### Video Transcript

Consider the initial value problem
π¦ prime is four π₯ minus three π¦ where π¦ at π₯ equal to zero is negative one. Use Eulerβs method with π equal to
five steps on the closed interval zero one to find the value of π¦ at π₯ is equal to
one.

Weβve actually been given a
differential equation π¦ prime is equal to four π₯ minus three π¦ and an initial
value π¦ at π₯ is equal to zero is negative one. This means that our starting point
for Eulerβs method is the point π₯ zero, π¦ zero which is equal to zero, negative
one. We also know that we have five
steps and that our closed interval zero, one means that zero is less than or equal
to π₯ is less than or equal to one. So we have π equal to five and
zero less than or equal to π₯ less than or equal to one.

Weβre asked to use Eulerβs method
to find the value of π¦ at π₯ is equal to one. To do this, the first thing we need
to do is to find the value of β. Thatβs our step size. Now, β is the interval width
divided by the number of steps, in our case, the interval where thereβs a maximum π₯
value one minus the minimum π₯ value zero which is one. The number of steps is five. So our step size is one over
five. Letβs just sketch our Interval so
that we can refer to it throughout our calculations. Our first π₯ value is zero, which
is π₯ zero. Our step size is one over five,
which is 0.2. So our next point π₯ one is
0.2. Our next π₯ value π₯ two is 0.4, π₯
three is 0.6, π₯ four is 0.8, and π₯ five is one.

And remember, weβre being asked to
calculate the value of π¦ at π₯ is equal to one. So thatβs π¦ of π₯ five, which is
π¦ five. So weβre going to use Eulerβs
method to calculate π¦ five. We canβt go directly to π¦ five
because the only value we have is π₯ zero, π¦ zero. So we start at this point and
calculate π¦ one, π¦ two, then π¦ three, then π¦ four, and then π¦ five. So to use the formula, we need to
know the value of π₯ zero, π¦ zero, which is zero, negative one. We need to know β, which is 0.2,
and we need to know what the function π of π₯, π¦ is. In our differential equation, π¦
prime, which is dπ¦ by dπ₯, is equal to four π₯ minus three π¦. So π of π₯ π¦, in our case, is
four π₯ minus three π¦.

We know that π¦ zero is equal to
negative one. And we can use this to find π¦
one. And in the formula, if π is equal
to one, then π minus one is equal to zero so that π¦ one is equal to π¦ zero plus β
times π of π₯ zero, π¦ zero. We know that π¦ zero was negative
one and that β is 0.2. So we need to find π of π₯ zero,
π¦ zero. In our differential equation, π of
π₯ zero, π¦ zero is four π₯ zero minus three π¦ zero. And thatβs equal to four times zero
minus three times negative one, which is equal to three, so that π¦ one is equal to
negative one plus 0.2 times three, which is negative 0.4. Now, we can use π¦ one to calculate
π¦ two.

We know that π¦ one is equal to
negative 0.4. We know that β is 0.2. So we need to work out π of π₯
one, π¦ one. And thatβs equal to four π₯ one
minus three π¦ one. And since π₯ one is 0.2, thatβs
four times 0.2 minus three times negative 0.4, which is 0.8 plus 1.2 which is
two. So π¦ two is equal to negative 0.4
plus 0.2 times two which is actually zero. Now, if we put π equal to three in
our formula, π¦ three is equal to π¦ two plus β times π of π₯ two π¦ two. We know that π¦ two is equal to
zero and β is 0.2. So we need to work out π of π₯ two
π¦ two. And thatβs equal to four π₯ two
minus three π¦ two. And since π₯ two is 0.4, thatβs
four times 0.4 minus three times zero, which is 1.6. So we have π¦ three equal to zero
plus 0.2 times 1.6, which is 0.32.

Now, with π equal to four, we have
π¦ four is equal to π¦ three plus β times π of π₯ three, π¦ three. π of π₯ three π¦ three is four π₯
three minus three π¦ three. So we have four times 0.6 minus
three times 0.32 which is 1.44, so that π¦ four is equal to 0.32 plus 0.2 times 1.44
and thatβs 0.608. And finally with π equal to five
in our formula, we have π¦ five equal to π¦ four plus β times π of π₯ four π¦
four. π of π₯ four π¦ four is equal to
four π₯ four minus three π¦ four. And thatβs equal to four times 0.8
minus three times 0.608 which is 1.376 so that we have π¦ five equal to 0.608 plus
0.2 times 1.376. And thatβs equal to 0.8832. And remember that itβs exactly π¦
five that weβre looking for because this is the value of π¦ at π₯ is equal to one,
so that π¦ of one is 0.8832.