Construct DE w/ Soln y(t) = e^t cos(3t)

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Discussion Overview

The discussion revolves around constructing a differential equation of the form y'' + by' + cy = 0 that has the solution y(t) = e^t cos(3t). Participants are exploring the relationship between the differential equation and its characteristic polynomial, as well as verifying the proposed coefficients.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant proposes that the coefficients b and c can be determined as b = -2 and c = -6, leading to the equation y'' - 2y' - 6y = 0.
  • Another participant suggests checking the answer by substituting the solution back into the differential equation and showing the working step-by-step.
  • There is a hint provided regarding the relationship between the differential equation and the corresponding quadratic equation, as well as an identity involving cosine that may assist in the analysis.

Areas of Agreement / Disagreement

The discussion does not indicate a consensus on the correctness of the proposed coefficients or the differential equation, as participants have not yet verified the solution through substitution.

Contextual Notes

Participants have not fully explored the implications of the characteristic polynomial or the steps involved in deriving the coefficients, leaving some assumptions and mathematical steps unresolved.

LocalStudent
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Hi

I'm working through some example question and the memo leaves out this question. I was hoping someone can help me out.

The question:
Construct a differential equation of the form y'' + by' + cy = 0 which has y(t) = e^t cos(3t) as one of its solutions.

What I did:
First I found the y'(t) and y''(t)
Plugged that into the DE.

My answer:
b = -2
c = -6

y'' - 2y' - 6y = 0
 
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Check your answer by plugging the solution into it from scratch.
It will also help to show your working step-by-step.
 
LocalStudent said:
Hi

I'm working through some example question and the memo leaves out this question. I was hoping someone can help me out.

The question:
Construct a differential equation of the form y'' + by' + cy = 0 which has y(t) = e^t cos(3t) as one of its solutions.

What I did:
First I found the y'(t) and y''(t)
Plugged that into the DE.

My answer:
b = -2
c = -6

y'' - 2y' - 6y = 0

Hint: What is the relationship between the ODE
[tex] y'' + by' + cy = 0[/tex]
and the quadratic equation
[tex] \lambda^2 + b\lambda + c = 0?[/tex]

You may also find the identity
[tex] \cos kt \equiv \frac{e^{ikt} + e^{-ikt}}{2}[/tex]
helpful.
 

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