d𝑦 by d𝑥 is equal to 𝑥 squared plus five. Otherwise, we would need to have positive slopes when 𝑥 is equal to zero.
We’ll do the same for option (B). We substitute in 𝑥 is equal to zero. We get five minus zero squared, which we can calculate is equal to five. Once again, we got a positive answer. And we know that the slopes on the line 𝑥 is equal to zero all are negative. So this can’t possibly be option (B).
Substituting 𝑥 is equal to zero into option (C), we get zero squared minus five, which is equal to negative five. This is a negative value, so option (C) might be the correct answer.
Substituting 𝑥 is equal to zero into option (D), we get zero plus five, which is equal to five. And just as with options (A) and (B), this is a positive value, but we know that this is supposed to be negative. So option (D) can’t possibly be the correct answer.
Finally, we substitute 𝑥 is equal to zero into option (E). This gives us zero minus five, which is equal to negative five. This is a negative value, so option (E) might be the correct answer.
And now we want to choose more values of 𝑥 to determine which is our correct answer. We could do this by guesswork. However, there is a better method. We can take a look at our two answers, 𝑥 squared minus five and 𝑥 minus five. If we can choose a value of 𝑥 where one of these is negative and the other one is positive, then we know we’ll be able to determine this from our sketch.
For example, we could choose 𝑥 is equal to three. Then we see that our answer in (C) is positive. However, our answer (E) will be negative. Another good option would’ve been to choose 𝑥 is equal to five because an option (E) would’ve given us a slope of zero. However, option (C) would’ve given us a positive slope. We’ll choose 𝑥 is equal to three. We can see from our sketch when 𝑥 is equal to three, our slopes are positive. And if the slopes when 𝑥 is equal to three are positive, then capital 𝐹 of three must be positive.
So once again we’ll substitute these values into our options. Substituting 𝑥 is equal to three into option (C), we get three squared minus five, which we can evaluate is equal to four. And this is a positive value, so option (C) might be the correct sketch. However, if we substitute 𝑥 is equal to three into option (E), as we expected, this gives us a negative value. So we know this can’t be a sketch of the slope field of the differential equation d𝑦 by d𝑥 is equal to 𝑥 minus five.
Otherwise, when 𝑥 is equal to three, our slope line should’ve been negative. And this tells us that this must be a slope field graph of option (C) d𝑦 by d𝑥 is equal to 𝑥 squared minus five. In this question, we were given the slope field graph of a differential equation. And we were able to use this sketch to determine information about our differential equation. We were able to deduce that this must be a sketch of the slope field graph of the differential equation d𝑦 by d𝑥 is equal to 𝑥 squared minus five, which was option (C).