Question

Use the method of Example 2 to computee^Atfor the co-efficient matrix. Use (1) to find the general solution of the given system

15.$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}4& 3\\ -4& -4\end{array}\right)\mathit{X}$

Short answer

The general solution for the given matrix is$X=C_3\left(\begin{array}{c}3\\ -2\end{array}\right)e^{2t}+C_4\left(\begin{array}{c}-1\\ 2\end{array}\right)e^{-2t}$

Step-by-step solution

Definition of Homogeneous system

• A matrix exponential e^Atso that X 5 e^At C (1) is a solution of the homogeneous system X' = AX.
• X = e^AtC
• Here A is an n × n matrix of constants, and C is an n × 1 column matrix of arbitrary constants.

Calculate the given matrix

Given that,

$X\text{'}=\left(\begin{array}{cc}4& 3\\ -4& -4\end{array}\right)X$

where

$A=\left(\begin{array}{cc}4& 3\\ -4& -4\end{array}\right)$

We first compute the matrix sI-A,

$$\begin{gathered} sI-A=\left(\begin{array}{cc}s-4& -3\\ 4& s+4\end{array}\right) \\ det(sI-A)=0\to \left|\begin{array}{cc}s-4& -3\\ 4& s+4\end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} (s-4)(s+4)+12=0 \\ {s}^2+4s-4s-16+12=0 \\ {s}^2-4=0 \\ (s-2)(s+2)=0 \end{gathered}$$

Now can find the inverse of (1),

${(sI-A)}^{-1}={\left(\begin{array}{cc}s-4& -3\\ 4& s+4\end{array}\right)}^{-1}=\left(\begin{array}{cc}\frac{s+4}{(s-2)(s+2)}& \frac{3}{(s-2)(s+2)}\\ -\frac{4}{(s-2)(s+2)}& \frac{s-3}{(s-2)(s+2)}\end{array}\right)$${(sI-A)}^{-1}={\left(\begin{array}{cc}s-4& -3\\ 4& s+4\end{array}\right)}^{-1}=\left(\begin{array}{cc}\frac{s+4}{(s-2)(s+2)}& \frac{3}{(s-2)(s+2)}\\ -\frac{4}{(s-2)(s+2)}& \frac{s-3}{(s-2)(s+2)}\end{array}\right)$

Find the inverse of sI-A

Let

$\frac{s+4}{(s-2)(s+2)}=\frac{A}{s-2}+\frac{B}{s+2}$

if,$s+4=A(s+2)+B(s-2)$

If s=2,

$2+4=A(2+2)+B(2-2)\to 6=4A$

A=3/2

If,s= -2,

$-2+4=A(-2+2)+B(-2-2)\to 2=-4B$

B= -1/2

Let

$\frac{4}{(s-2)(s+2)}=\frac{A}{s-2}+\frac{B}{s+2}$

If,

$-4=A(s+2)+B(s-2)$

If s = 2,

$-4=A(2+2)+B(2-2)\to -4=4A$

A = -1

If s = -2,

$$\begin{gathered} -4=A(-2+2)+B(-2-2)\to -4=-4B \\ B=1 \end{gathered}$$

Let

width="222">-3(s-2)(s+2)=As-2+Bs+2

If,

$3=A(s+2)+B(s-2)$

If s=2
$3=A(s+2)+B(s-2)\to 3=4A$,

A=3/4

If s=-2,

$$\begin{gathered} 3=A(-2+2)+B(-2-2)\to 3=-4B \\ B=-\frac{3}{4} \end{gathered}$$

Let

$\frac{s-4}{(s-2)(s+2)}=\frac{A}{s-2}+\frac{B}{s+2}$

If

$s-4=A(s+2)+B(s-2)$

If s=2,

$$\begin{gathered} 2-4=A(2+2)+B(2-2) \\ A=-\frac{1}{2} \end{gathered}$$

If s= -2,

$$\begin{gathered} -2-4=A(-2+2)+B(-2-2) \\ B=\frac{3}{2} \end{gathered}$$

Find the General solution of given matrix

So now,

${(sI-A)}^{-1}=\left(\begin{array}{cc}\frac{3/2}{s-2}-\frac{1/2}{s+2}& \frac{3/4}{s-2}+\frac{-3/4}{s+2}\\ \frac{-1}{s-2}+\frac{1}{s+2}& \frac{-1/2}{s-2}+\frac{3/2}{s+2}\end{array}\right)$

we know that

$e^{\mathbf{A}t}={\mathcal{L}}^{-1}\left\{{(s\mathbf{I}-\mathbf{A})}^{-1}\right\}$

It follows from (3) that the inverse Laplace transform of (2) gives the desired result

$e^{At}=\left(\begin{array}{cc}\frac{3}{2}{e}^{2t}-\frac{1}{2}{e}^{-2t}& \frac{3}{4}{e}^{2t}-\frac{3}{4}{e}^{-2t}\\ -e^{2t}+e^{-2t}& -\frac{1}{2}{e}^{2t}+\frac{3}{2}{e}^{-2t}\end{array}\right)$

Finally, the general solution of the system is

X=e^AtC

$=\left(\begin{array}{cc}\frac{3}{2}{e}^{2t}-\frac{1}{2}{e}^{-2t}& \frac{3}{4}{e}^{2t}-\frac{3}{4}{e}^{-2t}\\ -e^{2t}+e^{-2t}& -\frac{1}{2}{e}^{2t}+\frac{3}{2}{e}^{-2t}\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right)$

=localid="1664881731266" $C_1\left(\begin{array}{c}\frac{3}{2}\\ -1\end{array}\right)e^{2t}+C_1\left(\begin{array}{c}-\frac{1}{2}\\ 1\end{array}\right)e^{-2t}+C_2\left(\begin{array}{c}\frac{3}{4}\\ -\frac{1}{2}\end{array}\right)e^{2t}+C_2\left(\begin{array}{c}-\frac{3}{4}\\ \frac{3}{2}\end{array}\right)e^{-2t}$

Let localid="1664881719176" $C_3=\frac{1}{2}{C}_1+\frac{1}{4}{C}_{2}and$