Initial conditions for stability

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SUMMARY

The discussion focuses on determining the initial conditions for stability in the system of differential equations defined by x' = 2x - 3y and y' = x - 2y. The general solution is given as x = 4C1e^(2t) and y = C1e^(2t) - 3C2e^(-2t). For stability, the constants C1 and C2 must be chosen such that (x, y) approaches (0, 0) as t approaches infinity. Specifically, C1 must be set to 0 to eliminate the unstable term e^(2t), while C2 can be any value to ensure y approaches stability.

PREREQUISITES
  • Understanding of differential equations and their stability analysis
  • Familiarity with exponential functions and their behavior as t approaches infinity
  • Knowledge of the concepts of initial conditions in dynamic systems
  • Basic skills in solving systems of linear equations
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  • Study the stability criteria for linear systems of differential equations
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  • Learn about phase plane analysis for systems of differential equations
  • Investigate the role of initial conditions in determining the long-term behavior of dynamical systems
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Mathematics students, engineers, and researchers involved in dynamical systems analysis, particularly those studying stability in linear differential equations.

hbomb
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The general solution fo the following equations:

x'=2x-3y
y'=x-2y

Is, x=4C1e^2t, y=C1e^2t-3C2e^-2t

They ask for me to list a set of initial conditions (xo, yo) for which the solution is stable, i.e, (x, y)-->(0,0) for large t.

I don't understand this part of the problem.
 
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Basically, for what constants C1 and C2 do x and y go to infinity as t gets large? From that, solve for what x(0) and y(0) can be
 
Ok, I understand that for the general solution, the only unstable part of it is e^-2t. What does the setup of the solution look like?
 

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