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## Differential equations

### Unit 1: Lesson 3

Euler's Method# Worked example: Euler's method

AP.CALC:

FUN‑7 (EU)

, FUN‑7.C (LO)

, FUN‑7.C.4 (EK)

Finding the initial condition based on the result of approximating with Euler's method.

## Video transcript

- [Voiceover] Now that we are
familiar with Euler's method, let's do an exercise that
tests our mathematical understanding of it, or at
least the process of using it. So, it says consider the
differential equation: the derivative of y with respect to x is equal to three x minus two y. Let y is equal to g of x be a solution to the differential equation
with the initial condition g of zero is equal to
k where k is constant. Euler's method starting at x equals zero with the a step size of
one gives the approximation that g of two is approximately 4.5. Find the value of k. So once again, this is saying
hey, look, we're gonna start with this initial condition
when x is equal to zero, y is equal to k, we're
going to use Euler's method with a step size of one. So, we're essentially going
to use, we're going to step once from zero to one, and
then again from one to two. And then that approximation
is going to give us 4.5. And so, given that we started
at k, we should be able to figure out what k was to get us to g of two being approximated as 4.5. So with that, I encourage
you to pause the video, and try to figure this out on your own. I am assuming you have tried
to figure this out on your own. Now we can do it together. And I'll do the same thing that we did in the first video on Euler's method. I'll make a little table here
so let me make a little table. I can draw a straighter line than that. That's only marginally straighter, but it will get the job done. So let's make this column
x, I'm going to give myself some space for y, I might do some calculation here, y, and then dy/dx. Now, we can start at
our initial condition. When x is equal to zero, y is equal to k. When x is equal to zero, y is equal to k. And so, what's our derivative
going to be at that point? Well, dy/dx is equal
to three x minus two y. So in this case, it's three
times zero minus two times k, which is just equal to negative two k. And so now we can increment one more step. We have a step size of
one, so at each step we're going to increment x by one, and so we're now going to be at one. Now what's our new y going to be? Well, if we increment
x by one, and our slope is negative two k, that means
we're going to increment y by negative two k times
one, or just negative two k. So, negative two k. So k plus negative two k is negative k. So, our approximation using
Euler's method gets us the point one negative
k, and then what is going to be our slope starting at that point? So one negative k, our slope
is going to be three times our x, which is one, minus
two times our y, which is negative k now, and this is
equal to three plus two k. And now we'll do another step of one, because that's our step size. Another, whoops, I'm going to get to two. Now this is the one that
we care about right? Because we're trying to
approximate g of two. So we have to say, what
does our approximation give us for y when x is equal to two? And we're going to have
something expressed in k, but they're saying that's going to be 4.5, and then we can use that to solve for k. So what's this going to be? So if we increment by one in x, we should increment our y by
one times three plus two k. So we're going to increment
by three plus two k, or negative k plus three plus two k is just going to be three plus k. And they're telling us that our approximation gets that to be 4.5. So three plus k is equal to 4.5. So the k that we started
with must have been, if we just subtract three from both sides, this is a decimal here, it must have been k must be equal to 1.5,
and you can verify that. If this initial condition right over here, if g of zero is equal to 1.5,
then you put 1.5 over here. Then over here you would
get 4.5, and we're done.