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

16. $\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)\mathit{X}$

Short answer

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

Step-by-step solution

Definition of Homogeneous system

• A matrix exponential e^At so that X 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,

$X\text{'}=\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)X$

Where

$A=\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)$

First compute the matrix sI-A,

$det(sI-A)=0\to \left(\begin{array}{cc}s-4& 2\\ -1& s-1\end{array}\right)=0$

$$\begin{gathered} (s-4)(s-1)+2=0 \\ {s}^2-s-4s+4+2=0 \\ {s}^2-5s+6=0 \\ (s-3)(s-2)=0 \end{gathered}$$

Now can find the inverse of (1).

localid="1664955312902" $$\begin{gathered} {\left(sI-A\right)}^{-1}={\left(\begin{array}{cc}s-4& 2\\ -1& s-1\end{array}\right)}^{-1} \\ =\left(\begin{array}{cc}\frac{s-1}{(s-3)(s-2)}& -\frac{2}{(s-3)(s-2)}\\ \frac{1}{(s-3)(s-2)}& \frac{s-4}{(s-3)(s-2)}\end{array}\right) \end{gathered}$$

Find the inverse of sI-A

Let

$\frac{s-1}{(s-3)(s-2)}=\frac{A}{(s-3)}+\frac{B}{(s-2)}$

where,

s - 1 = A(s - 2) + B(s - 3)

when s=3,

3 - 1 = A(3 - 2) + B(3 - 3)

A = 2

when s=2,

2 - 1 = A(2 - 2) + B(2 - 3)

B = - 1

Let

$\frac{1}{(s-3)(s-2)}=\frac{A}{S-3}+\frac{B}{s-2}$

Where ,

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

when s=3.

1 = A(3 - 2) + B(3 - 3)

A = 1

when s=2,

1 = A(2 - 2) + B(2 - 3)

B = - 1

Let

$\frac{2}{(s-3)(s-2)}=\frac{A}{s-3}+\frac{B}{s-2}$

Where ,

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

when s=3,

$-2=A(3-2)+B(3-3)\to A=-2$

when s=2,

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

B = 2

Let

$\frac{s-4}{(s-3)(s-2)}=\frac{A}{s-3}+\frac{B}{s-2}$

where,

s - 4 = A(s - 2) + B(s - 3)

when s=3.

3 - 4 = A(3 - 2) + B(3 - 3)

A = - 1

when s=2,

2 - 4 = A(2 - 2) + B(2 - 3)

B = 2

Find the General solution of given matrix

So now,

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

know that

$e^{At}=L^{-1}\left\{{(sI-A)}^{-1}\right\}$

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

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

the general solution of the system is

X=e^AtC

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

$$\begin{gathered} =C_1\left(\begin{array}{c}2\\ 1\end{array}\right)e^{3t}+C_1\left(\begin{array}{c}-1\\ -1\end{array}\right)e^{2t}+C_2\left(\begin{array}{c}-2\\ -1\end{array}\right)e^{3t}+C_2\left(\begin{array}{c}2\\ 2\end{array}\right)e^{2t} \\ =\left[C_1-C_2\right]\left(\begin{array}{c}2\\ 1\end{array}\right)e^{3t}+\left[-C_1+2C_2\right]\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t} \\ let{C}_3=C_1-C_{2}and{C}_4=-C_1+2C_2 \\ =C_3\left(\begin{array}{c}2\\ 1\end{array}\right)e^{3t}+C_4\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t} \end{gathered}$$

Hence, the general solution for the given matrix is

$=C_3\left(\begin{array}{c}2\\ 1\end{array}\right)e^{3t}+C_4\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t}$