Math
posted by Jet .
Obtain the differential equation from the following:
1. y=C(subscript 1)e^x+ C(subscript 2)xe^x
2. y=[C(subscript 1)e^2x]cos3x+[C(subscript 2)e^2x]sin3x

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 D1 gives:
(D  1) x exp(x) = exp(x)
Applying D  1 again will annihilate the remaining exp(x):
(D1)^2 x exp(x) = (D1) exp(x) = 0
So, the differential operator in problem 1) is (D1)^2 = D^2  2 D + 1
In problem 2), you can write the function in terms of exp[(2+3i) x] and exp[(23i)x]. The operator that annihilates the term exp[(2+3i) x] is
O1 = D  (2+3i)
The operator that annihilates the term exp[(23i) x] is
O2 = D  (23i)
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