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Doubt on P.I of [itex]e^{ax} V [/itex]

  1. Oct 23, 2012 #1
    I have a doubt in finding out the Particular Integral of eaxV, where 'V' is a function of x.

    I saw the book but they seem to be somewhat unclear regarding this-

    Here's the derivation from the book:

    If u is a function of x, then
    D(eaxu)= eax Du + aeaxu= eax (D+a)u
    Similarly you can do for D2,D3... and so on

    and after substituting V=f(D+a)u you will get the final equation as -

    [itex]\frac{1}{f(D)}(e^{ax}V)[/itex]=[itex]e^{ax} \frac{1}{f(D+a)}[/itex]V

    Now my doubt is how do you decide where to put that f(D+a) in the right side of the above equation? I could have interchanged eax and V. This could have resulted in an entirely different answer. I couldn't get that

    Thanks a lot
    Last edited: Oct 23, 2012
  2. jcsd
  3. Oct 23, 2012 #2


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    If [itex]\frac{1}{f(D)} = Ʃb_n D^n[/itex], [itex]\frac{1}{f(D)}(e^{ax}V) = Ʃb_n D^n(e^{ax}V) = Ʃb_n e^{ax}(D+a)^n(V) = e^{ax} \frac{1}{f(D+a)}V[/itex]
  4. Oct 23, 2012 #3

    I still don't get how you take that [itex]e^ax[/itex] out of [itex]D^n(e^{ax}V)[/itex] and not 'V'.

    Thanks a lot :)
  5. Oct 23, 2012 #4


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    Just by applying D(eaxu)= eax Du + aeaxu= eax (D+a)u
    That took the eax out but not the u. Same with the higher powers.
  6. Oct 23, 2012 #5
    Sorry. I overlooked it. I asked a stupid question.

    Thanks a lot :)
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