What are the steps for finding u1 and u2 in the variation of parameters method?

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Discussion Overview

The discussion revolves around the variation of parameters method for solving a second-order linear differential equation. Participants explore the steps involved in finding the functions u1 and u2, particularly in the context of a specific problem involving a non-homogeneous term.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • Tom presents a specific problem and outlines his approach, including the characteristic equation and the functions derived from it.
  • Some participants suggest that the characteristic equation should be expressed differently, indicating a potential disagreement on the form of the solution.
  • Daniel advocates for the application of Lagrange's method, implying an alternative approach to the problem.
  • Another participant inquires about references for solving systems of equations involving trigonometric equations, indicating a broader interest in related topics.
  • Tom later mentions having found the answer by recalling Cramer's rule, suggesting a resolution to his initial query but not necessarily to the overall discussion.

Areas of Agreement / Disagreement

Participants express differing views on the correct form of the characteristic equation and the methods to apply. The discussion remains unresolved regarding the integration steps for finding u1 and u2, as well as the appropriateness of the methods suggested.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the forms of the solutions and the integration techniques. The integration steps for u1 and u2 are not fully explored, leaving some uncertainty in the method's application.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in differential equations, particularly those learning about variation of parameters and related integration techniques.

Cafka
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Hello,
I'm trying to understand this concept. Jere's the problem I'm doing.
I have to find the general solution for:
y'' + 36y = -4xsin(6x)

So you then solve for your characteristic equation and get lamda = +/- 6
so y1 = e^-6x and y2 = e^6x
You get your matrix for w, w1, and w2.
w = 12
w1 = 4xe^6xsin(6x)
w2 = -4xe^-6xsin(6x)

I have the problem at getting u1 and u2.
u1 = 1/3 the integral of xe^6xsin(6x) dx
u2 = -1/3 the integral of xe^-6xsin(6x) dx

How do you do that integration?

Thanks for your help,
Tom
 
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Your characteristic equation should be of the form y=Asin6x+Bcos6x since the the squareroot of -36 is 6i and -6i =)
 
Yeah,and Lagrange's method kicks ass.So apply it.

Daniel.
 
I was wondering if anyone knew of any good references on solving systems of equations involving trigonometric equations. Any information would be appreciated.
 
Found the answer - I had forgotten Cramer's rule.
 

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