Question

In Problems 15-18use the method of Example 2to compute e^Atfor the coefficient matrix. Use (1)to find the general solution of the given system.

$X\text{'}=\left(\begin{array}{cc}0& 1\\ -2& -2\end{array}\right)X$

Let Pdenote a matrix whose columns are eigenvectors K₁,K₂,.....,K_ncorresponding to distinct eigenvalues ${\lambda }_1,{\lambda }_2,......{\lambda }_n$of an nxnmatrix A. Then it can be show that A=PDP^-1,where Dis a diagonal matrix defined by

$D=\left(\begin{array}{ccc}{\lambda }_1& 0.....& 0\\ 0& {\lambda }_2.....& 0\\ 0& 0& {\lambda }_n\end{array}\right)$

Short answer

The solution for the given system $X\text{'}=\left(\begin{array}{cc}0& 1\\ -2& -2\end{array}\right)X$is $X=C_1\left(\begin{array}{c}\cos t+\sin t\\ -2\sin t\end{array}\right)e^{-t}+C_2\left(\begin{array}{c}\sin t\\ \cos t-\sin t\end{array}\right)e^{-t}$.

Step-by-step solution

Define matrix exponential.

Consider a square matrix A of size n*n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as

where I is the unit matrix of order n.

Therefore the infinite matrix power series is

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression e^At. It is given by the formula,

Find the matrix sI-A.

We are given,

$X\text{'}=\left(\begin{array}{cc}0& 1\\ -2& -2\end{array}\right)X$

where,

$A=\left(\begin{array}{cc}0& 1\\ -2& -2\end{array}\right)$

First compute the matrix sI-A.

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

and,

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

Hence the matrix sI-A is$\left(\begin{array}{cc}s& -1\\ 2& s+2\end{array}\right)$.

Find the inverse of (1).

Now we calculate the inverse of (1).

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

Now,

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

Hence the inverse of role="math" localid="1665058785755" $\left(\begin{array}{cc}s& -1\\ 2& s+2\end{array}\right)$is $\left(\begin{array}{cc}\frac{s+1}{{(s+1)}^2+1}+\frac{1}{{(s+1)}^2+1}& \frac{1}{{(s+1)}^2+1}\\ -\frac{2}{{(s+1)}^2+1}& \frac{s+1}{{(s+1)}^2+1}-\frac{1}{{(s+1)}^2+1}\end{array}\right)$.

Find the value of eAt and the general solution of X'=01-2-2X

We know that,

$e^{At}=L^{-1}\left\{{(sI-A)}^{-1}\right\}\left(3\right)$

Use the above formula to perform inverse Laplace transform on (2)

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

The solution is

X=e^AtC

$$\begin{gathered} =\left(\begin{array}{cc}{e}^{-t}\cos t+e^{-t}\sin t& e^{-t}\sin t\\ -2e^{-t}\sin t& e^{-t}\cos t-e^{-t}\sin t\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right) \\ =C_1\left(\begin{array}{c}\cos t+\sin t\\ -2\sin t\end{array}\right)e^{-t}+C_2\left(\begin{array}{c}\sin t\\ \cos t-\sin t\end{array}\right)e^{-t} \end{gathered}$$

Hence the general solution of the given system is $X=C_1\left(\begin{array}{c}\cos t+\sin t\\ -2\sin t\end{array}\right)e^{-t}+C_2\left(\begin{array}{c}\sin t\\ \cos t-\sin t\end{array}\right)e^{-t}$.