# Help with differential operators

1. Nov 14, 2007

### joker_900

1. The problem statement, all variables and given/known data
This is a problem about differential operators, but I don't really get the notation used. I have L1 = (d/dx + 2) and L2 = (d/dx - 1)

Find L1(xe^-2x)

Show that L1L2 = L2L1 and find L1L2 in terms of d/dx, d2/dx2, etc.

2. Relevant equations

3. The attempt at a solution
So i thought that L1 = (d/dx + 2) means L1(f) = (df/dx + 2), so

L1(xe^-2x) = -2xe^-2x + e^-2x + 2

And so L1L2 is just L1 acting on L2 (or L1L2(f) is just performing L2 on f and then performing L1 on that result) and so L1L2 is just the differential of L2 plus 2

L1L2 = d2/dx2 + 2

But by that logic, L2L1 = d2/dx2 - 1

So what's wrong

Thanks

2. Nov 14, 2007

### CompuChip

No, actually it means that
L1(f) = (d/dx + 2) f = df/dx + 2f.
In particular, you see that you should read a constant c as that constant times the zeroth derivative (= the function).

almost,
L1(xe^-2x) = -2xe^-2x + e^-2x + 2 x e^-2x (= e^-2x).

You better always act them on a test function and then throw that away afterwards. So for example, L1 L2 can be found by working out
L1 L2 f = (d/dx + 2) (d/dx - 1) f = (d/dx + 2) (df/dx - f) = (d2f/dx2 - df/dx + 2 df/dx - 2 f)
hence
L1 L2 = d2/dx2 - d/dx + 2 d/dx - 2,
and similarly for L2 L1 (didn't check for errors, you'll find out soon enough if I did it right )

3. Nov 14, 2007

### Kurdt

Staff Emeritus
If l1 operates on a function, f(x), you will have:

$$L_1 f=\frac{df}{dx}+2f$$