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

Consider the initial value problem π¦β² = 3π₯, π¦(0) = 1. Use Eulerβs method with π = 5 steps on the interval [0, 1] to find π¦(1).

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### Video Transcript

Consider the initial value problem π¦ prime is three π₯, where π¦ at π₯ is equal to zero is equal to one. Use Eulerβs method with π equal to five steps on the closed interval zero, one to find π¦ at π₯ is equal to one.

Weβve been given a differential equation π¦ prime is three π₯ so that π¦ prime, which is dπ¦ by dπ₯, is equal to three π₯. We have an initial value π¦ at π₯ is equal to zero is equal to one, which means that our starting point π₯ zero, π¦ zero is zero one. We have π equals to five steps and the closed interval zero, one. This means that zero is less than or equal to π₯ is less than or equal to one. And weβre asked to use Eulerβs method to find the value of π¦ at π₯ is equal to one. In order to use this formula, we need to know the value of β, which is the step size. And thatβs given by the interval width divided by the number of steps. In our case, the interval width is a maximum π₯ minus the minimum π₯ which is one minus zero divided by the number of steps which is five. Thatβs one over five, which is 0.2.

Letβs just sketch our interval, so we can refer back to it in our calculations. Since our step size is 0.2 and π₯ zero is equal to zero, π₯ one is equal to 0.2. Similarly, π₯ two is equal to 0.4, π₯ three is 0.6, and π₯ four is 0.8, and π₯ five, which is our maximum value is one. Remember, weβre trying to find the value of π¦ when π₯ is equal to one. And thatβs π¦ of π₯ five. And we call that π¦ five. In order to use the formula, we need to know what our function π is. And from our differential equation, we can see that dπ¦ by dπ₯ is equal to π of π₯ is three π₯. So if this is a function of π₯ only, this means we can delete the second variable π¦ in our function π in Eulerβs formula, so that our formula is π¦ π is equal to π¦ π minus one plus β times π of π₯ π minus one, where π of π₯ π is three π₯ π. So we have everything we need to use the formula.

We know that π¦ zero is equal to one, so we can use the formula to calculate π¦ one. And thatβs equal to π¦ zero plus β times π of π₯ zero. We know the value of π¦ zero, which is one, and we know the value of β, which is 0.2. So we need to work out π of π₯ zero. From our differential equation, thatβs equal to three times π₯ zero, which is three times zero, which is zero so that π¦ one is equal to one plus 0.2 times zero, which is one. And we can use this in our formula to calculate π¦ two, which is equal to π¦ one plus β times π of π₯ one. And we need to work out π of π₯ one. Thatβs equal to three times π₯ one from our differential equation. And from our interval, we know that π₯ one is equal to 0.2. So we have three times 0.2 is equal to π of π₯ one, and thatβs 0.6.

So π¦ two is equal to one plus 0.2 times 0.6 which is equal to 1.12. We use this to find π¦ three, which is π¦ two plus β times π of π₯ two, where π of π₯ two is three times π₯ two and π₯ two is 0.4, so that π of π₯ two is 1.2. π¦ three is then 1.12 plus 0.2 times 1.2, which is 1.36. We use this to calculate π¦ four which is π¦ three plus β times π of π₯ three and π of π₯ three is three times π₯ three. From our interval, π₯ three is 0.6. So thatβs three times 0.6, which is 1.8. And we have π¦ 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 π¦ five, where π¦ five is equal to π¦ four plus β times π of π₯ four which is three π₯ four. And since π₯ four is 0.8 from our interval, π of π₯ four is three times 0.8, which is 2.4. So π¦ five is equal to 1.72 plus 0.2 times 2.4, which is 2.2. Remember that π¦ five is equal to π¦ at π₯ is equal to one, which weβve just calculated as 2.2. So π¦ at π₯ is equal to one is 2.2.