Using laplace transforms to solve IVPplease check work thanks

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Homework Help Overview

The discussion revolves around solving an initial value problem (IVP) using Laplace transforms, specifically the equation y' - 3y = 13cos(2t) with the initial condition y(0) = 1. Participants are examining their work against a provided book solution.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Laplace transforms and expresses confusion regarding a decomposition step that may be affecting their results. They reference an external document containing their work and compare it to a book solution.

Discussion Status

Some participants have identified algebraic mistakes in the original poster's work, particularly related to the use of parentheses. The original poster acknowledges these errors and notes that their revised values now match the book solution. A follow-up question is raised about the implications of using a partially decomposed denominator in the Laplace transform process.

Contextual Notes

There is an ongoing discussion about the correctness of the decomposition step in the Laplace transform process and its impact on the final solution. The original poster is also exploring the relationship between different methods of solving the same equation.

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Homework Statement


y' - 3y = 13cos(2t)

y(0)=1

Homework Equations


y' = sY(s) - y(0)

The Attempt at a Solution



heres all my work.. i am confused as to why its not matching book solution.. i think (geussing) that I probably messing up the decomposition step..thanks for any help with this
https://docs.google.com/open?id=0BwJqUg33PgREVjdEN250WXVTaldod3NublhVc1pEdw

Homework Statement


Homework Equations


The Attempt at a Solution


sorry almost forgot to include this..book solution is:

y(t) = 4e^3t - 3cos(2t) + 2sin(2t)
 
Last edited:
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You made a couple of algebra mistakes, one of them due to an omission of parentheses where they were needed.
 
thanks for the help..yup i see it: (As+B)(s-3) makes all the difference, giving me C=3, B=4 and A = -3...now matches book solution.. but have follow-up question..

If I start solving equation with what I think is a fully decomposed denominator, when denominator could actually be decomposed further, then the answer I get should be identical to the one solved with fully decomposed denomiator, right? so, even though i probably had to solve with more steps because deenominator wasnt fully decomposed, the answers shouls still be the same, true?
 
Yes, you should ultimately get an equivalent answer. The two results may not be expressed in exactly the same way, but they will be equal to each other.
 

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