1. Ch 6.2: Solution of Initial Value Problems
The Laplace transform is named for the French mathematician
Laplace, who studied this transform in 1782.
The techniques described in this chapter were developed
primarily by Oliver Heaviside (1850-1925), an English
electrical engineer.
In this section we see how the Laplace transform can be used
to solve initial value problems for linear differential equations
with constant coefficients.
The Laplace transform is useful in solving these differential
equations because the transform of f' is related in a simple way
to the transform of f, as stated in Theorem 6.2.1.

2. Theorem 6.2.1
Suppose that f is a function for which the following hold:
(1) f is continuous and f' is piecewise continuous on [0, b] for all b > 0.
(2) | f(t) | ≤ Keat
when t ≥ M, for constants a, K, M, with K, M > 0.
Then the Laplace Transform of f' exists for s > a, with
Proof (outline): For f and f' continuous on [0, b], we have
Similarly for f' piecewise continuous on [0, b], see text.
{ } { } )0()()( ftfsLtfL −=′



 +−=



 −−=′
∫
∫∫
−−
∞→
−−
∞→
−
∞→
b
stsb
b
b
stbst
b
b
st
b
dttfesfbfe
dttfestfedttfe
0
000
)()0()(lim
)()()(lim)(lim

3. The Laplace Transform of f'
Thus if f and f' satisfy the hypotheses of Theorem 6.2.1, then
Now suppose f' and f '' satisfy the conditions specified for f
and f' of Theorem 6.2.1. We then obtain
Similarly, we can derive an expression for L{f (n)
}, provided f
and its derivatives satisfy suitable conditions. This result is
given in Corollary 6.2.2
{ } { }
{ }[ ]
{ } )0()0()(
)0()0()(
)0()()(
2
fsftfLs
fftfsLs
ftfsLtfL
′−−=
′−−=
′−′=′′
{ } { } )0()()( ftfsLtfL −=′

4. Corollary 6.2.2
Suppose that f is a function for which the following hold:
(1) f , f ', f '' ,…, f (n-1)
are continuous, and f (n)
piecewise continuous, on [0,
b] for all b > 0.
(2) | f(t) | ≤ Keat
, | f '(t) | ≤ Keat
, …, | f (n-1)
(t) | ≤ Keat
for t ≥ M, for constants
a, K, M, with K, M > 0.
Then the Laplace Transform of f (n)
exists for s > a, with
{ } { } )0()0()0()0()()( )1()2(21)( −−−−
−−−′−−= nnnnnn
fsffsfstfLstfL 

5. Example 1: Chapter 3 Method (1 of 4)
Consider the initial value problem
Recall from Section 3.1:
Thus r1 = -2 and r2 = -3, and general solution has the form
Using initial conditions:
Thus
We now solve this problem using Laplace Transforms.
( ) ( ) 30,20,065 =′==+′+′′ yyyyy
( )( ) 032065)( 2
=++⇔=++⇒= rrrrety rt
tt
ececty 3
2
2
1)( −−
+=
7,9
332
2
21
21
21
−==⇒



=−−
=+
cc
cc
cc
tt
eety 32
79)( −−
−=

6. Example 1: Laplace Tranform Method (2 of 4)
Assume that our IVP has a solution φ and that φ'(t) and φ''(t)
satisfy the conditions of Corollary 6.2.2. Then
and hence
Letting Y(s) = L{y}, we have
Substituting in the initial conditions, we obtain
Thus
( ) ( ) 30,20,065 =′==+′+′′ yyyyy
0}0{}{6}{5}{}65{ ==+′+′′=+′+′′ LyLyLyLyyyL
[ ] [ ] 0}{6)0(}{5)0()0(}{2
=+−+′−− yLyysLysyyLs
( ) ( ) 0)0()0(5)(652
=′−+−++ yyssYss
( ) ( ) 0352)(652
=−+−++ ssYss
( )( )23
132
)(}{
++
+
==
ss
s
sYyL

7. Using partial fraction decomposition, Y(s) can be rewritten:
Thus
( )( ) ( ) ( )
( ) ( )
9,7
1332,2
)32()(132
32132
2323
132
=−=
=+=+
+++=+
+++=+
+
+
+
=
++
+
BA
BABA
BAsBAs
sBsAs
s
B
s
A
ss
s
( ) ( )2
9
3
7
)(}{
+
+
+
−==
ss
sYyL
Example 1: Partial Fractions (3 of 4)

