Second order differential equation

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SUMMARY

The discussion centers on demonstrating that the functions y1(x) = e^(2+i)x and y2(x) = e^(2-i)x are linearly independent solutions to a second-order linear differential equation with constant coefficients. The Wronskian is calculated to confirm their independence. The participants seek guidance on deriving the auxiliary equation from the roots (r + 2 + i)(r + 2 - i) and subsequently formulating the corresponding differential equation. The goal is to verify that both functions satisfy the derived equation.

PREREQUISITES
  • Understanding of linear differential equations
  • Familiarity with the Wronskian determinant
  • Knowledge of complex numbers and their properties
  • Ability to derive auxiliary equations from characteristic roots
NEXT STEPS
  • Study the derivation of second-order linear differential equations from characteristic equations
  • Learn how to compute the Wronskian for verifying linear independence
  • Explore the application of complex roots in differential equations
  • Investigate methods for verifying solutions of differential equations
USEFUL FOR

Mathematics students, educators, and professionals dealing with differential equations, particularly those focusing on linear systems and complex analysis.

bobey
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show that y1(x) = e^(2+i)x and y2(x) = e^(2-i)x, i=sqrt(-1) are two linearly independent functions

hence obtain a second order linear differential equation with constant coefficients each that y1(x) and y2(x) are its two fundamental solutions.

my attempt :

for the first part, I use the definition of wroskian = y1y2'-y2y1' and show it not equal to zero... ok

the second part, I don't know how to do it...

how to get the second order differential equation?

is that setting : (r+2+i)(r+2-i) to get the auxillary equation? is that possible? can someone show me to solve this problem?:confused:
 
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If you want that auxiliary equation, what will be the differential equation? When you get it, you can check whether the two original functions satisfy it.
 
how to do that... still blurr...
 

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