Undetermined Coefficients to solve

Click For Summary

Homework Help Overview

The discussion revolves around solving a differential equation using the method of undetermined coefficients, specifically the equation y''' - 6y'' = 3 - cos(x). Participants are examining the differences between their solutions and the solution presented in a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their attempts to find a particular solution, with one participant providing their form of the particular solution and the derivatives calculated. Another participant questions the inclusion of a constant term in the particular solution, suggesting that it may not be linearly independent from the homogeneous solution.

Discussion Status

The discussion is ongoing, with participants exploring the implications of their approaches. One participant has received feedback regarding the need to adjust their particular solution to account for the constant term on the right-hand side of the equation. There is no explicit consensus yet, but guidance has been offered regarding the formulation of the particular solution.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information they can provide. The original poster has not shown all their work, which has led to some uncertainty in the discussion.

kuahji
Messages
390
Reaction score
2
Undetermined Coefficients to solve.

y'''-6y''=3-cosx

So I set everything up, get my system & I get an answer of

y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx

the book has y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx-1/4x^2.

So I'm not sure where the book gets the last term, any ideas?
 
Physics news on Phys.org
The book's answer is correct, but I have no idea where you are going wrong since you haven't shown your work...
 
Ok, I have yp= A + B cosx + c sinx
y'p=-Bsinx+C cosx
y''p=-B cosx - C sinx
y'''P=Bsinx - C cosx

After I have plugged these back into the LHS, I get a system of
B+6C=0
6B-C=-1

These give me my coefficients -6/37cosx & 1/37sinx.
 
Oh & yh=r^2(r-6), which is where I get c1+c2x+c3e^6c.
 
kuahji said:
Ok, I have yp= A + B cosx + c sinx

Well that's your problem right there, the constant term 'A' is not linearly independent to your homogeneous solution, which also contains a constant term... to account for the constant term (3) on the RHS of your DE, you need to add a term of the form Ax^r where r is the smallest non-negative integer such that no terms of the same order are present in your homogeneous solution...

kuahji said:
where I get c1+c2x+c3e^6x
 
Thanks. It works out nicely now.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K