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March 27, 2017

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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

  • Math - ,

    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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