Question 7.4.11: Solve the initial value problem x′ = [6 -17 8 -6]x, x(0) = [...

The coefficient matrix

A = \begin{bmatrix} 6 & -17 \\ 8 & -6 \end{bmatrix}         (47)

has characteristic equation

|\textbf{A} – \lambda \textbf{I}| = \begin{bmatrix} 6-\lambda & -17 \\ 8 & -6-\lambda \end{bmatrix} =\lambda ^{2}+100=0,

and hence has the complex conjugate eigenvalues λ_{1}, λ_{2} = ±10i . If v = [a     b]^{T} is an eigenvector associated with λ = 10i , then the eigenvector equation (A – λI)v = 0 yields

\left [ \begin{matrix}\textbf{A} -10i \cdot \textbf{I} \end{matrix} \right ] \textbf{v} = \begin{bmatrix} 6-10i & -17 \\ 8 & -6-10i \end{bmatrix} \left [ \begin{matrix} a \\ b \end{matrix} \right ] =\left [ \begin{matrix} 0 \\ 0 \end{matrix} \right ] .

Upon division of the second row by 2, this gives the two scalar equations

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

each of which is satisfied by a = 3 + 5i and b = 4. Thus the desired eigenvector is v = \left [ \begin{matrix} 3 + 5i   & 4 \end{matrix} \right ]^{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 ],            (49)

respectively. Taking q = 10 in Eq. (45) therefore gives the general solution of the system  x^{′} = Ax.

x(t) = c_{1} (a  cos qt  –  b  sin qt ) + c_{2} (b  cos qt  +  a  sin qt )            (45)

To solve the given initial value problem it remains only to determine values of the coefficients c_{1}  and  c_{2}. The initial condition x(0) = [4    2]^{T} readily yields c_{1} = c_{2} = \frac{1}{2} , and with these values Eq. (50) becomes (in scalar form)

x_{1}(t) = 4 \cos 10t  –  \sin 10t.

x_{2}(t) = 2 \cos 10t + 2 \sin 10t.                   (51)