00:01
So in this problem here, we need to find l1, l2, and then l2, l1.
00:06
So remember, this means that we're applying l2 first and then l1 to that.
00:11
So l1, l1, l2 of y, for example, right? so this is d plus x of d plus 2x minus 1 of y.
00:23
So we're going to do this first here.
00:27
So this is, okay, so bring that to d plus x.
00:31
And then we'll have d, y, plus 2x minus 1 of y, like that.
00:39
Now we're going to apply this operator here.
00:43
So we'll first do the d.
00:45
So we get d squared y.
00:48
Now we need to take d of this.
00:51
So that's a derivative of this whole thing.
00:53
So again, remember, we're going to need to use product rule.
00:55
So it's going to be first, first times derivative second.
01:00
So first 2x minus 1 and then derivative of the second is dy and then plus the derivative of this is just going to be 2, 2 and then times the second.
01:14
So 2y.
01:15
Now we're just going to take x and multiply it through by here.
01:20
So we get plus x the y and then 2x minus 1, right, plus so x times 2x minus 1.
01:31
1 y like so now let's group all the terms in this operator here so first we have the d squared and then all the d terms we have this term here and this term here so this becomes 2x minus 1 plus x so that becomes 3x minus 1 so we'll have 3x minus 1 b d and then lastly, we'll have our just y terms.
02:10
So that's going to be these two.
02:11
So we'll just take this plus two.
02:14
So plus, go ahead and multiply this out as well.
02:18
So we have 2x squared, 2x squared minus x plus 2.
02:24
That's all of y.
02:27
So this is our l1, l2 here.
02:33
L1, l2.
02:34
Now to find l2, l2, l2, of y...