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Homework Help: Simple Linear Differential Operator Problem

  1. May 28, 2012 #1
    1. (D+1)(D-x)(2e^x+cosx)



    2. None



    3. (D+1)(D-x)(2e^x+cos(x))=D(2e^x-sinx-2xe^x-xcosx)+1(2e^x-sinx-2xe^x-xcosx)=-4e^x+2e^x-xcosx-2cosx+xsinx-sinx

    The correct answer is 2e^x(xsinx-3sinx+2xcosx)
     
  2. jcsd
  3. May 28, 2012 #2

    HallsofIvy

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    It is the part that you don't show that is wrong. There is no way that "D"
    or "D- 1" can change "2e^x" to "2xe^x". I suggest you do that again.
     
  4. May 28, 2012 #3
    It's (D-x).
     
  5. May 28, 2012 #4
    Please open this again as the last poster closed it because he or she misread the equation.
     
  6. May 29, 2012 #5

    HallsofIvy

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    Yes, I misread the equation. No, I did not "close" this thread.
    [itex]D(2e^x+cos(x))= 2e^x- sin(x)[/itex]
    [itex]x(2e^x+cos(x))= 2xe^x+ xcos(x)[/itex]

    [itex](D- x)(2e^x+ cos(x))= 2e^x- 2xe^x- sin(x)- xcos(x)[/itex]

    [itex]D(2e^x- 2xe^x- sin(x)- xcos(x))= 2e^x- 2e^x- 2xe^x- cos(x)- cos(x)+ xsin(x)= -2xe^x- 2cos(x)+ xsin(x)[/itex]

    [itex](D+1)(2e^x- 2xe^x- sin(x)- xcos(x))= -2xe^x- 2cos(x)+xsin(x)+ 2e^x- 2xe^x- sin(x)- xcos(x)= 2e^x- 4xe^x- 2cos(x)- sin(x)+ xsin(x)- xcos(x)[/itex]

    What you state as the "correct answer" is impossible. You cannot get [itex]e^x[/itex] multiplied by the trig functions.
     
  7. May 29, 2012 #6
    Thanks Ivy. This was from the first set of problems on the topic I have worked so I was not sure I was solving it correctly, but looks like I did. The solution in the back of the book must be incorrect. What do the open and closed envelopes next to the threads signify?
     
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