Complex Eigenvalues

  • Thread starter Totalderiv
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  • #1
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Homework Statement


Apply the eigenvalue method to find a general solution of the given system.
[tex]x_1' = 5x_1 - 9x_2[/tex]
[tex]x_2' = 2x_1 - x_2[/tex]


Homework Equations


(A-λI)v=0


The Attempt at a Solution


[tex]x_1' = 5x_1 - 9x_2[/tex]
[tex]x_2' = 2x_1 - x_2[/tex]

[tex]\left[
\begin{array}{cc}
5-λ & -9\\
2 & -1-λ
\end{array}
\right]=(5-λ)(-1-λ)+18=0[/tex]
[tex]λ^2-4λ-13=0[/tex]
[tex](λ-2)^2 -9=0[/tex]
[tex]λ=2+3i,\overline{λ}=2-3i[/tex]
So I plugged λ into the matrix;
[tex]\left[
\begin{array}{cc}
3-3i & -9\\
2 & -3-3i
\end{array}
\right]
\left[
\begin{array}{cc}
a\\
b
\end{array}
\right]=(3-3i)a-9b=0
2a-(3+3i)b=0[/tex]
This is where I'm stuck...the answer is;
[tex]x_1(t)=3e^{2t}(c_1cos2t - 5c_2sin2t)[/tex]
[tex]x_2(t)=e^{2t}[(c_1+c_2)cos3t + (c_1-c_2)sin3t)][/tex]
 

Answers and Replies

  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
967

Homework Statement


Apply the eigenvalue method to find a general solution of the given system.
[tex]x_1' = 5x_1 - 9x_2[/tex]
[tex]x_2' = 2x_1 - x_2[/tex]


Homework Equations


(A-λI)v=0


The Attempt at a Solution


[tex]x_1' = 5x_1 - 9x_2[/tex]
[tex]x_2' = 2x_1 - x_2[/tex]

[tex]\left[
\begin{array}{cc}
5-λ & -9\\
2 & -1-λ
\end{array}
\right]=(5-λ)(-1-λ)+18=0[/tex]
[tex]λ^2-4λ-13=0[/tex]
[tex](λ-2)^2 -9=0[/tex]
[tex]λ=2+3i,\overline{λ}=2-3i[/tex]
This is your error. [itex](\lambda- 2)^2- 9= 0[/itex] is equivalent to [itex](\lambda- 2)^2= 9[/itex] so [itex]\lambda- 2= \pm 3[/itex]. There is no "i".

So I plugged λ into the matrix;
[tex]\left[
\begin{array}{cc}
3-3i & -9\\
2 & -3-3i
\end{array}
\right]
\left[
\begin{array}{cc}
a\\
b
\end{array}
\right]=(3-3i)a-9b=0
2a-(3+3i)b=0[/tex]
This is where I'm stuck...the answer is;
[tex]x_1(t)=3e^{2t}(c_1cos2t - 5c_2sin2t)[/tex]
[tex]x_2(t)=e^{2t}[(c_1+c_2)cos3t + (c_1-c_2)sin3t)][/tex]
There are no eigenvectors because 2+ 3i and 2- 3i are not eigenvalues.
 

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