Question 7.4.12: Solve the initial value problem x′ = [5 -17 8 -7]x, x(0) = [...

The coefficient matrix

A = \begin{bmatrix} 5 & -17 \\ 8& -7 \end{bmatrix}               (55)

has characteristic equation

|A – λI| = \begin{vmatrix} 5-\lambda & -17 \\ 8 & -7-\lambda \end{vmatrix} = (\lambda +1)^{2} + 100 = 0.

and hence has the complex conjugate eigenvalues λ_{1}, λ_{2} = -1 ± 10i . If v = [a   b]^{T} is an eigenvector associated with λ = -1 + 10i , then the eigenvector equation (A – λI)v = 0 yields the same system (48) of equations found in Example 11:

(6 – 10i )a – 17b = 0.
4a – (3 + 5i)b = 0.                              (48)

As in  Example 11, each of these equations is satisfied by a = 3 + 5i and b = 4. Thus the desired eigenvector, associated with λ_{1} = -1 + 10i, is once again v = [3 + 5i       4]^{T}, with real and imaginary parts

a =  \left [ \begin{matrix} 3 \\ 4 \end{matrix} \right ]      and        b = \left [ \begin{matrix} 5 \\ 0 \end{matrix} \right ],            (56)

respectively. Taking p = -1 and q = 10 in Eq. (5) therefore gives the general solution of the system x^{′} = Ax:

x(t) = c_{1} e^{pt} (a  \cos qt  –  b  sin qt ) + c_{2} e^{pt} (b  \cos qt  +  a  sin qt )            (5)

x(t) = c_{1} e^{-t} \left ( \begin{matrix} \left [ \begin{matrix} 3 \\ 4 \end{matrix} \right ] \cos 10t – \left [ \begin{matrix} 5 \\ 0 \end{matrix} \right ] \sin 10t \end{matrix} \right ) + c_{2} e^{-t} \left ( \begin{matrix} \left [ \begin{matrix} 5 \\ 0 \end{matrix} \right ]\cos 10t + \left [ \begin{matrix} 3 \\ 4 \end{matrix} \right ] \sin 10t \end{matrix} \right )

The initial condition x(0) = [4     2]^{T} gives  c_{1} = c_{2} = \frac{1}{2} once again, and with these values Eq. (57) becomes (in scalar form)

x_{1}(t) = e^{-t} (4 \cos 10t  –  \sin 10t),

x_{2}(t) = e^{-t} (2 \cos 10t + 2 \sin 10t).                   (58)