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 1 −3 t 4 t 2 2 t 2 2 −3 t
e + e e − e 
e A t =  5 5 5 5
 2 2 4 1 
 e − e −3 t e −3 t + e 
t 2
t 2
 5 5 5 5 
Therefore, the solution of the system of equations is

t 2
2 e − e −3 t 
Xt() =  t
t 2
 e + 2 e −3 
dX
8.9.3 Solution of Equations of the Form = AX + B(t)
dt
Multiplying this equation on the left by e , we obtain:
–At

dX −A t −A t
t
−A
e = e AX + e B() (8.36)
t
dt
Rearranging terms, we write this equation as:

dX −A t −A t
−A
t
t
e − e AX = e B() (8.37)
dt
We note that the LHS of this equation is the derivative of e –At X. Therefore, we
can now write Eq. (8.37) as:

d −A t −A t
e [ X t ( )] = e B t ( ) (8.38)
dt

This can be directly integrated to give:

t t −Aτ
−A
t
[e X ( )]t = ∫ e B ( )dττ (8.39)
0
0
or, written differently as:

d )
t − X( )0
e −A t X() = ∫ t e −Aτ B(ττ (8.40a)
0
which leads to the standard form of the solution:

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