Answer to Question #265393 in Differential Equations for mossa

Question

dx/dt=-4x+y+z

dy/dt=x+5y-z

dz/dt=y-3z

Answer 1

he given system can be expressed as

$X'=AX$ , where $X' = \begin{bmatrix} \frac{dx}{dt} \\[0.3em] \frac{dy}{dt} \\[0.3em] \frac{dz}{dt} \end{bmatrix},A = \begin{bmatrix} -1 & 4 & 2 \\[0.3em] 4 & -1 & 2 \\[0.3em] 0 & 0 & 6 \end{bmatrix},X = \begin{bmatrix} x \\[0.3em] y \\[0.3em] z \end{bmatrix}$

Find the eigenvalues and eigenvectors of the coefficient matrix $A$.

Characteristic equation of the matrix $A$ is given by

$det(A-\lambda I)=0$

$\Rightarrow \left | \begin{matrix} \lambda +1 & 4 & 2 \\ 4 & \lambda +1 & 2 \\ 0 & 0 & 6-\lambda \end{matrix} \ \right |=0$

$\Rightarrow -\lambda ^3+4\lambda ^2+27\lambda -90=0$

$\Rightarrow (\lambda +5)(\lambda -3)(\lambda -6)=0$

$\Rightarrow \lambda =-5,3,6$

So, the eigenvalues of the matrix $A$ are: $\Rightarrow \lambda =-5,3,6$

Now let us find the eigenvectors corresponding to the eigenvalues.

Eigenvector corresponding to the eigenvalue $\lambda =-5$ :

Let $v_{1}$ be the eigenvector corresponding to the eigenvalue $\lambda=-5$

Then, $(A+5*I)v=0$

$\Rightarrow \begin{bmatrix} 4 & 4 & 2 \\[0.3em] 4 & 4 & 2 \\[0.3em] 0 & 0 & 11 \end{bmatrix}v=0$

Solve the homogeneous system by Gaussian elimination.

$\Rightarrow \begin{bmatrix} 2 & 2 & 1 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 1 \end{bmatrix}v=0$

$\Rightarrow x_{1}+x_{2}=0,x_{3}=0$

$\Rightarrow v_{1}=\begin{bmatrix} -1 \\[0.3em] 1 \\[0.3em] 0 \end{bmatrix}$

Similarly, the eigenvectors corresponding to the eigenvalues $\lambda=3,6$ are

$v_{2}=\begin{bmatrix} 1 \\[0.3em] 1 \\[0.3em] 0 \end{bmatrix},v_{3}=\begin{bmatrix} 2 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}$

Therefore, the general solution of the given system is

$X(t)= c_{1}e^{-5t}\begin{bmatrix} -1 \\[0.3em] 1 \\[0.3em] 0 \end{bmatrix}+c_{2}e^{3t}\begin{bmatrix} 1 \\[0.3em] 1 \\[0.3em] 0 \end{bmatrix}+c_{3}e^{6t}\begin{bmatrix} 2 \\[0.3em] 2 \\[0.3em] 3 \end{bmatrix}$

That is, $\begin{bmatrix} x(t) \\[0.3em] y(t) \\[0.3em] z(t) \end{bmatrix}= \begin{bmatrix} - c_{1}e^{-5t} +c_{2} e^{3t}+2c_{3}e^{6t} \\[0.3em] c_{1}e^{-5t} +c_{2} e^{3t}+2c_{3}e^{6t} \\[0.3em] 3c_{3}e^{6t} \end{bmatrix}$