Solving Transcendental Equations (and Laplace Transforms)

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SUMMARY

This discussion focuses on solving the differential equation H'(t) + u H(t - T) = 0, where u > 0, using solutions of the form e^{rt}. It establishes that these solutions are exponentially damped if e^{-1} > uT > 0 and identifies conditions for oscillatory behavior when r is complex. The use of Laplace Transforms is suggested to simplify the problem, leading to the expression Y(s) = H(0)/(s + e^{-sT}). The participants express uncertainty about the application of Laplace Transforms in this context.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with complex numbers and their representation in exponential form.
  • Knowledge of Laplace Transforms and their properties.
  • Ability to analyze oscillatory behavior in mathematical functions.
NEXT STEPS
  • Study the application of Laplace Transforms in solving linear differential equations.
  • Explore the concept of stability in solutions of differential equations.
  • Investigate the relationship between damping and oscillation in complex solutions.
  • Learn about the interpretation of the Laplace Transform results in the context of system behavior.
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Students and professionals in mathematics, engineering, and physics who are dealing with differential equations and seeking to understand the application of Laplace Transforms in solving such equations.

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Homework Statement


Given the equation H'(t) + u H(t - T) = 0 u > 0
Look for solutions of the form e^{rt}

Show that these solutions are exponentially damped if e^{-1} > uT > 0
Find uT for which these solutions for r complex are oscillatory with growing, decaying, or constant amplitude.

The book also hints that a Laplace Transform would be helpful (albeit not necessary), but I'm not sure how to do these.

Homework Equations


r = - u e^{-rT}
Let r = x + yi
x = -u e^{-xT} cos(yT)
y = u e^{-xT} sin(yT)

The Attempt at a Solution


I find found the real solutions.
Set y = 0, then cos(yT) = 1, so
x = -u e^{-xT}

define F(x) = x + u e^{-xT}
F(0) = u >0
F(-1/T) = \frac{-1 + uTe}{T} < \frac{-1 + 1}{T} = 0
when e^{-1} > uT > 0

For oscillatory, I just set x = 0, so cos(yT) = 0, which means sin(yT) = +-1
So y = +-u, which means cos(uT) = 0 sin(uT) = +-1, which we know happens when uT = pi/2, 3pi/2, etc...

I'm not sure how to do the rest.

It says a Laplace Transform would make it faster, well, I took the transform from the DE and got
Y(s) = \frac{H(0)}{s + e^{-sT}}

How do I interpret this?
 
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Although I don't like bumping threads, I want to make sure everyone sees this.

In particular, I'm really curious to how I would be using Laplace Transforms to solve this problem.
 

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