Question

In Problems 23and 24use the results of problems 19-20to solve the given system.

$X\text{'}=\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right)X$.

Short answer

The general solution of the system $X\text{'}=\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right)X$is $X=C_1\left(\begin{array}{cc}-\frac{1}{2}{e}^{5t}& +\frac{3}{2}{e}^{3t}\\ -\frac{3}{2}{e}^{5t}& +\frac{3}{2}{e}^{3t}\end{array}\right)+C_2\left(\begin{array}{cc}\frac{1}{2}{e}^{5t}& -\frac{1}{2}{e}^{3t}\\ \frac{3}{2}{e}^{5t}& -\frac{1}{2}{e}^{3t}\end{array}\right)$.

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 value of eAt.

It is given that,

$X\text{'}=\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right)X$

From the previous problem we know that,

$$\begin{gathered} PD{P}^{-1}=\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right)\left(\begin{array}{cc}5& 0\\ 0& 3\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right) \\ =\left(\begin{array}{cc}5& 3\\ 15& 3\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right) \\ =\left(\begin{array}{cc}2& 1\\ -3& 6\end{array}\right) \\ =A \end{gathered}$$

We also know that,

$$\begin{gathered} e^{At}=P{e}^{Dt}{P}^{-1} \\ {e}^{Dt}=\left(\begin{array}{cc}{e}^{5t}& 0\\ 0& e^{3t}\end{array}\right) \end{gathered}$$

Therefore,

$$\begin{gathered} e^{At}=\left(\begin{array}{cc}1& 1\\ 3& 1\end{array}\right)\left(\begin{array}{cc}{e}^{5t}& 0\\ 0& e^{3t}\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right) \\ =\left(\begin{array}{cc}{e}^{5t}& e^{3t}\\ 3e^{5t}& e^{3t}\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right) \\ =\left(\begin{array}{cc}-\frac{1}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}& \frac{1}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\\ -\frac{3}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}& \frac{3}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\end{array}\right) \end{gathered}$$

Hence the value of e^Atis $\left(\begin{array}{cc}-\frac{1}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}& \frac{1}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\\ -\frac{3}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}& \frac{3}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\end{array}\right)$.

Find the general solution of the system.

To find the general solution of the system, we apply the formula,

X=e^AtC

$$\begin{gathered} =\left(\begin{array}{cc}-\frac{1}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}& \frac{1}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\\ -\frac{3}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}& \frac{3}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\end{array}\right) \\ =C_1\left(\begin{array}{c}-\frac{1}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}\\ -\frac{3}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}\end{array}\right)+C_2\left(\begin{array}{c}\frac{1}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\\ \frac{3}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\end{array}\right) \end{gathered}$$
Hence the general solution is $X=C_1\left(\begin{array}{c}-\frac{1}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}\\ -\frac{3}{2}{e}^{5t}+\frac{3}{2}{e}^{3t}\end{array}\right)+C_2\left(\begin{array}{c}\frac{1}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\\ \frac{3}{2}{e}^{5t}-\frac{1}{2}{e}^{3t}\end{array}\right)$.