Question 7.6.3: Find a general solution of the system x′ = [1 -3 3 7]x. (20)
Find a general solution of the system
\textbf{x}^{′} = \begin{bmatrix} 1 & -3 \\ 3 & 7 \end{bmatrix}\textbf{x}. (20)
In Example 2 we found that the coefficient matrix A in Eq. (20) has the defective eigenvalue λ = 4 of multiplicity 2. We therefore begin by calculating
(\textbf{A} – 4\textbf{I})² = \begin{bmatrix} -3 & -3 \\ 3 & 3 \end{bmatrix} \begin{bmatrix} -3 & -3 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 0& 0 \\ 0 & 0\end{bmatrix}.
Hence Eq. (16) is
(\textbf{A} – λ\textbf{I})^{2}\textbf{v}_{2} = 0 (16)
\begin{bmatrix} 0& 0 \\ 0 & 0\end{bmatrix} \textbf{v}_{2}= \textbf{0}.
and therefore is satisfied by any choice of \textbf{v}_{2}. In principle, it could happen that (A – 4I)\textbf{v}_{2} is nonzero (as desired) for some choices of \textbf{v}_{2} though not for others. If we try \textbf{v}_{2} = [1 0]^{T} we find that
(A – 4I)v_{2} = \begin{bmatrix} 3 & -3 \\ 3 & 3 \end{bmatrix} \left [ \begin{matrix} 1 \\ 0 \end{matrix} \right ] = \left [ \begin{matrix} -3 \\ 3 \end{matrix} \right ] = \textbf{v}_{1}
is nonzero, and therefore is an eigenvector associated with λ = 4. (It is -3 times the eigenvector found in Example 2.) Therefore, the two solutions of Eq. (20) given by Eqs. (18) and (19) are
\textbf{x}_{1}(t) =\textbf{v}_{1} e^{4t} = \left [ \begin{matrix} -3 \\ 3 \end{matrix} \right ] e^{4t}.
\textbf{x}_{2}(t) = (\textbf{v}_{1} t + \textbf{v}_{2}) e^{4t} = \left [ \begin{matrix} -3t + 1 \\ 3t \end{matrix} \right ] e^{4t}.
The resulting general solution
x(t) = c_{1} \textbf{x}_{1}(t) + c_{2} \textbf{x}_{2}(t)
has scalar component functions
x_{1}(t) = (-3c_{2} t + c_{2} – 3c_{1})e^{4t},
x_{2}(t) = (3c_{2} t + 3c_{1})e^{4t}.
With c_{2} = 0 these solution equations reduce to the equations x_{1}(t) = -3c_{1} e^{4t} , x_{2}(t) = 3c_{1}e^{4t} , which parametrize the line x_{1} = – x_{2} in the x_{1} x_{2} -plane. The point (x_{1}(t), x_{2}(t)) then recedes along this line away from the origin as t → + \infty, to the northwest if c_{1} > 0 and to the southeast if c_{1} < 0. As indicated in Fig. 7.6.1, each solution curve with c_{2} ≠ 0 is tangent to the line x_{1} = -x_{2} at the origin; the point (x_{1}(t), x_{2}(t)) approaches the origin as t → -1 and approaches + \infty along the solution curve as t → + \infty.
