Differential Operator to prove identity

In summary, the homework statement is trying to solve a differential equation with a complex RHS. The first equation states that the left hand side of the equation is equal to D^2 - bD - aD + ab. To solve for a, the student found an a such that the left hand side of both equations held. To get b, the student found an a such that the left hand side of the first equation held and then substituted in b on the RHS of the first equation. With this information, the student was able to solve for f(x) in terms of Ax^2 + Bx + C. However, there are still terms with a and b attached that do not cancel. The student was advised to check
  • #1
CAF123
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Homework Statement


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 $$(D^2 + D +1)[(Ax^2 + Bx + C)e^{ix}] = (iAx^2 + [iB + (4i + 2)A]x + 2A + (2i + 1)B + iC)e^{ix}$$

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?
 
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  • #2
Is my method okay? I can post my work if required.
 
  • #3
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
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
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.
 
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  • #6
Subbing into 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
 
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  • #7
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
DimReg said:
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.

Got it! Thanks for your advice.
 

FAQ: Differential Operator to prove identity

What is a differential operator?

A differential operator is a mathematical operation that combines differentiation with other mathematical operations. It is used to solve differential equations and prove identities in various branches of mathematics, such as calculus and differential geometry.

How is a differential operator used to prove identities?

A differential operator is used to manipulate and simplify expressions in order to show that they are equivalent. By applying the operator to both sides of an equation, the expression can be transformed in a way that makes the identity apparent.

What are some common differential operators used in proving identities?

Some common differential operators used in proving identities include the derivative, gradient, divergence, and curl operators. These operators can be applied to functions, vectors, and tensors to simplify expressions and prove identities.

Can differential operators be used in any branch of mathematics?

Yes, differential operators can be used in various branches of mathematics, including calculus, differential geometry, and physics. They are an important tool in solving differential equations and proving identities in these fields.

Are there any limitations to using a differential operator to prove an identity?

While differential operators can be a powerful tool in proving identities, they may not be applicable to all types of equations or expressions. It is important to carefully consider the properties and limitations of the operator being used in order to successfully prove an identity.

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