There are three cases to consider.
Case 1 The roots of characteristic equation (9.136) are real and unique. If r 1 , r 2
are these roots, then the scalar homogeneous differential equation (9.137) has
the fundamental set of solutions {er^1 t, er^2 t}and the general of equation (9.137)is
yc=c 1 er^1 t+c 2 er^2 twhere c 1 and c 2 are arbitrary scalar constants.
Case 2 The roots of the characteristic equation (9.136) are real and equal, say
r 1 =r 2. In this case the fundamental set of solutions is given by {er^1 t, ter^1 t}and
the general solution of equation (9.137) is yc=c 1 er^1 t+c 2 ter^1 twhere c 1 and c 2 are
arbitrary scalar constants.
Case 3 The roots of the characteristic equation (9.136) are complex roots, say
r 1 =a+ib and r 2 =a−ib. In this case the fundamental set of solutions can be
represented in the form {e(a+ib)t, e(a−ib)t}or one can make use of Euler’s equation
eibt = cos bt +isin bt and take appropriate linear combination of solutions to write
the fundamental set of solutions in the form {eat cos bt, e atsin bt}.The general so-
lution to the scalar homogeneous equation can then be expressed in either of the
forms
y=c 1 e(a+ib)t+c 2 e(a−ib)t
or yc=eat(c 1 cos bt +c 2 sin bt)
where c 1 and c 2 are arbitrary constants.
If {y 1 (t), y 2 (t)}is a fundamental set of solutions to the homogeneous scalar differ-
ential equation (9.137), then
yc=c 1 y 1 (t) + c 2 y 2 (t) (9 .138)
where c 1 and c 2 are arbitrary vector constants, is the representation of the general
solution to the vector differential equation (9.135). Substitute equation (9.138) into
equation (9.135) and show there results the vector equation
c 1
(
d^2 y 1
dt^2 +α
dy 1
dt +βy^1
)
+c 2
(
d^2 y 2
dt^2 +α
dy 2
dt +βy^2
)
= 0 (9 .139)
Observe that if c 1 and c 2 are arbitrary independent vectors, then in order for equation
(9.139) to be satisfied, the scalar components of the arbitrary vectors c 1 and c 2 must
equal zero.
The solution of the nonhomogeneous vector differential equation
d^2 y
dt^2 +α
dy
dt +βy=
F(t) (9 .125)
