Homogenous constant coefficient linear differential equations

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SUMMARY

The discussion centers on solving the homogeneous constant coefficient linear differential equation Y'' + 6Y' + 9Y = 0. The characteristic equation X^2 + 6X + 9 = 0 yields a repeated root of X = -3. The correct general solution for this equation is Y = C1 e^(-3x) + C2 x e^(-3x), which incorporates both the exponential and polynomial terms due to the repeated root.

PREREQUISITES
  • Understanding of homogeneous linear differential equations
  • Familiarity with characteristic equations and their solutions
  • Knowledge of exponential functions and their derivatives
  • Concept of repeated roots in differential equations
NEXT STEPS
  • Study the method of solving second-order linear differential equations
  • Learn about the implications of repeated roots in characteristic equations
  • Explore the general solution forms for different types of roots
  • Practice solving additional examples of homogeneous differential equations
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are working with differential equations and require a solid understanding of their solutions.

kmikias
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Hello
I am kinda confused when it comes to finding a general solution of equation.
Here is the question.

Y'' + 6Y' +9Y = 0
Solution
Y'' + 6Y' +9Y = 0
I used "x" instead of Lamda...ANYWAY

X^2 + 6X + 9 = 0
FACTOR IT WHICH IS (X+3) (X+3) = 0

X IS EQUAL TO -3.

HERE IS WHERE I GET CONFUSED

WHEN I WRITE THE GENERAL SOLUTION ...DO I SUPPOSE TO WRITE
Y= C1 e-3x + C2 e-3x

or

Y = C1 e-3x

WHICH ONE IS CORRECT?
 
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kmikias said:
Hello
I am kinda confused when it comes to finding a general solution of equation.
Here is the question.

Y'' + 6Y' +9Y = 0
Solution
Y'' + 6Y' +9Y = 0
I used "x" instead of Lamda...ANYWAY

X^2 + 6X + 9 = 0
FACTOR IT WHICH IS (X+3) (X+3) = 0

X IS EQUAL TO -3.

HERE IS WHERE I GET CONFUSED

WHEN I WRITE THE GENERAL SOLUTION ...DO I SUPPOSE TO WRITE
Y= C1 e-3x + C2 e-3x

or

Y = C1 e-3x

WHICH ONE IS CORRECT?

Neither is completely correct. A repeated root of r gives solution pair {erx,xerx}.
 

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