Example 10-13. Represent the differential equation d
(^2) y
dt^2
+ω^2 y= sin 2 tin matrix
form.
Solution Let y 1 =yand y 2 =
dy 1
dt =dydt, then
dy 2
dt =d^2 y 1
dt^2 =d^2 y
dt^2 =−ω(^2) y+ sin 2 t=−ω (^2) y 1 + sin2 t
The given scalar differential equation can then be represented in the matrix form
(dy 1
dydt 2
dt)
=(
0 1
−ω^20)(
y 1
y 2)
+(
0
sin 2 t)or
dy
dt=Ay+f(t) where A=
(
0 1
−ω^20)
, f(t) =(
0
sin2t)and y=col(y 1 , y 2 )
Example 10-14. Associated with the initial value problem
d ̄y
dt =A(t) ̄y+
f ̄(t), y ̄(0) = ̄c (10.9)is the matrix differential equation
dX
dt=A(t)X, X (0) = I (10 .10)where X= (xij)n×n and Iis the n×nidentity matrix. Associated with the matrix
differential equation (10.10) is the adjoint differential equation
dZ
dt =−ZA (t), Z (0) = I, Z = (zij)n×n (10 .11)The relationship between the three differential equations given by equations (10.9),
(10.10), and (10.11), is as follows. Left-multiply equation (10.10) by Z and right-
multiply equation (10.11) by Xto obtain
ZdX
dt=ZA (t)X (10 .12)
dZ
dt X=−ZA (t)X (10 .13)and then add the equation (10.12) and (10.13) to obtain
ZdX
dt +dZ
dt X=d
dt(ZX ) = [0] (10 .14)