8. Recall from Section 6.1:
Thus
Recalling Y(s) = L{y}, we have
and hence
( ) ( )
,2},{9}{7
2
9
3
7
)( 23
−>+−=
+
+
+
−= −−
seLeL
ss
sY tt
{ } as
as
dtedteesFeL tasatstat
>
−
==== ∫∫
∞
−−
∞
−
,
1
)(
0
)(
0
}97{}{ 23 tt
eeLyL −−
+−=
tt
eety 23
97)( −−
+−=
Example 1: Solution (4 of 4)

9. General Laplace Transform Method
Consider the constant coefficient equation
Assume that this equation has a solution y = φ(t), and that
φ'(t) and φ''(t) satisfy the conditions of Corollary 6.2.2. Then
If we let Y(s) = L{y} and F(s) = L{ f }, then
)(tfcyybya =+′+′′
)}({}{}{}{}{ tfLycLybLyaLcyybyaL =+′+′′=+′+′′
[ ] [ ]
( ) ( )
( )
cbsas
sF
cbsas
yaybas
sY
sFyaybassYcbsas
sFycLyysLbysyyLsa
++
+
++
′++
=
=′−+−++
=+−+′−−
22
2
2
)()0()0(
)(
)()0()0()(
)(}{)0(}{)0()0(}{

10. Algebraic Problem
Thus the differential equation has been transformed into the
the algebraic equation
for which we seek y = φ(t) such that L{φ(t)} = Y(s).
Note that we do not need to solve the homogeneous and
nonhomogeneous equations separately, nor do we have a
separate step for using the initial conditions to determine the
values of the coefficients in the general solution.
( )
cbsas
sF
cbsas
yaybas
sY
++
+
++
′++
= 22
)()0()0(
)(

11. Characteristic Polynomial
Using the Laplace transform, our initial value problem
becomes
The polynomial in the denominator is the characteristic
polynomial associated with the differential equation.
The partial fraction expansion of Y(s) used to determine φ
requires us to find the roots of the characteristic equation.
For higher order equations, this may be difficult, especially if
the roots are irrational or complex.
( )
cbsas
sF
cbsas
yaybas
sY
++
+
++
′++
= 22
)()0()0(
)(
( ) ( ) 00 0,0),( yyyytfcyybya ′=′==+′+′′

12. Inverse Problem
The main difficulty in using the Laplace transform method is
determining the function y = φ(t) such that L{φ(t)} = Y(s).
This is an inverse problem, in which we try to find φ such
that φ(t) = L-1
{Y(s)}.
There is a general formula for L-1
, but it requires knowledge
of the theory of functions of a complex variable, and we do
not consider it here.
It can be shown that if f is continuous with L{f(t)} = F(s),
then f is the unique continuous function with f(t) = L-1
{F(s)}.
Table 6.2.1 in the text lists many of the functions and their
transforms that are encountered in this chapter.

13. Linearity of the Inverse Transform
Frequently a Laplace transform F(s) can be expressed as
Let
Then the function
has the Laplace transform F(s), since L is linear.
By the uniqueness result of the previous slide, no other
continuous function f has the same transform F(s).
Thus L-1
is a linear operator with
)()()()( 21 sFsFsFsF n+++= 
{ } { })()(,,)()( 1
1
1
1 sFLtfsFLtf nn
−−
== 
)()()()( 21 tftftftf n+++= 
{ } { } { })()()()( 1
1
11
sFLsFLsFLtf n
−−−
++== 

14. Example 2
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1
{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
s
sY
2
)( =






==
ss
sY
1
2
2
)(
{ } ( ) 212
1
2
2
)( 111
==






=






= −−−
s
L
s
LsYL
2)( =ty

15. Example 3
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1
{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
5
3
)(
−
=
s
sY






−
=
−
=
5
1
3
5
3
)(
ss
sY
{ } t
e
s
L
s
LsYL 5111
3
5
1
3
5
3
)( =






−
=






−
= −−−
t
ety 5
3)( =

16. Example 4
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1
{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
4
6
)(
s
sY =
44
!36
)(
ss
sY ==
{ } 3
4
11 !3
)( t
s
LsYL =






= −−
3
)( tty =

17. Example 5
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1
{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus






=











== 333
!2
4
!2
!2
88
)(
sss
sY
{ } 2
3
1
3
11
4
!2
4
!2
4)( t
s
L
s
LsYL =






=












= −−−
2
4)( tty =
3
8
)(
s
sY =

18. Example 6
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1
{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus




+
+



+
=
+
+
=
9
3
3
1
9
4
9
14
)( 222
ss
s
s
s
sY
{ } tt
s
L
s
s
LsYL 3sin
3
1
3cos4
9
3
3
1
9
4)( 2
1
2
11
+=






