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I've been working on this problem for the better part of a day and could use some help.

Express the general solution of the given system of equations in terms of real-valued functions:

$\mathbf{X}'=\left[\begin{array}{ccc}1 & 0 & 0\\ 2 & 1 & -2\\ 3 & 2 & 1\end{array}\right]\mathbf{X}$

The answer in the back of the book is as follows:

$\mathbf{X}=c_1\left[\begin{array}{c}2 \\3\\ 2\end{array}\right]e^t+c_2e^t\left[\begin{array}{c}0\\ \cos(2t)\\ \sin(2t)\end{array}\right]+c_3e^t\left[\begin{array}{c}0\\ \sin(2t)\\ -\cos(2t)\end{array}\right]$

(Note that the bracketed entries in the answer are vectors, and $c_1$, $c_2$, and $c_3$ are arbitrary constants.)

After numerous attempts, I have not managed to reach an answer that looked remotely close to this. Any help would be appreciated.

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We are given:

$$X' = A x = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \\ \end{bmatrix}x$$

We find the eigenvalues of this matrix using $|A - \lambda I| = 0$, which gives us:

$$-\lambda^3 + 3\lambda^2 -7 \lambda + 5 = -(\lambda + 1)(\lambda^2-2\lambda +5) = 0 \rightarrow \lambda_1 = 1, \lambda_2 = 1+ 2i, \lambda_3 = 1 - 2i$$

We see that we have three distinct eigenvalues, so we can find the eigenvectors using Gaussian Elimination (or many methods) for $[A-\lambda_i I]v_i = 0$.

This leads to the three eigenvectors:

$$v_1 = (2,-3,2), v_2 = (0, i, 1), v_3 = (0,-i,1)$$

We can now write out our general solution as:

$$X(t) = c_1 \lambda_1 v_1 + c_2 Re(\lambda_2 v_2) + c_3 Im(\lambda_2 v_2)$$

Note, since we have a complex conjugate eigenvalue, we can find the solution for that as $e^{\lambda t}v$ and just split the real and imaginary parts as two independent solutions (I will assume you understand this). Lets do one:

$$e^{\lambda t}v = e^{(1+2i)t} \begin{bmatrix} 0 \\ i \\ 1\end{bmatrix} = e^te^{2i}\begin{bmatrix} 0 \\ i \\ 1\end{bmatrix} = e^t(\cos 2t + i \sin 2t) \begin{bmatrix} 0 \\ i \\ 1\end{bmatrix} = e^t \begin{bmatrix} 0 \\ -\sin 2t + i \cos 2t \\ \cos 2t + i \sin 2t\end{bmatrix}$$

From this, we can write two linearly independent solutions:

$$e^t\left(c_2 \begin{bmatrix} 0 \\ \cos 2t \\ \sin 2t\end{bmatrix} + c_3 \begin{bmatrix} 0 \\ -\sin 2t \\ \cos 2t \end{bmatrix}\right)$$

Finally, we arrive at:

$$X(t) = e^t\left(c_1 \begin{bmatrix} 2 \\ -3 \\ 2\end{bmatrix} + c_2 \begin{bmatrix} 0 \\ \cos 2t \\ \sin 2t\end{bmatrix} + c_3 \begin{bmatrix} 0 \\ - \sin 2t \\ \cos 2t \end{bmatrix}\right)$$

The sign difference is okay in the third eigenvector given we can swap signs around on eigenvectors as they are not unique. Also, I think you missed a sign in the first vector (check that).

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  • $\begingroup$ Thanks for this! It made great sense. $\endgroup$ – Heath Huffman Nov 3 '13 at 22:34

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