Laplace Transform and Di↵erential Equations
HarishKumar
hkumar@iitd.ac.in
Department of Mathematics,
Indian Institute of Technology,
Delhi
Harish Kumar
Laplace Transform and Di↵erential Equations
September 20, 2020
1 / 55
Definition
Consider a real valued function f (t) on [0, 1). We define Laplace
transform of f (denoted by L[f (t)](s)) as,
Z 1
L[f (t)](s) =
e st f (t)dt,
(1)
0
provided the integral makes sense.
Definition (Exponential Order:)
A function f is said to be of exponential order if there exist constants M
and ↵ such that,
|f (t)|  Me ↵t .
(2)
For example: Any polynomial is of exponential order. Similarly, any
2
bounded function is also of exponential order. However f (t) = e t is not
of exponential order.
Harish Kumar
Laplace Transform and Di↵erential Equations
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Definition (cont.)
Also Consider:
Definition
Piecewise Continuous: A function f is said to be piecewise continuous
on domain D if f is continuous on D except on a countable set S on
which it has jump discontinuity.
Combining both the definition, we have the following result:
Theorem
Suppose f is piecewise continuous on [0, 1) and satisfies (2). Then
L[f ](s) exists for s > ↵.
( Ga )
-
[ stlfctydt.cm/e-stedtdt--Mf?e-CEdtdt
=
I f*e Stf
-
# de
Harish Kumar
)
s
it
e-
s > a.
Lcf] is
)
converges
Laplace Transform and Di↵erential Equations
.
September 20, 2020
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Important Remark
Remark
In the following, we will not explicitly specify that f is exponential order
but we will always assume it, whenever we need to define its Laplace
transform.
-
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Example:
Consider f (t) = e ↵t which is exponential order. Laplace transform of it is
calculated as follows:
Z 1
1
↵t
L[e ](s) =
e st e ↵t dt =
, s > ↵.
s
↵
0
S > L
For ↵ = 0 we have
L[1](s) =
1
.
s
=
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Laplace Transform and Di↵erential Equations
570
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Linearity
Theorem
The operator L is linear i.e.,
-
L[af (t) + bg (t)](s) = aL[f ](s) + bL[g ](s)
where
→
LED
&
LEFT
is
defined
for all s and constants a and b.
-
-
Proof.
L[af (t) + bg (t)](s) =
=a
Z 1
0
Harish Kumar
e st f (t)dt + b
Z 1
Z 1
e st (af (t) + bg (t))dt
0
e st g (t)dt = aL[f ](s) + bL[g ](s).
0
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Linearity (cont.)
Example
Laplace transform of hyperbolic functions cosh(at) and sinh(at):
cosh(at) =
1 at
(e + e at ),
2
sinh(at) =
1
1
(L[e at ] + L[e at ]) =
2
2
✓
1
1
L[sinh(at)] = (L[e at ]oL[e at ]) =
2
2
✓
L[cosh(at)] =
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II.
1
s
a
s
a
Laplace Transform and Di↵erential Equations
e at )
1
s +a
◆
=
1
s +a
◆
=
+
w
1
1 at
(e
2
,
a
17-0
s
s2
2
☒
a
s2
a2
.
September 20, 2020
8 / 55
Linearity (cont.)
Example
Laplace transform of Trigonometric function sin(at) and cos(at):
Define Lc = L[cos(at)] and Ls = L[sin(at)], then using integration by
parts,
Z
-
-
-
-
-
-
Lc = L[cos(at)] =
-
=
=

e
st
s
cos(at)
1
e st cos(at)dt
Z I
a 1 st
1
e
sin(at)dt =
s 0
s
1
-
0
0
soo
a
Ls
s
⑤
=
Similarly,
Ls =
a
Lc
s
Solving these two linear equations for Lc and Ls we get,
T
Lc =
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s
s 2 + a2
,
Ls =
a
s 2 + a2
Laplace Transform and Di↵erential Equations
.
September 20, 2020
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Di↵erentiation of Laplace transform
Theorem
( Skl
Suppose f is of exponential order, then in its region of convergence
F (s) = L[f (t)](s) is di↵erentiable infinitely many times at each point and
-
=
F (n) (s) = ( 1)n L[t n f (t)](s).
""
Proof.
F 0 (s) =
=
d
ds
=
Z 1
0
e st f (t)dt =
Z 1
0
;
f (t)
=
d
(e st )dt =
ds
=
L[tf (t)](s).
=
Then the result follow from induction argument.
Harish Kumar
Laplace Transform and Di↵erential Equations
September 20, 2020
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Integration of Laplace transform
Theorem
If F = L[f ] then,
-
-
L

f (t)
=
t
Z 1
F (s1 )ds1 .
s
Proof.
Z 1
F (s1 )ds1 =
s
-
Z 1 Z 1
s
f (t)dt ds1 =
0
-
=
Z 1
0
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e
s1 t
.
e st
Z 1
f (t)
0

