8/2/2019 La Place Transform Solution of Odes

1/2

LaplaceTransformSolutionOfODEs (c) 2006 K.E. Holbert Page 1 of 2

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)

8/2/2019 La Place Transform Solution of Odes

2/2

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)