Begin2.DVI

where αand β are scalar constants is obtained by first solving the homogeneous

equation

d^2 
y

dt^2 +α

d
y

dt +β
y=

 0

given by equation (9.138) and then finding any particular solution
yp=
yp(t)which

satisfies

d^2 
y p

dt^2

+αd
yp

dt

+β
yp=F
(t)

The general solution to the vector differential equation (9.125) is then given by

y=
yc+
yp=
c 1 y 1 (t) + 
c 2 y 2 (t) + 
y p(t) (9 .126)

Example 9-9. Solve the vector differential equation

d^2 
y

dt^2 + 2α

d
y

dt +β

(^2)
y= sin 3tˆe 3 (9 .133)

Solution First solve the homogeneous vector differential equation

y ′′+ 2 α
y′+β^2 
y=
 0 (9 .128)

If
y =
c 1 y 1 (t) +
c 2 y 2 (t)is the general solution of equation (9.128), then y 1 (t)and y 2 (t)

must be independent solutions of the scalar differential equation

d^2 y

dt^2

+ 2αdy

dt

+β^2 y= 0 (9 .129)

This is an equation with constant coefficients. The general procedure to solve a

differential equation with constant coefficients is to assume an exponential solution

y=eλt. Substituting the assumed exponential solution into the differential equation

(9.129) produce the characteristic equation

λ^2 + 2αλ +β^2 = 0 (9 .130)

for determining values of λ to be substituted into the assumed solution. Solving

equation (9.130) for λgives the characteristic roots

λ=

− 2 α±

√

(2 α)^2 − 4 β^2

2 =−α±

√

α^2 −β^2 (9 .131)

Case 1 If α^2 −β^2 = ω^2 > 0 , then a fundamental set of solutions is given by

{e−(α−ω)t, e−(α+ω)t}and the general solution to equation (9.128) is

y=
c 1 e−(α−ω)t+
c 2 e−(α+ω)t