Nonhomogeneous diff equations method of undeterined coeff.

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Homework Help Overview

The discussion revolves around finding the general solution to a nonhomogeneous differential equation using the method of undetermined coefficients. The specific equation under consideration is y'' - 2y' - 3y = 3te^-t.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the form of the particular solution, yp, suggesting variations such as (At^2 + Bt + C)e^-t and questioning the necessity of including the constant term C. There is also a focus on ensuring that the proposed solution does not duplicate existing solutions.

Discussion Status

Some participants have provided guidance on the form of the particular solution and have pointed out potential errors in the differentiation process. There is an ongoing exploration of the correct approach to finding yp, with participants sharing their attempts and corrections.

Contextual Notes

Participants are working under the constraints of the method of undetermined coefficients and are addressing the specific challenge of ensuring the proposed solution does not overlap with the homogeneous solution.

iamtrojan3
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Homework Statement


Find the general solution to the diff equation using undetermined coefficients
y''-2y'-3y = 3te^-1


Homework Equations





The Attempt at a Solution


r^2 - 2r -3 = 0
r = -1, 3
so y = c1 e^-t + c2e^3t + yp
since e^-t already exists as a solution, i have to multiply my Yp by t to make sure i don't' end up with the same solution.
So my Yp with the unknown coeffcient should be,
yp = At^2e^-t
should it be (At^2+bt+c)e^-t instead or something else? I've done it both ways and came up with wrong answers.
I know what to do after figuring out yp, its just i can't get it right with ideas i have right now.
Any help is greatly appreciated.
Thank you!
 
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iamtrojan3 said:

Homework Statement


Find the general solution to the diff equation using undetermined coefficients
y''-2y'-3y = 3te^-1


Homework Equations





The Attempt at a Solution


r^2 - 2r -3 = 0
r = -1, 3
so y = c1 e^-t + c2e^3t + yp
since e^-t already exists as a solution, i have to multiply my Yp by t to make sure i don't' end up with the same solution.
So my Yp with the unknown coeffcient should be,
yp = At^2e^-t
should it be (At^2+bt+c)e^-t instead or something else? I've done it both ways and came up with wrong answers.
I know what to do after figuring out yp, its just i can't get it right with ideas i have right now.
Any help is greatly appreciated.
Thank you!

Yes, go with (At2 + Bt + C)e-t for your particular solution. Show us your work, and we'll figure out what's going wrong.
 
thanks for responding, Let's see if i have the yp's correct
yp = (At^2+Bt+c)e^-t
y'p = (2At + B)e^-t -(At^2+Bt+c)e^-t
y''p = 2Ae^-t - (2At + B)e^-t - (2At+B)e^-t -(At^2+Bt+c)e^-t
 
In general, it you have a power of t on the right side, you will need to try a polynomial up to that power. If your right side were te^{at} and e^{at} were not already a solution, you would try (At+ B)e^{at}. Since, here, e^{at} is a solution, multiply that by t: try (At^2+ Bt)e^{at}.

You should find that you do NOT need that "c". (Using it will just give C= 0.)

iamtrojan3 said:
thanks for responding, Let's see if i have the yp's correct
yp = (At^2+Bt+c)e^-t
y'p = (2At + B)e^-t -(At^2+Bt+c)e^-t
y''p = 2Ae^-t - (2At + B)e^-t - (2At+B)e^-t -(At^2+Bt+c)e^-t
You can add those two middle terms and the last term should be "+".
y"/= 2Ae^-t- 2(2At+ B)e^-t+ (At^2+ Bt+ C)e^-t
 
Thanks a lot HallsofIvy, it seems i just subbed in wrong for y''p and the "c" ended up canceling out anyways.
 

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