Question 7.4.13: Solve the initial value problem x′ = [-5 17 -8 7]x = x(0) = ...
Solve the initial value problem
\textbf{x}^{′} = \begin{bmatrix} -5 & 17 \\ -8 & 7 \end{bmatrix} \textbf{x}, \textbf{x}(0) =\left [ \begin{matrix} 4 \\ 2 \end{matrix} \right ] . (59)
Although we could directly apply the eigenvalue/eigenvector method as in previous cases (see Problem 40), here it is more convenient to notice that the coefficient matrix
A = \begin{bmatrix} -5 & 17 \\ -8 & 7 \end{bmatrix} (60)
is the negative of the matrix in Eq. (55) used in Example 12. By the principle of time reversal, therefore, the solution of the initial value problem (59) is given by simply replacing t with -t in the right-hand sides of the solution (58) of the initial value problem in that example:
A = \begin{bmatrix} 5 & -17 \\ 8& -7 \end{bmatrix} (55)
x_{1}(t) = e^{t} (4 \cos 10t + \sin 10t),
x_{2}(t) = e^{t} (2 \cos 10t – 2 \sin 10t). (61)