### Video Transcript

Consider the initial value problem
π¦ prime is equal to π¦ plus one where π¦ at π₯ is equal to zero is zero. Use Eulerβs method with π equal to
five steps on the interval zero, one to find π¦ at π₯ is equal to one.

Weβve been given a first-order
differential equation, π¦ prime is equal to π¦ plus one. In other words, the derivative of
π¦ with respect to π₯ is equal to π¦ plus one. We have an initial value π¦ at π₯
is equal to zero is equal to zero so that our starting point π₯ zero, π¦ zero is the
point zero, zero. Weβre asked to find π¦ at π₯ is
equal to one using Eulerβs method.

Eulerβs method is a way of
numerically integrating the first-order differential equation. How this works is that from a given
starting point over a range of π₯-values β in our case, the closed interval zero,
one β we use Eulerβs formula to calculate successive approximate π¦-values for the
range of π₯-values. To do this, we split our
π₯-interval into equal-size steps. In our case, we have π equal to
five steps. So, we need to split our interval
zero, one into five steps.

And to find the step size β, we
divide the interval width by the number of steps. Since our maximum π₯ is one and the
minimum is zero, our interval width is one minus zero, which we divide by the number
of steps, five, so that our step size is one over five, which is 0.2. In our interval then, if zero is π₯
zero with a step size of 0.2, π₯ one is equal to 0.2, π₯ two is equal to 0.4, π₯
three is 0.6, π₯ four is 0.8, and π₯ five is one.

Remember, weβre asked to find the
value of π¦ when π₯ is equal to one. So, weβre looking for π¦ at π₯ is
equal to π₯ five, which weβll call π¦ five. In order to reach π¦ five, we need
to calculate π¦ one, which is π¦ of π₯ one. We use that to calculate π¦ two,
which is π¦ of π₯ two. We use that to calculate π¦ three,
which is π¦ of π₯ three, and so on until we get to π¦ five, which is π¦ of π₯
five.

Looking at our formula, to
calculate π¦ π, we need π¦ π minus one, β, and a function π. We know that β is 0.2 and a
function π is actually our derivative. In our case, this is a function of
π¦. So, we have π of π¦ is equal to π¦
plus one. Since this function doesnβt depend
on π₯, we can eliminate the π₯ term from our equation. So, we have π¦ π is equal to π¦ π
minus one plus β times π of π¦ π minus one. We now have everything we need to
start calculating our values.

Our starting point is π¦ zero is
equal to zero. We use this in our formula to
calculate π¦ one. If π is equal to one in our
formula, then π minus one is equal to zero. And we have π¦ one is equal to π¦
zero plus β times π of π¦ zero. π¦ zero is equal to zero. β is equal to 0.2. And π of π¦ zero is π¦ zero plus
one. Thatβs equal to zero plus one,
which is one, so that π¦ one is equal to zero plus 0.2 times one, which is 0.2.

Our next step is to calculate π¦
two using π¦ one. And in our formula, π¦ two is equal
to π¦ one plus β times π of π¦ one. That is 0.2 plus 0.2 times π of π¦
one. And π of π¦ one is π¦ one plus
one, which is 0.2 plus one, which is 1.2. So that π¦ two is equal to 0.2 plus
0.2 times 1.2. And thatβs equal to 0.44. π¦ two is, therefore, 0.44. We use this to calculate π¦
three. So, π¦ three is π¦ two plus β times
π of π¦ two. And thatβs equal to 0.44 plus β,
which is 0.2 times π of π¦ two. π of π¦ two is π¦ two plus one,
which is 0.44 plus one. And thatβs 1.44 so that π¦ three is
equal to 0.44 plus 0.2 times 1.44. And thatβs equal to 0.728.

And we use this to calculate π¦
four. And π¦ four is π¦ three plus β
times π of π¦ three. And thatβs 0.728 plus 0.2 times π
of π¦ three. π of π¦ three is π¦ three plus
one, which is 0.728 plus one, which is 1.728. We, therefore, have π¦ four is
equal to 0.728 plus 0.2 times 1.728. Thatβs equal to 1.0736 so that π¦
four is equal to 1.0736.

Finally, we can use π¦ four to
calculate π¦ five, which is equal to π¦ four plus β times π of π¦ four. And thatβs equal to 1.0736 plus 0.2
times π of π¦ four. π of π¦ four is π¦ four plus one,
which is 1.0736 plus one, which is equal to 2.0736. So that π¦ five is equal to 1.0736
plus 0.2 times 2.0736 which is equal to 1.48832. π¦ five is, therefore, 1.48832. And remember that π¦ five is π¦ of
π₯ five, which is equal to π¦ of one. So π¦ of one is equal to
1.48832.

The graph shown is a
computer-generated solution to a differential equation. We found the π¦-values for the five
steps within our interval. And if we plot these points on our
graph, we can see that our solution curve is not too far away from the
computer-generated solution. The computer-generated solution
would have had a much smaller step size and so would be closer to the true
solution. This demonstrates that the smaller
the step size, the more accurate the approximation.