7—Operators and Matrices 198Write this as a pair of equations, using the velocity as an independent variable.
dx
dt=vx anddvx
dt=−
k
mx−b
mvxIn matrix form, this is a matrix differential equation.
d
dt(
x
vx)
=
(
0 1
−k/m −b/m)(
x
vx)
This is a linear, constant-coefficient differential equation, only now the constant coefficients are matrices. Don’t
let that slow you down. The reason that an exponential form of solution works is that the derivative of an
exponential is an exponential. Assume such a solution here.
(
x
vx)
=
(
A
B
)
eαt, giving α(
A
B
)
eαt=(
0 1
−k/m −b/m)(
A
B
)
eαt (35)When you divide the equation byeαt, you’re left with an eigenvector equation, where the eigenvalue isα. As
usual, to get a non-zero solution set the determinant of the coefficients to zero and the characteristic equation is
det(
0 −α 1
−k/m −b/m−α)
=α(α+b/m) +k/m= 0with familiar roots
α=
(
−b±√
b^2 − 4 km)
/ 2 mIf the two roots are equal youmaynot have distinct eigenvectors, and in this case youdonot. No matter, you
can solve any such problem for the case thatb^2 − 4 km 6 = 0and then take the limit as this approaches zero.
The eigenvectors come from the equationαA=B, the simpler of the two linear equations represented by
Eq. ( 35 ). Really one linear equation because of the determinant condition.
(
x
vx)
(t) =A+(
1
α+)
eα+t+A−(
1
α−)
eα−t