Solving Complex Eigenvalues Homework

[Totalderiv]

Homework Statement

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

Homework Equations

(A-λI)v=0

The Attempt at a Solution

$$x_1' = 5x_1 - 9x_2$$
$$x_2' = 2x_1 - x_2$$

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

 

[Totalderiv]

Any one?

 

[HallsofIvy]

Homework Statement

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

Homework Equations

(A-λI)v=0

The Attempt at a Solution

$$x_1' = 5x_1 - 9x_2$$
$$x_2' = 2x_1 - x_2$$

$$\left[ <br /> \begin{array}{cc} <br /> 5-λ & -9\\ <br /> 2 & -1-λ <br /> \end{array} <br /> \right]=(5-λ)(-1-λ)+18=0$$
$$λ^2-4λ-13=0$$
$$(λ-2)^2 -9=0$$
$$λ=2+3i,\overline{λ}=2-3i$$

This is your error. $(\lambda- 2)^2- 9= 0$ is equivalent to $(\lambda- 2)^2= 9$ so $\lambda- 2= \pm 3$. There is no "i".

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

There are no eigenvectors because 2+ 3i and 2- 3i are not eigenvalues.