590 CHAPTER 30. LAPLACE TRANSFORM METHODS
Example 30.2.7 Find the solution to
x′ =
(−4 −36 5
)x+
(cos t
et
),x(0) =
(11
)
First find the fundamental matrix. One goes backwards in the table to find the following.
(sI−A)−1 =
(s
(1 00 1
)−
(−4 −36 5
))−1
=
− s−5−s2+s+2
3−s2+s+2
− 1− 1
6 s2+ 16 s+ 1
3−
16 s+ 2
3− 1
6 s2+ 16 s+ 1
3
Then using going backwards in the table and writing in terms of cosh and sinh,
Φ(t) =
(e
12 t(cosh 3
2 t−3sinh 32 t)
−2(sinh 3
2 t)
e12 t
4(sinh 3
2 t)
e12 t e
12 t(cosh 3
2 t +3sinh 32 t) )
Then the solution is
x(t) =
(e
12 t(cosh 3
2 t−3sinh 32 t)
−2(sinh 3
2 t)
e12 t
4(sinh 3
2 t)
e12 t e
12 t(cosh 3
2 t +3sinh 32 t) )( 1
1
)
+∫ t
0Φ(t−u)f (u)du
x(t) =
(e
12 t(cosh 3
2 t−5sinh 32 t)
e12 t(cosh 3
2 t +7sinh 32 t) )+
∫ t
0Φ(t− s)
(coss
es
)ds
Doing the integrations, one obtains
∫ t
0Φ(t− s)
(coss
es
)ds
=
(110 e−t
(15e2t −14e3t +14(cos t)et +8et sin t−15
)− 1
10 e−t(25e2t −28e3t +18(cos t)et +6et sin t−15
) )
It follows that x(t) =32 et + 4
5 sin t− 32 e−t − 7
5 e2t +(cosh 3
2 t)
e12 t
−5(sinh 3
2 t)
e12 t + 7
5 (cos t)ete−t
32 e−t − 3
5 sin t− 52 et + 14
5 e2t +(cosh 3
2 t)
e12 t
+7(sinh 3
2 t)
e12 t − 9
5 (cos t)ete−t
Using the table as just described really is a pretty good way to solve these kinds of
equations, but there is a much easier way to do it. You let the computer algebra system dothe tedious work for you. Here is the general idea for a first order system. Be patient. I will