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Laplace Transform Solution of Ordinary Differential Equations
The Laplace transform converts differential equations in the time domain to algebraic equations in the
frequency domain. There are three important steps to the process:
(1) transform ODE from the time domain to the frequency domain;
(2) manipulate the algebraic equations to form a solution; and
(3) inverse transform the solution from the frequency to the time domain.
Perhaps, the most common Laplace transform pairs are those appearing in the table below:
f(t) (t) u(t) tae t
F(s) 1s
1
as +
1
2
1
s
When applied to differential equations, Laplace transforms automatically account for initial conditions via
)0()0()()(
)0()()(
2
2
2
yysssdt
tyd
xss
dt
txd
L
L
=
=
Y
X
(1)
Inverse Laplace Transform
Performing the inverse transform is straightforward when using partial fractions expansion with the
method of residues. There are three possible cases: (1) simple roots (real and unequal), (2) simple
complex roots, and (3) repeated roots (of the same value).
1. Simple roots
The simple roots are placed in a partial fractions expansion
( ) ( )( )( ) ( ) n
n
n
m
psK
psK
psK
pspspszszsKs
+++
++
+=
+++++= KL
L
2
2
1
1
21
10)(F (2)
The constants,Ki, can be found using the method of residues
ipsii spsK =+= )()( F (3)
Finally, the tabulated Laplace transform pairs are used to invert the expression. This is a nice form since the
solution is
tpn
tptp neKeKeKtf +++= L21 21)( (4)
2. Complex conjugate rootsComplex roots result in a Laplace transform of the form
L
L
+++
+
+
=
+++
++
=
)()(
)()()(
11
*11
js
K
js
K
js
K
js
KsF
(5)
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LaplaceTransformSolutionOfODEs (c) 2006 K.E. Holbert Page 2 of 2
TheK1 can then be found using the same method as for simple roots; however, the positive root of the
form j+ must be the one used to findK1, that is
jssjsK
+=+= )()(1 F (6)
The corresponding time domain function is then
( ) L++= teKtf t cos2)( 1 (7)
3. Repeated roots
When F(s) has a root of multiplicity r, then F(s) is written as
( ) ( ) ( )LL +
+++
++
+=
+=
r
r
rps
K
ps
K
ps
K
pss
ss
1
1
2
1
12
1
11
11
1
)(
)()(
Q
PF (8)
where the time domain function is then
( )
LL +
+++=
tpr
rtptp e
r
tKetKeKtf 111
!1
)(1
11211 (9)
That is, we obtain the exponentials multiplied by ts, where theK1j terms are evaluated from
( )( )[ ]
1
)(!
111
ps
r
jr
jr
j spsds
d
jrK
=
+
= F (10)
This actually simplifies nicely until you reachs3 terms, that is, for a double root 21)( ps
( )
( )[ ]1
1
)(
)(
2
111
2
112
ps
ps
sps
ds
dK
spsK
=
=
+=
+=
F
F
(11)
ThusK12 is found just like for simple roots. Note this backward order of solving for theKs, that is, we
solve for theKs of the higher order terms first.
Initial-Value and Final-Value Theorems
Sometimes, the initial and final value theorems provide a quick manner in which to partial check ones
overall answer. The initial value theorem states
)(lim)(lim sstfs0t
F
= (12)
Almost inevitably, we must use L'Hopitals Rule, which states that ifg(x)/h(x) has the indeterminate form
0/0 or/ atx = c, then
)('
)('lim
)(
)(lim
xh
xg
xh
xg
cxcx = (13)
The final value theorem is
)(lim)(lim sstf0st
F
= (14)