f (t)
f (t)
dt = L
.
t
t
-
=
Z 1
e s1 t ds1 dt
s
.
=
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Laplace Transform of Derivatives of a Function
Theorem
Suppose
f is di↵erentiable for t
#
well defined then,
0 and of exponential order. If L[f 0 ] is
=
I
L[f 0 (t)](s) = sL[f (t)](s)
Lcf ] us
'
Proof.
By definition we have,
=
S
-
-
?
fest
-
f (0).
=
f- lol
F-
LEFT
=
=
L[f 0 (t)](s) =
=
Z 1
f. Gifts ]
e st f 0 (t)dt
0
:
o
-
Moo)
x④¥I=s=
Integrating by parts we get,
⇥
⇤1
L[f 0 ](s) = e sr f (r ) 0 + s
-
-
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€-0
Z 1
0
e st f (t)dt =
sL[f ](s)
④
=
f (0).
=
.
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Laplace Transform of Derivatives of a Function
(cont.)
Corollary
Now repeating this process n times we get,
L[f (n) (t)](s) = s n L[f ]
=
s n 1 f (0)
s n 2 f 0 (0)
···
f (n 1) (0).
O↵ course we have to assume that L[f (n) ] exists.
Harish Kumar
Laplace Transform and Di↵erential Equations
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Laplace Transform of Integration of a Function
Theorem
L
Z t
f (⌧ )d⌧ =
0
L[f ]
s
Proof.
Rt
Define g (t) = 0 f (⌧ )d⌧ , then g (0) = 0 and g is di↵erentiable with
g 0 = f . Furthermore if f satisfies (2), then
-
=
|g (t)| 
-
.
.
Z t
#
0
|f (⌧ )|d⌧  M
Z t
0
e ↵⌧
d⌧ =
==
#
M ↵t
(e
↵
1) 
M ↵t
e .
↵
So, g is also exponential order. So, L[g ] exists. Now,
L[g 0 ] = sL[g ]
g (0),
gives the desired equality.
Harish Kumar
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September 20, 2020
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From these result we will now derive the following properties of Laplace
transform:
LCH
Theorem
-
↵t
L[eO
f ](s) = L[f ](s
-
Fcs )
L
t
fit D= Fcs
-
H
↵)
O
Proof.
L[e ↵t f ](s) =
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Z 1
0
e ↵t e st f (t)dt =
Z 1
0
e (s ↵)t f (t)dt = L[f ](s
Laplace Transform and Di↵erential Equations
↵).
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Laplace Transform of Various Functions
S. No.
1
2
3
4
5
6
7
8
9
10
11
12
f (t)
1
t
t2
tn
a
t (a positive )
e at
cos(at)
sin(at)
cosh(at)
sinh(at)
at
e sin(!t)
e at sin(!t)
-
↳
( wt )
L[f ]
1/s
1/s 2
O
2!
s3
n!
s n+1
(a+1)
s a+1
1
s a
s
s 2 +a2
a
s 2 +a2
s
s 2 a2
a
s 2 a2
s a
(s a)2 +! 2
!
(s a)2 +a2
Table 1: Laplace transform of some functions
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Heaviside function
Consider the Heaviside function at a,
(
0, if t < a,
Ha (t) =
1, if t a.
=
-
-
then
L[Ha (t)] =
-
Z 1
0
ta
e st Ha (t)dt =
Z 1
A
e st dt =
a
as
e
s
-
o
Holt )
YI
Hao CHI L
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Ho
*
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Characteristic function
Similarly, consider the Characteristic function,
8
>
<0, if t < a,
(t)
=
1, if a  t  b,
[a,b]
>
:=
0, if t > b.
x#
f-
{ to
Is
-
.
. .
-
na
.
-
a c
b
Laplace transform:
L[ [a,b] (t)](s) =
=
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Z 1
e st
[a,b] dt =
0
-
Z b
i
a
e st dt =
-
Laplace Transform and Di↵erential Equations
e bs
e as
+
s
s
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Shifted function
f˜(t) = f (t
a)Ha (t) =
(
t.EE
.
f- It
-
!!
0,
f (t
=
if t < a,
if t a,
=
a)
F- ⇐
Hi
Ho )
e
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Shifted function (cont.)
Theorem
Let f (t) be an function which satisfied (2) and F (s) = L[f ](s). Consider
the shifted function
(
0,
if t < a,
˜
f (t) =°
f (t a)Ha (t) =
f (t a) if t a,
has Laplace transform L[f˜] = e as F (s), i.e.
L[f (t
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a)Ha (t)] = e as F (s)
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Shifted function (cont.)
Proof.
L[f (t
a)Ha (t)] =
Z 1
0
-
=D
e as f (t
a)Ha (t)dt =
Z 1
a
o
e st f (t
-
-
a)dt.
=
Now introduce variable t1 = t a and change the variable in the integral.
We get,
Z 1
Z 1
s(t1 +a)
as
L[f (t a)Ha (t)] =
e→ f (t1 )dt1 = e
e st1 f (t1 )dt = e as F (s).
=
,
0
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=
Laplace Transform and Di↵erential Equations
0
#
September 20, 2020
=
22 / 55
Shifted function (cont.)
This theorem is very important as this allows us to calculate Laplace
transform of functions with jump discontinuities. Furthermore, from this
we can also deduce that
L[f (t)Ha (t)] = e as L[f (t + a)],
which is more useful form of the above result.
t
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tf hit
✓
I
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Example:
Find the Laplace transform of the following function:
8
>
<2, if 0 < t < 1,
f (t) = 12 t 2 , if 1 < t < ⇡/2,
>
:
cos(t), if t > ⇡/2.
Using the Heaviside function we will rewrite f (t) as,
XCTIK 'D
Xa i
1
#
H1 (t)) + t 2 (H1 (t) H⇡/2 (t)) + H⇡/2 (t) cos(t)
2
Xco , D
f (t) = 2(1
=
¥
.÷a
,
=
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Now using the Linearity of Laplace transform, we get,
L[f ] = 2L [1
=
Now,
⇤
1 ⇥
H1 (t)] + L t 2 H1 (t)
2
⇤
⇥
⇤
1 ⇥ 2
L t H⇡/2 (t) + L H⇡/2 cos(t)
2
=\,
2L [1
=
H1 (t)] =
-
and
⇤
1 ⇥ 2
L t H1 (t) =e TL
2
-
Similarly,
Now,

