# Toronto Math Forum

## MAT244-2014F => MAT244 Math--Tests => MT => Topic started by: Victor Ivrii on October 29, 2014, 09:01:30 PM

Title: MT problem 5
Post by: Victor Ivrii on October 29, 2014, 09:01:30 PM
Find the general solution of the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t=x\ +\ y + 2z\ , \\
&y'_t=x\ + 2y +\ z\ , \\
&z'_t=2x +\ y +\ z\ .
\end{aligned}\right.
\end{equation*}
Title: Re: MT problem 5
Post by: Roro Sihui Yap on October 29, 2014, 11:07:37 PM
First, find the eigenvalues of the matrix A.
\begin{equation*} A = \begin{bmatrix}
1 & 1 & 2
\\1 & 2 & 1
\\ 2 & 1 & 1\end{bmatrix} \end{equation*}

\begin{equation*}
\left| \begin{matrix} 1-\lambda & 1 & 2 \\1 &  2-\lambda & 1 \\ 2& 1 & 1-\lambda\end{matrix}\right| = 0  \notag
\end{equation*}

\begin{gather}
(1 - \lambda)( (2-\lambda)(1-\lambda) - 1) - (1 - \lambda - 2 ) + 2 (1 - 4 + 2\lambda) = 0,\label{eq-1} \\
(1 - \lambda)(\lambda^2 - 3\lambda + 1) + (1 + \lambda) +  (-6 + 4\lambda) = 0 \label{eq-2} \\
-\lambda^3 + 4\lambda^2 + \lambda - 4 = 0 \label{eq-3} \\
-(\lambda - 4)(\lambda - 1)(\lambda + 1) = 0 \label{eq-4} \\
\end{gather}
The eigenvalues are $\lambda = 1, \lambda = -1,\ \lambda = 4$.

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

When $\lambda = -1$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
2 & 1 & 2
\\1 & 3 & 1
\\ 2 & 1 & 2\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & 1
\\0 & 1 & 0
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(1, 0, -1)^T$ and corresponding solution is  $x^{(2)} = c_2e^{-t}(1, 0, -1)^T$.

When $\lambda = 4$,
\begin{equation*}
A-\lambda I= \begin{bmatrix}
-3 & 1 & 2
\\1 & -2 & 1
\\ 2 & 1 & -3\end{bmatrix} \cong
\begin{bmatrix}
1 & 0 & -1
\\0 & 1 & -1
\\ 0 & 0 & 0\end{bmatrix}
\end{equation*}
The eigenvector is $(1, 1, 1)^T$ and corresponding solution is  $x^{(3)} = c_3e^{4t}(1, 1, 1)^T$.

The general solution is $(x, y, z)^T = c_1e^t(1, -2, 1)^T + c_2e^{-t}(1, 0, -1)^T + c_3e^{4t}(1, 1, 1)^T$
Title: Re: MT problem 5
Post by: Victor Ivrii on October 30, 2014, 06:50:56 AM
Roro, give chance to others :D
Title: Re: MT problem 5
Post by: Roro Sihui Yap on October 30, 2014, 09:50:59 AM
I will in the future :)