Simple Linear Differential Operator Problem

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Homework Help Overview

The discussion revolves around a problem involving a linear differential operator applied to the expression (D+1)(D-x)(2e^x+cosx). Participants are analyzing the steps taken to simplify or solve the expression and questioning the correctness of the original poster's calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the application of differential operators and questioning the transformations applied to the terms, particularly the transition from "2e^x" to "2xe^x". There are multiple interpretations of the operations involved.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning. Some participants are suggesting re-evaluations of specific steps and expressing doubts about the correctness of the original answer provided by the poster. There is a recognition of misinterpretations in the problem setup.

Contextual Notes

One participant notes uncertainty regarding the correctness of the solution provided in the textbook, indicating that this is part of their learning process with the topic.

sriracha
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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)
 
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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.
 
HallsofIvy said:
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.

It's (D-x).
 
Please open this again as the last poster closed it because he or she misread the equation.
 
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.
 
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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