Posted by Jet on Sunday, June 17, 2012 at 5:17pm.
Denoting the diferential operator d/dx by D, we can write:
D exp(a x) = a exp(a x) ------->
(D - a) exp(x) = 0
If we have a term x exp(x), then applying D-1 gives:
(D - 1) x exp(x) = exp(x)
Applying D - 1 again will annihilate the remaining exp(x):
(D-1)^2 x exp(x) = (D-1) exp(x) = 0
So, the differential operator in problem 1) is (D-1)^2 = D^2 - 2 D + 1
In problem 2), you can write the function in terms of exp[(2+3i) x] and exp[(2-3i)x]. The operator that annihilates the term exp[(2+3i) x] is
O1 = D - (2+3i)
The operator that annihilates the term exp[(2-3i) x] is
O2 = D - (2-3i)
The perator that will annihikate both terms can then be taken to be O1O2. Obviously if O1 f = 0, then O1O2f = O2O1f = 0.
We have with z = 2+3i
O1O2 = (D - z ) (D - z*) =
D^2 - 2 Re(z) D + |z|^2 =
D^2 - 4 D + 13
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