Solving ODE Problems with e^{2s} + 2e^s +2 = 0

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SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) represented by the expression e^{2s} + 2e^s + 2 = 0. Participants suggest substituting e^s with x to simplify the problem into a polynomial form. Additionally, the conversation addresses finding the complex logarithm for given complex numbers, specifically determining λ + wi for z = -2i, where |z| = 2 and arg(z) = -1. The solution involves calculating λ and w using the logarithmic properties of complex numbers.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with complex numbers and their properties
  • Knowledge of logarithmic functions in the context of complex analysis
  • Experience with polynomial equations and substitutions
NEXT STEPS
  • Study the substitution method for solving ODEs, particularly exponential forms
  • Learn about the properties of complex logarithms, including principal values
  • Explore the geometric interpretation of complex numbers in the Argand plane
  • Investigate polynomial root-finding techniques for exponential equations
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations and complex analysis, will benefit from this discussion. It is also relevant for anyone looking to deepen their understanding of solving ODEs and working with complex logarithms.

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ODE-related problems

How do I find all the solutions of
[tex]e^{2s} + 2e^s +2 = 0[/tex]
?
 
Last edited:
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That looks like a regular polynomial to me. Replace [itex]e^{s}~with~x[/itex] and see if you can pick it up.
 
okay thanks, i got that.

This is the last part of a problem for which i completed all the other parts. The question asks to find [itex]\lambda + wi[/itex] such that a given [itex]z = e^{\lambda + wi}[/itex].
For example, given [itex]z = 1-i[/itex], |z| = [itex]\sqrt{2}[/itex], arg(z) = -pi/4, so [itex]\lambda + wi = \ln{\sqrt{2}} - \frac{\pi}{4}[/itex]

The last part of the problem asks for the same information, except with z = -2i. I have found |z| = 2, arg(z) = -1, but I can't find out what [itex]\lambda + wi[/itex] is supposed to be.
 

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