Systems of ODE: Converting complex solution to real

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SUMMARY

The discussion focuses on converting a complex solution of a system of linear first-order ordinary differential equations (ODEs) into its real equivalent. The general solution provided is X = c1v1e^(-1+2i)t + c2v2e^(-1-2i)t, where v1 = [-1+2i, 5] and v2 = [-1-2i, 5]. To achieve a real solution, the user seeks a step-by-step explanation of how to separate the real and imaginary components without losing the solution's integrity. The key takeaway is that resolving the solution into real and imaginary parts is essential for maintaining the solution space.

PREREQUISITES
  • Understanding of linear first-order ordinary differential equations (ODEs)
  • Familiarity with complex numbers and their properties
  • Knowledge of exponential functions involving complex exponents
  • Ability to manipulate and separate real and imaginary parts of complex expressions
NEXT STEPS
  • Study the method of converting complex solutions to real solutions in ODEs
  • Learn about Euler's formula and its application in solving differential equations
  • Explore the concept of linear combinations of solutions in the context of ODEs
  • Investigate the implications of complex eigenvalues in systems of differential equations
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations, as well as educators seeking to clarify the conversion of complex solutions to real solutions in ODEs.

Shaybay92
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Homework Statement



So, I have found a general solution to a system of linear first order ODE's and this is what I got:

X = c1v1e^(-1+2i)t + c2v2e^(-1-2i)t

where v1 = [-1+2i, 5], v2=[-1-2i,5]. The question is, how do I now change this solution into its real equivalent? i.e. I don't want any complex numbers in my solution.

I have a textbook which explains but it still doesn't make sense to me how they manage to go from a complex to real solution. Could someone explain step by step? Thanks.
 
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I think they're saying that X=x+yi, just resolve in real and imaginary parts...
 
I don't think so, because it has to still span the solution space, and merely dropping the imaginary parts will not ensure this.
 

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