Answer to Question #145342 in Differential Equations for Luisa chastel Messner

Question

ind the general solution of the given system.
dx /dt = 3x − y − z
dy /dt = x + y − z
dz /dt = x − y + z x(t), y(t), z(t) =

Answer 1

$$Solution$$

In matrix from, the system is

$$\frac{d}{dt}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}3 &-1 & -1 \\ 1 & 1 & -1 \\ 1 &-1 & 1 \end{bmatrix} \begin{bmatrix}x \\ y \\ z \end{bmatrix}$$

The coefficient matrix has eigenvalues $\lambda$ such that

$\begin{vmatrix} 3-\lambda & -1 & -1 \\ 1 & 1-\lambda & -1 \\ 1 & -1 & 1-\lambda \\ \end{vmatrix} =-(\lambda-2)^2(\lambda-1)=0 \implies 2_{(2)}, \lambda=1$

(where the subscript denotes multiplicity of the eigenvalue).

$\lambda =2$ :

$\begin{bmatrix}1 &-1 & -1 \\ 1 & -1 & -1 \\ 1 &-1 & -1 \end{bmatrix} \vec n_1=\vec 0\\ \implies n_{1,1}-n_{1,2}-n_{1,3}=0 \implies n_{1,1}=n_{1,2}+n_{1,3}$

By picking $n_{1,1}=1$ , we can then set $n_{1,2}=1$ and $n_{1,3}=0$ , and vice versa, to find two corresponding eigenvectors,

$\vec n_1=\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}, \vec n_2=\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}$

$\lambda=1$ :

$\begin{bmatrix}2 &-1 & -1 \\ 1 & 0 & -1 \\ 1 &-1 & 0 \end{bmatrix} \vec n_3=\vec 0\\ \implies 2n_{3,1}-n_{3,2}-n_{3,3}=0 \implies 2n_{3,1}=n_{3,2}+n_{3,3}$

We obtain $\vec n_3$ independent of $\vec n_1, \vec n_2$ by picking $\vec n_{3,2}=\vec n_{3,3}=1$, so that the third corresponding eigenvector is;

$\vec n_3=\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}$

Then the general solution to this system is

$\begin{bmatrix}x \\ y \\ z \end{bmatrix}=c_1e^{2t} \vec n_1+ c_2te^{2t} \vec n_2+ c_3e^{t} \vec n_3\\ \begin{cases} x=c_1e^{2t} + c_2te^{2t} + c_3e^{t}\\ y=c_1e^{2t} + c_3e^{t}\\ z=c_2te^{2t} + c_3e^{t}\\ \end{cases}$