In this segment, we’ll talk about Euler’s method of solving ordinary differential equations,
we’ll look at an example.
So, the example we’re asked to look at is this
first order ordinary differential equation right here. We're asked to calculate the value of y
at 3 when y sub 0 is already given to us as 5 and we'll take a step size of take a step size of 1.5,
which simply implies that we’ll take two steps in order to do this particular problem. So,
let's go through the two steps and see what we get as the value of y at x equal to three.
So, the differential equation is given in this format here,
and what we have to do is try to rewrite it in the format which is needed for Euler's method
and that is this format. And that's what gives us the function f here;
slope is equal to some function f of x and y, and of course subject to y sub 0 equal to 5.
So, let's go ahead and apply the Euler's method formula. The general formula is given as this,
where h is the step size and let's choose i equal to zero. So, if we choose i equal to zero, I get y
one is equal to y naught plus the value of the function f at x naught comma y naught times h.
So, this is what you do, is to think about the steps and you'll be all right. X naught is equal
to 0 because that's what the initial condition is given y naught is 5, that's the value of the y
at this value of x, and then h is given to us as 1.5. So, all we have to do is say y
1 is equal to 5 plus the value of the function at 0 comma 5 so having said that what is the
value of the function at 0 comma 5 times the h value which is 1.5. So, having said that,
you have to substitute x equal to 0 and y equal to 5 there, so let’s see what we get.
And this one gives you 5 plus 1 times 1.5, and that is 6.5.
So, 6.5 is the value of y1, which is the approximate y at 1.5 because the step size is,
x1 is x naught plus h right, so x naught is zero and h is 1.5 so it’s 1.5.
So, that's where we are finding out whatever value of y is, but that’s not what we’re after.
We're after the y value of y at x equal to 3, so that means that we have to take
one more step. Please don't call it one more iteration, because we are not refining the value,
we are going a step ahead from 0 to 1.5, and the next step will go from 1.5 to 3. So,
if I use i equal to 1 now, I’ll get y2 is equal to y1 plus the value of the function at x1 comma y1
times h. And what is the value of, let's write down x1, we just said is 1.5, the corresponding
value of y at that particular point will now be 6.5, and of course our h is still 1.5.
Let's go and calculate what x 2 is. X 2 is nothing but x 1 plus h so, so it would be 1.5 plus 1.5,
that gives you 3. So, when we calculate the value of y 2, it will be in fact,
the approximate value of y at x equal to 3. So, we get y2 is equal to what is y1, it is 6.5,
plus the value of the function calculated at 1.5 comma 6.5, times the step size, 1.5. So,
what we have is what is the value of the function at these two values 3 e to the power minus 1.5,
minus 0.4 times the value of y which is 6.5, times 1.5, so this gives me 6.5 plus minus 1.93061
times 1.5, and this value here turns out to be 3.604.
So, this is the approximate value of y at x2 which we know that x2 is 3, and that's what we're after.
So, y sub 3 is approximately equal to 3.604,
y sub 3 can be found exactly to be 2.763, that is the exact value guys. And in the description
of this YouTube video, you will find a link to how I found 2.763, quite a number of students ask me
that question, how did I get 2.763? And then if you want to calculate your relative true error,
in this case, it will turn out to be 30.44 percent. The relative true error is the
difference, is the relative difference, between the approximate value and the exact value.
So, what you’re finding out is that, even a step size of 1.5 gives you a very large
relative true error. So, one way of doing this is to decrease the step size, and of course that
will take more time to do it by hand, but you can try to write a computer program and see that if
you can see that when you reduce the step size, that you get your relative two errors reduced,
and that is the end of this segment.