Video: Finding the Monotonicity of a Function given Its Derivative Graph

Video Transcript

Consider the initial value problem 𝑦 prime is three 𝑥, where 𝑦 at 𝑥 is equal to zero is equal to one. Use Euler’s method with 𝑛 equal to five steps on the closed interval zero, one to find 𝑦 at 𝑥 is equal to one.

We’ve been given a differential equation 𝑦 prime is three 𝑥 so that 𝑦 prime, which is d𝑦 by d𝑥, is equal to three 𝑥. We have an initial value 𝑦 at 𝑥 is equal to zero is equal to one, which means that our starting point 𝑥 zero, 𝑦 zero is zero one. We have 𝑛 equals to five steps and the closed interval zero, one. This means that zero is less than or equal to 𝑥 is less than or equal to one. And we’re asked to use Euler’s method to find the value of 𝑦 at 𝑥 is equal to one. In order to use this formula, we need to know the value of ℎ, which is the step size. And that’s given by the interval width divided by the number of steps. In our case, the interval width is a maximum 𝑥 minus the minimum 𝑥 which is one minus zero divided by the number of steps which is five. That’s one over five, which is 0.2.

Let’s just sketch our interval, so we can refer back to it in our calculations. Since our step size is 0.2 and 𝑥 zero is equal to zero, 𝑥 one is equal to 0.2. Similarly, 𝑥 two is equal to 0.4, 𝑥 three is 0.6, and 𝑥 four is 0.8, and 𝑥 five, which is our maximum value is one. Remember, we’re trying to find the value of 𝑦 when 𝑥 is equal to one. And that’s 𝑦 of 𝑥 five. And we call that 𝑦 five. In order to use the formula, we need to know what our function 𝑓 is. And from our differential equation, we can see that d𝑦 by d𝑥 is equal to 𝑓 of 𝑥 is three 𝑥. So if this is a function of 𝑥 only, this means we can delete the second variable 𝑦 in our function 𝑓 in Euler’s formula, so that our formula is 𝑦 𝑛 is equal to 𝑦 𝑛 minus one plus ℎ times 𝑓 of 𝑥 𝑛 minus one, where 𝑓 of 𝑥 𝑛 is three 𝑥 𝑛. So we have everything we need to use the formula.

We know that 𝑦 zero is equal to one, so we can use the formula to calculate 𝑦 one. And that’s equal to 𝑦 zero plus ℎ times 𝑓 of 𝑥 zero. We know the value of 𝑦 zero, which is one, and we know the value of ℎ, which is 0.2. So we need to work out 𝑓 of 𝑥 zero. From our differential equation, that’s equal to three times 𝑥 zero, which is three times zero, which is zero so that 𝑦 one is equal to one plus 0.2 times zero, which is one. And we can use this in our formula to calculate 𝑦 two, which is equal to 𝑦 one plus ℎ times 𝑓 of 𝑥 one. And we need to work out 𝑓 of 𝑥 one. That’s equal to three times 𝑥 one from our differential equation. And from our interval, we know that 𝑥 one is equal to 0.2. So we have three times 0.2 is equal to 𝑓 of 𝑥 one, and that’s 0.6.

So 𝑦 two is equal to one plus 0.2 times 0.6 which is equal to 1.12. We use this to find 𝑦 three, which is 𝑦 two plus ℎ times 𝑓 of 𝑥 two, where 𝑓 of 𝑥 two is three times 𝑥 two and 𝑥 two is 0.4, so that 𝑓 of 𝑥 two is 1.2. 𝑦 three is then 1.12 plus 0.2 times 1.2, which is 1.36. We use this to calculate 𝑦 four which is 𝑦 three plus ℎ times 𝑓 of 𝑥 three and 𝑓 of 𝑥 three is three times 𝑥 three. From our interval, 𝑥 three is 0.6. So that’s three times 0.6, which is 1.8. And we have 𝑦 four equal to 1.36 plus 0.2 times 1.8, which is 1.72. Now, we use this in our final step to calculate 𝑦 five, where 𝑦 five is equal to 𝑦 four plus ℎ times 𝑓 of 𝑥 four which is three 𝑥 four. And since 𝑥 four is 0.8 from our interval, 𝑓 of 𝑥 four is three times 0.8, which is 2.4. So 𝑦 five is equal to 1.72 plus 0.2 times 2.4, which is 2.2. Remember that 𝑦 five is equal to 𝑦 at 𝑥 is equal to one, which we’ve just calculated as 2.2. So 𝑦 at 𝑥 is equal to one is 2.2.