### Video Transcript

Consider the initial value problem
π¦ prime is equal to two π¦ plus five π₯ minus four, where π¦ of zero is equal to
two. Use Eulerβs method with six steps
to estimate the value of π¦ evaluated at three to two decimal places.

Weβre given an initial value
problem, and we need to use Eulerβs method with six steps to estimate the value of
π¦ evaluated at three. We should give our answer to two
decimal places. Letβs start by recalling Eulerβs
method. Eulerβs method gives us a way of
approximating outputs of our function π¦. In our case, since we want to use
six steps, our approximation will be π¦ six. So to use this formula to find the
value of π¦ six, we need to know how to find π¦ five, π₯ five, β, and π.

Letβs start with the function
π. This is actually just the function
in our initial value problem. In our case, since we want to solve
the initial value problem, π¦ prime is equal to two π¦ plus five π₯ minus four,
weβll set our function π of π₯, π¦ to be equal to two π¦ plus five π₯ minus
four. So weβve now found our function
π. Remember, in our initial value
problem, weβre also told that π¦ of zero is equal to two. In other words, when π₯ is equal to
zero, we know π¦ is equal to two. So weβll use this to give us our
initial values. π₯ zero is equal to zero and π¦
zero is equal to two.

The last thing we need to explain
is the value of β. Remember, weβre estimating the
value of π¦ when π₯ is equal to three. In this case, we want our values of
π₯ to start at zero and get closer and closer to three. We want this to happen in six
equal-width steps. So our initial value of π₯, π₯
zero, is equal to zero. And our final value of π₯ is equal
to three. And we want to split this into six
sections of equal width. We call these steps. And the width of these steps is
equal to β.

From this diagram, itβs now easy to
calculate the value of β. β will be equal to the length of
our line divided by the number of steps. In this case, thatβs three minus
zero divided by six. And we can calculate this. Itβs equal to one-half or 0.5.

Remember, to find the value of π₯
sub π, we need to go up π steps from π₯ zero. So to go from π₯ zero to π₯ one, we
need to add our value of β. We know β is equal to 0.5. So we get π₯ one is equal to
0.5. We can then do the same to find the
value of π₯ two. We need to add 0.5. So we get π₯ two is equal to
one. We can then do the same to find π₯
three is 1.5, π₯ four is two, and π₯ five is 2.5.

Weβre now ready to use Eulerβs
method. We know the value of β. We know our function π. We know all of our values of
π₯. And we know the value of π¦
zero. This means we now have all the
information to find π¦ one. Itβs equal to π¦ zero plus β times
two π¦ zero plus five π₯ zero minus four.

Substituting in π¦ zero is equal to
two, π₯ zero is equal to zero, and β is equal to 0.5, we get π¦ one is equal to two
plus 0.5 times two times two plus five multiplied by zero minus four. And we can then just calculate this
expression; itβs equal to two. Now that we found the value of π¦
one, we can use this to find the value of π¦ two.

This time, we get π¦ two is equal
to π¦ one plus β times two π¦ one plus five π₯ one minus four. This time, our value of π¦ one is
two and our value of π₯ one is 0.5. So we get π¦ two is equal to two
plus 0.5 times two multiplied by two plus five times 0.5 minus four. And if we evaluate this expression,
we get 3.25.

Now that we found π¦ two, we can
find an expression for π¦ three. Remember, we need to go all the way
to π¦ six to find our approximation by using Eulerβs method. This time, π¦ two is 3.25 and π₯
two is equal to one. So we get π¦ three is equal to 3.25
plus 0.5 multiplied by two times 3.25 plus five times one minus four. And we can then just calculate that
this expression is equal to seven. We can then find π¦ four by using
π¦ three is equal to seven and π₯ three is equal to 1.5. And we can then calculate this;
itβs equal to 15.75.

Doing the same again with π¦ four
equal to 15.75 and π₯ four equal to two, we get that π¦ five is equal to the
following expression, which we can calculate is equal to 34.5. We just need to do this one more
time, since Eulerβs method tells us π¦ six will be our approximation of π¦ evaluated
at three. This time, we have π¦ five is 34.5
and π₯ five is equal to 2.5. Using these values, we get that π¦
six is equal to 73.25, which is our final answer.

Therefore, given the initial value
problem π¦ prime is equal to two π¦ plus five π₯ minus four where π¦ of zero is
equal to two, we were able to use Eulerβs method with six steps to show that π¦ of
three is approximately equal to 73.25.