Homework Help: Simple Linear Differential Operator Problem

1. May 28, 2012

sriracha

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

2. May 28, 2012

HallsofIvy

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.

3. May 28, 2012

sriracha

It's (D-x).

4. May 28, 2012

sriracha

Please open this again as the last poster closed it because he or she misread the equation.

5. May 29, 2012

HallsofIvy

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

$(D- x)(2e^x+ cos(x))= 2e^x- 2xe^x- sin(x)- xcos(x)$

$D(2e^x- 2xe^x- sin(x)- xcos(x))= 2e^x- 2e^x- 2xe^x- cos(x)- cos(x)+ xsin(x)= -2xe^x- 2cos(x)+ xsin(x)$

$(D+1)(2e^x- 2xe^x- sin(x)- xcos(x))= -2xe^x- 2cos(x)+xsin(x)+ 2e^x- 2xe^x- sin(x)- xcos(x)= 2e^x- 4xe^x- 2cos(x)- sin(x)+ xsin(x)- xcos(x)$

What you state as the "correct answer" is impossible. You cannot get $e^x$ multiplied by the trig functions.

6. May 29, 2012

sriracha

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?