Question

Q2. Consider the system described by the following differential equation x(t) Ax(t) where

Answer 1

Given the differential equation is

$\dot{x}(t)=Ax(t)$ ........ Eq.1

$A=\begin{bmatrix} 0 &1 \\ -8& -6 \end{bmatrix}$

$x(0)=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Applying Laplace transform on Eq.1

$sX(s)-X(0)=AX(s)$

$\Rightarrow sIX(s)-AX(s)=X(0)$

where I is a 2x2 identity matrix  $I=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$\Rightarrow (sI-A)X(s)=X(0)$

$\Rightarrow X(s)=(sI-A)^{-1}X(0)$ ........ Eq.2

$sI-A=s\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} 0 & 1\\ -8& -6 \end{bmatrix}=\begin{bmatrix} s &-1 \\ 8& s+6 \end{bmatrix}$

$(sI-A)^{-1}=\frac{1}{s(s+6)+8}\begin{bmatrix} s+6 &1 \\ -8& s \end{bmatrix}$

$(sI-A)^{-1}=\frac{1}{s^2+6s+8}\begin{bmatrix} s+6 &1 \\ -8& s \end{bmatrix}$

$(sI-A)^{-1}=\frac{1}{(s+2)(s+4)}\begin{bmatrix} s+6 &1 \\ -8& s \end{bmatrix}$...... Eq.3

From Eq.2 and Eq.3

$X(s)=\frac{1}{(s+2)(s+4)}\begin{bmatrix} s+6 &1 \\ -8& s \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix}$

$\Rightarrow X(s)=\frac{1}{(s+2)(s+4)}\begin{bmatrix} s+7 \\ s-8 \end{bmatrix}$

$\Rightarrow X(s)=\begin{bmatrix} \frac{(s+7)}{(s+2)(s+4)} \\ \frac{(s-8)}{(s+2)(s+4)} \end{bmatrix}$

$\Rightarrow X(s)=\begin{bmatrix} \frac{5}{2(s+2)}-\frac{3}{2(s+4)} \\ -\frac{5}{(s+2)}+\frac{6}{(s+4)} \end{bmatrix}$

Applying Inverse Laplace transform on the above equation,

$\Rightarrow x(t)=\begin{bmatrix} \frac{5}{2}e^{-2t}-\frac{3}{2}e^{-4t} \\ --5e^{-2t}+6e^{-4t} \end{bmatrix}$

$\Rightarrow x(t)=\begin{bmatrix} 2.5e^{-2t}-1.5e^{-4t} \\ --5e^{-2t}+6e^{-4t} \end{bmatrix}$

Solving $\dot{x}(t)=Ax(t)$

we get

$x(t)=\begin{bmatrix} 2.5e^{-2t}-1.5e^{-4t} \\ --5e^{-2t}+6e^{-4t} \end{bmatrix}$