### Video Transcript

Consider the initial value problem π¦ prime is equal to two minus five π¦, where π¦ of zero is equal to negative one. Use Eulerβs method with four steps to estimate the value of π¦ of one to four significant figures.

The question gives us an initial value problem. Weβre told π¦ prime is equal to two minus five π¦ and π¦ of zero is equal to negative one. We need to use Eulerβs method with four steps to estimate the value of π¦ of one to four significant figures.

Letβs start by recalling Eulerβs method with π steps. If weβre given the initial value problem π¦ prime is equal to some function π of π₯, π¦ and weβre told that π¦ of π₯ zero is equal to π¦ zero. Then Eulerβs method tells us we can approximate π¦ evaluated at π₯ π with π¦ π, where π¦ π is equal to π¦ sub π minus one plus β times π evaluated at π₯ sub π minus one π¦ sub π minus one. Where β is equal to π₯ sub π minus π₯ sub zero over π. This is often referred to as the interval width divided by the number of steps. And each of our π₯ π is equal to π₯ sub zero plus π times β.

So, letβs start taking some information from our question. First, weβre given the differential equation π¦ prime is equal to two minus five π¦. So, our function π of π₯, π¦ is equal to two minus five π¦. Of course, we can see this doesnβt rely on our variable π₯. So, we can just call this π of π¦. Next, the question wants us to use four steps, so our value of π is equal to four. Next, since π¦ of zero is equal to negative one, we know that π₯ zero is equal to zero and π¦ zero is equal to negative one.

Finally, remember, the value weβre asked to approximate is π¦ evaluated at π₯ π. In our case, π is equal to four. And weβre asked to approximate π¦ evaluated at one. So, weβll set π₯ four equal to one. This is all the information we need to start answering our question. Our approximation will be π¦ four. So, we need to calculate π¦ four. And weβre given a formula to calculate π¦ π. Itβs equal to π¦ π minus one plus β times π of π₯ π minus one π¦ π minus one. At this point, we only know the values of π₯ zero and π¦ zero. So, letβs find the value of β.

We have that β is π₯ π minus π₯ zero all divided by π. We know that π is the number of steps, which is four. We know that π₯ four is equal to one and π₯ zero is equal to zero. So, this gives us β is one minus zero divided by four. We can evaluate this, and weβll write this as a decimal. Weβll write it as 0.25.

Now that we found the value of β, we can find the values of our π₯ π. Letβs start with π₯ one. Thatβs equal to π₯ zero plus one times β. We found β is 0.25. So, we have π₯ one is zero plus one times 0.25. And of course, we can calculate this is equal to 0.25. We can do exactly the same to find π₯ two. We see that itβs equal to 0.5. We could have also found this by just adding β to our value of π₯ one. And we can do exactly the same to find the value of π₯ three. We get that itβs equal to 0.75. And of course, we know π₯ zero is zero and π₯ four is equal to one. So, we found all of our values use of π₯ π.

Now that we found expressions for each of our π₯ π and β, weβre ready to try and apply our formula. However, the only value of π¦ which we know is π¦ zero. So, weβll need to start by finding an expression for π¦ one. Using our formula, we have π¦ one is equal to π¦ zero plus β times π of π₯ zero, π¦ zero. However, remember, in this case, we showed that our function π is only a function in π¦, so we donβt need to input our value of π₯.

So, in this case, we didnβt actually need to calculate our values of π₯ π. So, letβs now find an expression for π¦ one. We know π¦ zero is negative one, β is 0.25, and π of π¦ is two minus five π¦. So, substituting these into our expression, we get negative one plus 0.25 times two minus five times negative one. And if we calculate this expression, we get three divided by four. Weβll write this in decimal form as 0.75. Now that weβve found π¦ one, we can substitute this into our formula to find π¦ two.

This time, we get π¦ two is equal to π¦ one plus β times π of π¦ one. And we just calculated π¦ one to be 0.75. So, using this, we get π¦ two is equal to 0.75 plus 0.25 times two minus five multiplied by 0.75. And if we calculate this expression, we get five over 16, which weβll write as a decimal 0.3125. We can do exactly the same to find π¦ three. We find itβs equal to 0.421875. And we can then do the same to find π¦ four. To four significant figures, we get π¦ four is equal to 0.3945. And by Eulerβs method, weβre told this will approximately be equal to π¦ evaluated at one.

Therefore, by using Eulerβs method with four steps on the initial value problem π¦ prime is equal to two minus five π¦, where π¦ of zero is equal to negative one. We found that our estimate of π¦ one, to four significant figures, was 0.3945.