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Solving a system of diffy q's with complex eigenvalues

  • Thread starter Jamin2112
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Homework Statement



Express the general solutoins of the system of equations in terms of real-valued functions.

x'= [1 0 0; 2 1 -2; 3 2 1]x (I wrote the matrix MATLAB-style)

Homework Equations



The coolest equation ever: eib=cosb + isinb

The Attempt at a Solution



Assume x=Rert (the underlined r is an eigenvector)

Determinate[1-r 0 0; 2 1-r -2; 3 2 1-r]=0 --> r = 1, 1+2i, 1-2i

r = 1 --> [0 0 0; 2 1 -2; 3 2 0](r1 r2 r3)T=(0 0 0)T

---> R1= (2 -3 2)T

I do the same with the other eigenvalues, and come up with 3 eigenvectors: R1= (2 -3 2)T, R2= (0 1 -i)T, R3= (0 1 i)T.

By Superpsition, the full solution will be

x(t)=C1et(2 -3 2)T + C2ete2it(0 1 -i)T + C3ete-2it(0 1 i)T


= et [ C1(2 -3 2)T + C2(cos(2t)+isin(2t))(0 1 -i)T + C3(cos(-2t)+isin(-2t))(0 1 i)T ]

................................. This somehow simplifies to the answer in the back of the book, C1et(2 -3 2)T + C2et(0 cos2t sin2t)T + C3et(0 sin2t -cos2t)T. I don't understand the simplification process. Yes, I know the imaginary numbers just get absorbed into the constants; but I can't figure out the rest.
 

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