Why Not Use Laplace Transforms for This ODE?

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Discussion Overview

The discussion revolves around the application of Laplace transforms to solve a specific ordinary differential equation (ODE) involving a delta function. Participants are examining the initial conditions and the correctness of the solution provided in a textbook.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant asserts that the textbook solution is incorrect due to a discrepancy in the initial condition for x'(0).
  • Another participant presents their own solution using Laplace transforms, suggesting that the textbook may have omitted the Heaviside function in their answer.
  • A question is raised about the interpretation of the initial condition x'(0) and whether it refers to x'(0+) or x'(0-).
  • A later reply acknowledges a misunderstanding regarding the initial condition and suggests that x'(0-) is the relevant value when applying the Laplace transform.
  • It is noted that the presence of the delta function implies that x'(0-) and x'(0+) cannot be equal, leading to the conclusion that x'(0+) is x'(0-) plus an additional increment.
  • A participant questions the rationale behind using Laplace transforms for this problem.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the textbook solution and the interpretation of initial conditions. The discussion remains unresolved regarding the proper application of Laplace transforms in this context.

Contextual Notes

There are uncertainties regarding the interpretation of initial conditions in the presence of a delta function, as well as the potential omission of the Heaviside function in the textbook solution.

Who May Find This Useful

Readers interested in differential equations, Laplace transforms, and the implications of initial conditions in mathematical modeling may find this discussion relevant.

AlonsoMcLaren
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x''+2x'+x=t+delta(t) x(0)=0 x'(0)=1

The textbook, "Elementary differential equations" by Edwards and Penney, gives the answer as -2+t+2exp(-t)+3t exp(-t)

It is clearly wrong, as in this case x'(0)=2, not x'(0)=1.
 
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I got -2u(t)+t+2exp(-t)+3t exp(-t), with u(t) being the Heaviside function. I got this using a Laplace transform and then doing partial fractions. I think they just left off the u(t) for their solution on accident.
 
What do mean by saying that x'(0)=1、 Do we mean x'(0+)=1 or x'(0-)=1?
 
EDIT: Whoa, I was way off for my reasoning for x'(0)! Sorry about that. I'd suspect it would be x'(0-) since that is what you use when you take the transform.
 
Last edited:
But there is a delta(t) in the input so x'(0-) and x'(0+) cannot be the same. x'(0+)=x'(0-)+1
 
Why not take Laplace transforms?
 

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