Nonhomogeneous Linear Differential Equations with Constant Coefficients

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Discussion Overview

The discussion revolves around finding particular solutions to the nonhomogeneous linear differential equation y''' + 3y'' + 3y' + y = e^(-x) + 1 + x. Participants explore different methods for solving the equation, specifically focusing on the method of undetermined coefficients and variation of parameters.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their approach to finding a particular solution by splitting it into two parts, y_p1 and y_p2, and concludes with y_p = -2 + x.
  • Another participant points out that since e^(-x), xe^(-x), and x^2e^(-x) are solutions to the homogeneous equation, y_p1 = e^(-x) yields no new information and suggests trying y_p1 = x^3e^(-x).
  • A subsequent participant calculates y_p using y_p1 = (x^3)(e^(-x)) and arrives at (1/6)(x^3)e^(-x) - 2 + x, seeking confirmation of their result.
  • One participant confirms the correctness of the calculation provided by the previous participant.
  • Another participant expresses a preference for variation of parameters over undetermined coefficients, citing anxiety during exams and a belief that variation of parameters is more methodical and involves less room for error.
  • A participant challenges the notion of "guessing" the form of the solution, arguing that the next logical step after the known solutions is to try x^3e^(-x).
  • There is a mention that variation of parameters can involve less work in many cases.

Areas of Agreement / Disagreement

Participants express differing opinions on the methods used for solving the differential equation, with some favoring undetermined coefficients and others preferring variation of parameters. There is no consensus on the best approach, as participants share their preferences and reasoning.

Contextual Notes

Participants discuss the challenges of selecting appropriate forms for particular solutions, particularly when existing solutions to the homogeneous equation are present. The discussion reflects varying levels of comfort with different solution methods.

Who May Find This Useful

Students and practitioners interested in solving nonhomogeneous linear differential equations, particularly those exploring different solution methods and their applications in mathematical contexts.

highlander2k5
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I was wondering if anyone could check my work on this to make sure I'm doing this right for finding a particular solution to y''' + 3y'' + 3y' + y = e^(-x) + 1 + x. First I split the problem into 2 halfs y_p1 and y_p2.

y_p1 = Ce^(-x)
-Ce^(-x) + 3Ce^(-x) - 3Ce^(-x) + Ce^(-x) = e^(-x)
0*Ce^(-x) = e^(-x)
y_p1 = 0

y_p2 = C1 + (C2)x
0 + 3(0) + 3(C2) + C1 + (C2)x = 1 + x
get like terms together so...
(C2)x = x and 3(C2) + C1 = 1 therefore C2 = 1 and C1 = -2
y_p2 = -2 + x

particular solution = y_p1 + y_p2 = -2 + x
and then the general solution to the problem would be:
C1e^(-x) + c2(x)e^(-x) + C3(x^2)e^(-x)
 
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e-x, xe-x and x2e-x are solutions to the homogeneous solution so of course y_p1= e-x gives you nothing. Try y_p1= x3e-x.
 
If I use (x^3)(e^-x) I get y_p = (1/6)(x^3)e^(-x) - 2 + x does that sound right?
 
Yes, that is correct.
 
Thanks... I understand why now.
 
i find in this situation variation of parameters works better than undetermined coefficients, when you get to having to guess the form of somin like x^3e^-x i just freak out :P
 
I don't see why you would have to "guess". Obviously, something of the form e-x is called for but e-x, xe-x, and x2e-x are already "taken". x3e-x is obviously "next in line".
 
ok fair point :P I am lazy though, i freak out in exams and make mistakes, variation of parameters, for me, leaves less room for error as its more methodical
 
it involves less work in a lot of cases as well
 

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