Understanding Non-Real Eigenvalues to Solving Homework Problems

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Homework Help Overview

The discussion revolves around understanding the implications of non-real eigenvalues in the context of solving differential equations, particularly focusing on the general forms of solutions involving trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the general forms of solutions and express confusion regarding the incorporation of certain variables, particularly 'v' and constants in the equations. There is an attempt to clarify the relationship between the constants and the functions.

Discussion Status

Some participants are exploring different forms of the solutions and questioning the consistency of the constants used. There is an ongoing examination of the implications of non-real eigenvalues, with no clear consensus reached yet.

Contextual Notes

Participants note discrepancies in the expected forms of the solutions and express uncertainty about how to proceed without additional information regarding variable 'A'.

cowmoo32
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Homework Statement


et0re.jpg




The Attempt at a Solution


I know the general form should be

x1(t)=-C1sin(3t) + C2cos(3t)
x2(t)=C1sin(3t) + C2cos(3t)

but there's something going on with v that I'm not getting. I'm not sure how to incorporate it without knowing A
 
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cowmoo32 said:

Homework Statement


et0re.jpg




The Attempt at a Solution


I know the general form should be

x1(t)=-C1sin(3t) + C2cos(3t)
x2(t)=C1sin(3t) + C2cos(3t)
The constants shouldn't be the same in both functions.
cowmoo32 said:
but there's something going on with v that I'm not getting. I'm not sure how to incorporate it without knowing A
 
Whoops, I wrote that wrong. All of the problems I have worked so far have had the form:x1(t)=-C1sin(3t) + C1cos(3t)
x2(t)=C2sin(3t) + C2cos(3t)
 
After looking through some more examples, the answer will have the form


x1(t)=-C1sin(3t)*a + C1cos(3t)*b
x2(t)=C2sin(3t)*a + C2cos(3t)*b

where v=a+ib

v = [-1-i,1] = [-1,1]+i[-1,0]

a=[-1,1]
b=[-1,0]

I get:
C1sin(3t)
-C2cos(3t)

But it's telling me that's incorrect.
 

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