Question 7.3.1: Find a general solution of the system x′1 = 4x1 + 2x2, x′2 =...
Find a general solution of the system
x^{′}_{1} = 4x_{1} + 2x_{2},
x^{′}_{2} = 3x_{1} – x_{2}. (9)
The matrix form of the system in (9) is
x^{′} = \begin{bmatrix} 4 & 2 \\ 3 & -1 \end{bmatrix} x. (10)
The characteristic equation of the coefficient matrix is
\begin{vmatrix} 4-\lambda & 2 \\ 3 & -1-\lambda \end{vmatrix} = (4-\lambda )(-1-\lambda )-6
=\lambda² – 3\lambda – 10 = (\lambda + 2)(\lambda – 5)=0.
so we have the distinct real eigenvalues λ_{1} = -2 and λ_{2} = 5. For the coefficient matrix A in Eq. (10), the eigenvector equation (A – λI)v = 0 takes the form
\begin{bmatrix} 4-\lambda & 2 \\ 3 & -1-\lambda \end{bmatrix} \left [ \begin{matrix} a \\ b \end{matrix} \right ] =\left [ \begin{matrix} 0 \\ 0 \end{matrix} \right ] (11)
for the associated eigenvector v = [a b ]^{T}.
CASE 1: λ_{1} = -2. Substitution of the first eigenvalue λ_{1} = -2 in Eq. (11) yields the system
\begin{bmatrix} 6 & 2 \\ 3 & 1 \end{bmatrix} \left [ \begin{matrix} a \\ b \end{matrix} \right ] =\left [ \begin{matrix} 0 \\ 0 \end{matrix} \right ]
—that is, the two scalar equations
6a + 2b = 0,
3a + b = 0. (12)
In contrast with the typical nonsingular (algebraic) linear system that has a unique solution, the homogeneous linear system in (12) is singular—the two scalar equations obviously are equivalent (each being a multiple of the other). Therefore, Eq. (12) has infinitely many nonzero solutions—we can choose a arbitrarily (but nonzero) and then solve for b. Substitution of an eigenvalue λ in the eigenvector equation (A – λI)v = 0 always yields a singular homogeneous linear system, and among its infinity of solutions we generally seek a “simple” solution with small integer values (if possible). Looking at the second equation in (12), the choice a = 1 yields b = -3, and thus
v_{1 } = \left [ \begin{matrix} 1 \\ -3 \end{matrix} \right ]
is an eigenvector associated with λ_{1} = -2 (as is any nonzero constant multiple of v_{1}).
Remark If, instead of the “simplest” choice a = 1, b = -3, we had made another choice a = c, b = -3c, we would have obtained the eigenvector
v_{1} = \left [ \begin{matrix} c \\ -3c \end{matrix} \right ] = c \left [ \begin{matrix} 1 \\ -3 \end{matrix} \right ] .
Because this is a constant multiple of our previous result, any choice we make leads to (a constant multiple of ) the same solution
x_{1}(t) = \left [ \begin{matrix} 1 \\ -3 \end{matrix} \right ] e^{-2t}.