Differential Operator to prove identity

1. Mar 10, 2013

CAF123

1. The problem statement, all variables and given/known data
Use $D = \frac{d}{dx}$as a differential operator and the following $$(D - a)(D -b)[f(x)e^{\lambda x}] = e^{\lambda x} (D + \lambda -a)(D + \lambda -b)f(x)$$ to obtain $$(D^2 + D +1)[(Ax^2 + Bx + C)e^{ix}] = (iAx^2 + [iB + (4i + 2)A]x + 2A + (2i + 1)B + iC)e^{ix}$$

3. The attempt at a solution

This question is part of a bigger question about solving a differential equation with a complex RHS. I proved the first equality in another exercise. To get it in the same form so I could use it I found an a,b such that the left hand sides of both equalities hold. That is, :$$D^2 - bD - aD + ab = D^2 + D + 1 \Rightarrow a = -1/2 \pm \sqrt{3}/2 i, \,\,b = 1/(-1/2 \pm \sqrt{3}/2 i).$$
I then subbed in these a,b on the RHS of the first equality, $\lambda = i$ and $f(x) = (Ax^2 + Bx + C)$. With this I proceeded and multiplied out terms etc.. and in the end I recover some terms but others have the value for a calculated on the denominator. E.g I want a single iBx term but in my answer I have $iBx/(-1/2 \pm \sqrt{3}/2 i)$ etc..

So, I just want to check: Is my method okay?

2. Mar 11, 2013

CAF123

Is my method okay? I can post my work if required.

3. Mar 11, 2013

DimReg

I haven't checked your expressions for a and b yet, but it looks like you have the right idea. The second formula is a specific case of the first, so you just have to plug in the right numbers/equations.

4. Mar 11, 2013

CAF123

My expression for $a$ is$-1/2 \pm \sqrt{3}/2 i$ and that for $b$ is $-1/2 \mp \sqrt{3}/2 i$. When I sub in these to the first equality I don't recover the terms.

5. Mar 11, 2013

DimReg

Well, I checked your a and b, and they seem to be correct. I would guess you are making an algebra mistake somewhere, it might be useful for you to post your work if that's the case. Otherwise, I don't see anything wrong with your approach, I can only suggest being more careful.

Edit: I worked it out myself, and your approach works (unless I made some miraculous mistakes that made it come out correctly). Just make sure you made no mistakes and you should be ok. I found it was a lot of work to do in one go, so I split it up into easy to work with terms then recombined at the end.

Last edited: Mar 11, 2013
6. Mar 11, 2013

CAF123

Subbing in to the second equality I get (Take a = $-1/2 + \sqrt{3}/2\,i$ and $b = -1/2 - \sqrt{3}/2\,i$ for convenience to type: $$e^{ix}(D + i - (\frac{-1}{2} + \frac{\sqrt{3}}{2}i)(D + i + \frac{1}{2} + \frac{\sqrt{3}}{2}i)$$ Multiplying out gives: $$e^{ix}(D^2 + iD + bD + iD + i^2 +ib -aD -ai + 1)(Ax^2 + Bx + C)$$ Expand: $$e^{ix}(2A + 4iAx + b2Ax + biAx^2 - a2Ax -\sqrt{3}/2\, Ax^2 + i/2 Ax^2 + 2iB +bB + ibBx - aB + i/2Bx - \sqrt{3}/2 \,Bx + bCi + iC/2 -\sqrt{3}/2\,C$$

I get some terms (like 4iAx) but the others have a and b attached to them and they are not cancelling.
Thanks.

Edit: Nearly there.. just got one stray term

Last edited: Mar 11, 2013
7. Mar 11, 2013

DimReg

Ok, expressions like that are just terrible to work with. That's 16 terms in the last expression...

Try this: when you multiply out the expression, group the terms into powers of D. So you should have a D^2 term, a D term, and a D^0 term. Each is fairly simple. That's what I did.

8. Mar 11, 2013