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

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SUMMARY

The discussion focuses on constructing a differential equation of the form y'' + by' + cy = 0, with y(t) = e^t cos(3t) as a solution. The user derived the first and second derivatives, resulting in the coefficients b = -2 and c = -6. The final differential equation is y'' - 2y' - 6y = 0. The discussion emphasizes verifying the solution by substituting it back into the equation and suggests using WolframAlpha for confirmation.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear ODEs.
  • Knowledge of derivatives and how to compute y'(t) and y''(t).
  • Familiarity with complex exponentials and their relationship to trigonometric functions.
  • Ability to use computational tools like WolframAlpha for verification of mathematical results.
NEXT STEPS
  • Study the method of characteristic equations for solving linear differential equations.
  • Learn about the relationship between differential equations and their characteristic polynomials.
  • Explore the use of complex numbers in solving differential equations involving trigonometric functions.
  • Investigate additional examples of constructing differential equations from given solutions.
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Students and professionals in mathematics, engineering, and physics who are working with differential equations, particularly those involving exponential and trigonometric functions.

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
<br /> y&#039;&#039; + by&#039; + cy = 0<br />
and the quadratic equation
<br /> \lambda^2 + b\lambda + c = 0?<br />

You may also find the identity
<br /> \cos kt \equiv \frac{e^{ikt} + e^{-ikt}}{2}<br />
helpful.
 

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