+
+






+
= −−−
ttty 3sin
3
1
3cos4)( +=
9
14
)( 2
+
+
=
s
s
sY

19. Example 7
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1
{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus






−
+





−
=
−
+
=
9
3
3
1
9
4
9
14
)( 222
ss
s
s
s
sY
{ } tt
s
L
s
s
LsYL 3sinh
3
1
3cosh4
9
3
3
1
9
4)( 2
1
2
11
+=






−
+






−
= −−−
ttty 3sinh
3
1
3cosh4)( +=
9
14
)( 2
−
+
=
s
s
sY

20. Example 8
Find the inverse Laplace Transform of the given function.
To find y(t) such that y(t) = L-1
{Y(s)}, we first rewrite Y(s):
Using Table 6.2.1,
Thus
{ }
( )
t
et
s
LsYL −−−
−=






+
−= 2
3
11
5
1
!2
5)(
t
etty −
−= 2
5)(
( )3
1
10
)(
+
−=
s
sY
( ) ( ) ( ) 





+
−=





+
−=
+
−= 333
1
!2
5
1
!2
!2
10
1
10
)(
sss
sY

21. Example 9
For the function Y(s) below, we find y(t) = L-1
{Y(s)} by using
a partial fraction expansion, as follows.
tt
eety
ss
sY
BA
ABBA
ABsBAs
sBsAs
s
B
s
A
ss
s
ss
s
sY
34
2
7
10
7
11
)(
3
1
7
10
4
1
7
11
)(
7/10,7/11
134,3
)34()(13
)4()3(13
34)3)(4(
13
12
13
)(
+=⇒





−
+





+
=
==
=−=+
−++=+
++−=+
−
+
+
=
−+
+
=
−+
+
=
−

22. Example 10
For the function Y(s) below, we find y(t) = L-1
{Y(s)} by
completing the square in the denominator and rearranging the
numerator, as follows.
Using Table 6.1, we obtain
( ) ( )
( )
( ) ( ) ( ) 





+−
+





+−
−
=
+−
+−
=
+−
+−
=
++−
−
=
+−
−
=
13
1
2
13
3
4
13
234
13
2124
196
104
106
104
)(
222
222
ss
s
s
s
s
s
ss
s
ss
s
sY
tetety tt
sin2cos4)( 33
+=

23. Example 11: Initial Value Problem (1 of 2)
Consider the initial value problem
Taking the Laplace transform of the differential equation, and
assuming the conditions of Corollary 6.2.2 are met, we have
Letting Y(s) = L{y}, we have
Substituting in the initial conditions, we obtain
Thus
( ) ( ) 60,00,0258 =′==+′−′′ yyyyy
[ ] [ ] 0}{25)0(}{8)0()0(}{2
=+−−′−− yLyysLysyyLs
( ) ( ) 0)0()0(8)(2582
=′−−−+− yyssYss
258
6
)(}{ 2
+−
==
ss
sYyL
( ) 06)(2582
=−+− sYss

24. Completing the square, we obtain
Thus
Using Table 6.2.1, we have
Therefore our solution to the initial value problem is
( ) 9168
6
258
6
)( 22
++−
=
+−
=
ssss
sY
( ) 





+−
=
94
3
2)( 2
s
sY
{ } tesYL t
3sin2)( 41
=−
tety t
3sin2)( 4
=
Example 11: Solution (2 of 2)

25. Example 12: Nonhomogeneous Problem (1 of 2)
Consider the initial value problem
Taking the Laplace transform of the differential equation, and
assuming the conditions of Corollary 6.2.2 are met, we have
Letting Y(s) = L{y}, we have
Substituting in the initial conditions, we obtain
Thus
( ) ( ) 10,20,2sin =′==+′′ yytyy
[ ] )4/(2}{)0()0(}{ 22
+=+′−− syLysyyLs
( ) )4/(2)0()0()(1 22
+=′−−+ sysysYs
)4)(1(
682
)( 22
23
++
+++
=
ss
sss
sY
( ) )4/(212)(1 22
+=−−+ sssYs

26. Using partial fractions,
Then
Solving, we obtain A = 2, B = 5/3, C = 0, and D = -2/3. Thus
Hence
Example 12: Solution (2 of 2)
41)4)(1(
682
)( 2222
23
+
+
+
+
+
=
++
+++
=
s
DCs
s
BAs
ss
sss
sY
( )( ) ( )( )
)4()4()()(
14682
23
2223
DBsCAsDBsCA
sDCssBAssss
+++++++=
+++++=+++
4
3/2
1
3/5
1
2
)( 222
+
−
+
+
+
=
sss
s
sY
tttty 2sin
3
1
sin
3
5
cos2)( −+=