### Author Topic: Q3 problem 2 (night sections)  (Read 2363 times)

#### Victor Ivrii

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##### Q3 problem 2 (night sections)
« on: October 23, 2014, 12:25:58 AM »
7.6 p 417,# 7
Express the general solution of the system of ODEs  $\ x'= Ax\$ in terms of real-valued functions, where
\begin{equation*}
A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{pmatrix}
\end{equation*}

#### Allan Lam

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##### Re: Q3 problem 2 (night sections)
« Reply #1 on: October 25, 2014, 12:17:29 AM »
It doesn't match the solution in the textbook but the two eigenvectors (edit: from the complex roots) are linearly independent so I think it should still be correct.

#### Victor Ivrii

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##### Re: Q3 problem 2 (night sections)
« Reply #2 on: October 25, 2014, 05:19:11 PM »
Portions are unreadable. This solution is of no use.

#### Roro Sihui Yap

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##### Re: Q3 problem 2 (night sections)
« Reply #3 on: October 25, 2014, 08:35:32 PM »
First, find the eigenvalues of the matrix A.

\begin{equation*}
\left| \begin{matrix} 1-\lambda & 0 & 0 \\2 &  1-\lambda & -2 \\ 3& 2 & 1-\lambda\end{matrix}\right| = 0  \notag
\end{equation*}
\begin{gather}
(1 - \lambda)( (1-\lambda)(1-\lambda) + 4) = 0,\label{eq-1} \\
(1 - \lambda)( \lambda^2 - 2\lambda + 5) = 0. \label{eq-2}
\end{gather}
The eigenvalues are $\lambda = 1, \lambda = 1 + 2i,\ \lambda = 1- 2i$.

When $\lambda = 1$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
0 & 0 & 0
\\2 & 0 & -2
\\ 3 & 2 & 0\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & -1
\\0 & 2 & 3
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(2, -3, 2)^T$ and corresponding solution is  $x^{(1)} = c_1e^t(2, -3, 2)^T$.

When $\lambda = 1 + 2i$,
\begin{equation*}
A-\lambda I=\begin{bmatrix}
-2i & 0 & 0
\\2 & -2i & -2
\\ 3 & 2 & -2i\end{bmatrix}\cong
\begin{bmatrix}
1 & 0 & 0
\\0 & i & 1
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(0, i, 1) ^T$ and corresponding solution is
\begin{equation*}
(0, i, 1)^T e^{(1+2i)t} = (0, i, 1)^T (e^t)( e^{2it})  = (0, i, 1)^T (e^t)(\cos  2t + i\sin 2t) = (0, i\cos 2t - \sin  2t, \cos 2t + i\sin 2t)^T (e^t) $=(e^t)[ (0, -\sin 2t, \cos 2t)^T +i(0, \cos 2t, \sin 2t)^T] \end{equation*} The general solution is$x = e^t\left[ c_1 (2, -3, 2)^T + c_2 (0, -\sin  2t, \cos  2t)^T +c_3 (0, \cos  2t, \sin 2t)^T\right]\$
« Last Edit: October 25, 2014, 10:26:39 PM by Victor Ivrii »