Question

(1 point) Solve the system 14-36 dx dt 6-16 with the initial value -20 x(0) = -8 x(t) =

Answer 1

Solution:

The characteristic equation is

$\small \lambda ^{2}-\left ( \mathrm{sum\: of\: diagonal\: elements} \right )\lambda +\left | A \right |=0$

$\small \lambda ^{2}+2\lambda -8=0$

$\small \therefore \lambda =2,-4$

For $\small \lambda_{1} =2,\: \: \: \left [ A-\lambda _{1} I\right ]X=0$

$\small \begin{bmatrix} 12 & -36\\ 6& -18 \end{bmatrix}\begin{bmatrix} x\\ y\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$

By $\small R_{2}-\frac{1}{2}R_{1}$

$\small \begin{bmatrix} 12 & -36\\ 0& 0 \end{bmatrix}\begin{bmatrix} x\\ y\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$

$\small \therefore 12x-36y=0\Rightarrow x=3y$

$\small \therefore v_{1}=\begin{bmatrix} 3\\ 1\\ \end{bmatrix}$

For $\small \lambda_{2} =-4,\: \: \: \left [ A-\lambda _{2} I\right ]X=0$

$\small \begin{bmatrix} 18 & -36\\ 6& -12 \end{bmatrix}\begin{bmatrix} x\\ y\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$

By $\small R_{2}-\frac{1}{3}R_{1}$

$\small \begin{bmatrix} 18 & -36\\ 0& 0 \end{bmatrix}\begin{bmatrix} x\\ y\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$

$\small \therefore 18x-36y=0\Rightarrow x=2y$

$\small \therefore v_{2}=\begin{bmatrix} 2\\ 1\\ \end{bmatrix}$

The general solution is

$\small X\left ( t \right )=c_{1}v_{1}e^{\lambda _{1}t}+c_{2}v_{2}e^{\lambda _{2}t}\small =c_{1}\begin{bmatrix} 3\\ 1\\ \end{bmatrix}e^{2t}+c_{2}\begin{bmatrix} 2\\ 1\\ \end{bmatrix}e^{-4t}$

$\small \therefore X\left ( t \right )=\begin{bmatrix} 3c_{1}e^{2t}+2c_{2}e^{-4t}\\ c_{1}e^{2t}+c_{2}e^{-4t} \end{bmatrix}$

$\small \therefore X\left ( 0 \right )=\begin{bmatrix} 3c_{1}+2c_{2}}\\ c_{1}+c_{2} \end{bmatrix}$

$\small \therefore \begin{bmatrix} -20\\ -8\\ \end{bmatrix}=\begin{bmatrix} 3c_{1}+2c_{2}}\\ c_{1}+c_{2} \end{bmatrix}$

$\small 3c_{1}+2c_{2}=-20\: \: ,\: \: c_{1}+c_{2} =-8$

Solving simultaneously both the equation we get

$\small c_{1}=c_{2}=-4$

$\small \therefore X\left ( t \right )=\begin{bmatrix} -12e^{2t}-8e^{-4t}\\ -4e^{2t}-4e^{-4t} \end{bmatrix}$