00:01
For this problem, we're going to be using oilers method to approximate the y value that corresponds to an x of 1.
00:08
And so we're told that we want to use five intervals.
00:13
So that means that n is equal to 5.
00:17
And because our range goes from, or our domain goes from 0 to 1, that tells us that our h value, or our step size is going to be 1 divided by 5, which will be 0 .2.
00:30
So we can go ahead and use this to fill out our column for x of n values.
00:36
So we know that essentially each time we increase our n by 1, we're going to increase our x by 0 .2.
00:44
So that means that our x values will be 0 .2, 0 .4, 0 .6, 0 .8, and 1.
00:54
And now we're going to use oilers equation or oilers method in order to fill out the right side of our column.
01:02
And so we know that in order to approximate our y value, we're going to use this formula right here.
01:10
Our y sub n is equal to y sub, okay, so i'll write this as y sub 1 is equal to y sub n minus 1, which will be y sub 0, plus h, which is 0 .2, times f of, x -0 y -0 which is going to be three times x -0 which is zero minus y -s of zero so minus one and when we plug in all of our values here we get y sub 1 is equal to 1 plus 0 .2 times negative 1 and we're going to get y sub 1 is equal to 0 .8 so we'll fill that in right there and we can do the same thing to calculate y -sub 2 y 2 will be equal to y sub 1 plus 0 .2 times 3 times x of 1 minus y sub 1.
02:07
So when we fill in our values here, we get y sub 2 is equal to 0 .8 plus 0 .2 times 2 times 0 .2.
02:17
So i'll go ahead and write that.
02:19
Just write that out right there.
02:20
Oops...