Differential problem Confused :S

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Discussion Overview

The discussion revolves around solving a differential initial value problem using different methods, specifically the method of undetermined coefficients and the Laplace transform. Participants explore the implications of potentially obtaining different answers from these methods for the given equation.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether it is possible to arrive at different answers when solving the same initial value problem using different methods.
  • Another participant provides the characteristic equation and suggests that the particular integral for the exponential term should be of the form xe-2t.
  • A different participant suggests that the Laplace transform method may be excessive for this problem.
  • One participant proposes a specific form for the particular solution that includes a constant term to address the non-homogeneous part of the equation.

Areas of Agreement / Disagreement

Participants express differing opinions on the appropriateness of the Laplace transform method and the correct form of the particular solution, indicating that multiple competing views remain without a clear consensus.

Contextual Notes

Participants have not fully resolved the implications of the constant term in the non-homogeneous part of the equation, nor have they reached a definitive agreement on the methods to be used.

malee006
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Hey ! i am having a problem in differential initial value problem, can it happen that solving an initial value problem with two different methods (e.g. method of undetermined coefficient and using Laplace) give us two different answer!

the equation is : y"-4'=(5e^-2t)+1 y(0)=0 y'(0)=10


and if possible can u do it by method of undetermined coefficient bcoz i think i am doing a mistake in tht method becoz of that '1' on right hand side and no comparing coefficient on the other side!
 
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[tex]\frac{d^2y}{dt^2}-4=5e^{-2t}+1[/tex]

Aux.Eq'n=[tex]r^2-4=0 \Rightarrow (r+2)(r-2)=0 \Rightarrow r= \pm2[/tex]

[tex]y_{CF}=Ae^{2t}+Be^{-2t}[/tex]

I think you got this far right?

Well since -2 is one root of the aux. eq'n. The PI for the exponential should be xe-2t
 
Yeah, I agree. Try laplace transform method and for the second method find the roots of the characteristic equation.
 
I personally think the "Laplace Transform method" is overkill.

Malee006, I presume the equation is y"- 4y'= 5e2t+ 1. Then, as Rockfreak667 said, the general solution to the homogeneous equation is Ae2t[/sub]+ Be-2t[/itex].

Since e2t is already a solution, you need to look for a specific solution to the entire equation of the form Cte2t+ B. The "B" is to get the "1" on the right hand side of the original equation.
 

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