I
e s)
2(1
s
ta
1
(t + 1)2 = e s
2
✓
1
1
1
+ 2+
s2
s
2s
Ox
◆
.
I ①← Ice
⇤
1 ⇥
L H⇡/2 t 2 = e ⇡s/2
2
O
✓
1
⇡
⇡2
+
+
s3
2s 2
8s
L[H⇡/2 (t) cos(t)] = e ⇡s/2 L[cos(t + ⇡/2)] =
-
-
◆
.
e ⇡s/2
=
1
.
1 + s2
'
-
ans
-
=
Combing all the terms we get the Laplace transform of f .
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Dirac Delta ”Function”I
=
Consider the function,
-
fk (t a) =
-0
÷
.
it
÷
.
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-
a'
-
-
f:
(
I
1
k
0,
if a  t  a + k,
otherwise.
P%
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Dirac Delta ”Function” (cont.)
Note that,
Ik =
Z 1
0
fk (t
a)dt = 1,
-
We define Dirac delta function (t
-
(t
(t
and
a) as follows,
a) = lim fk (t
a) =
Z 1
a),
k!0
=
=
which gives,
8k
(
1, if t = a,
0, otherwise.
(t
a)dt = 1.
g¥
0
-
-
imf ful
t
Hdt
-
a)
f- Mdt
le
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Dirac Delta ”Function” (cont.)
One of the important property of Dirac delta function is,
H ful t af
Z 1
II
g (t) (t a)dt = g (a)
-
=
=
Ig
0
gca ,
,
for any continuous function g . This follows from
→
Z 1
0
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g (t) (t
a)dt = lim
k!0
Z 1
fk (t
a)g (t)dt.
0
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L. T. of Dirac ”Function”
f. [ felt HI
-
§
-
stfu It a) dt
a)] = e as
=
Jagathy
-
st
at
a
Note that
L[fk (t
=
-
e
=
1
e ks
ks
are
#€
So, we define,
L[ (t
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a)] = lim L[fk (t
k!0
a)] = e as .
Laplace Transform and Di↵erential Equations
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Consider a IVP with constant coefficients,
DO
x 00 (t) + ax 0 (t) + bx(t) = g (t),
x(0) = x0 .
x 0 (0) = x1 .
Let X (s) = L[x(t)](s), then
s 2 X (s)
sx0
x1 + a(sX (s)
x0 ) + bX (s) = G (s).
Solving it for X , we get,
A
X (s) =
w¥d_
sx0 + x1 + ax0 G (s)
.
s 2 + as + b
→
1
f- [✗ I
¥
=
-
if
Solution x(t) is,
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✗ is)
s⇐÷
¥#
gkfH1
x(t) = L
=
Lexx]
=
1
[X (s)] = L
1

sx0 + x1 + Aax0 G (s)
.
s 2 + as + b
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Inverse L. T.: Well Defined?
Lcf ]
=
Icg]
Important Question:
L
is
well defined
if true , f
f- I G
Let f and g are two functions such that there Laplace transform is same.
What can be wait about these functions ? are they equal. This question
is answered in the following result:
-
.
Theorem
Let f , g are two piecewise continuous on [0, 1) and satisfy (2). If
L[f ] = L[g ] on their common region of convergence, then f (t) = g (t),
for all t 0 except at countable number of points. Furthermore, if f and
g are continuous then f (t) = g (t) for all t 0.
assume
e- Stf its adt
we
Sg e stay # It
=
§
{
(
fit
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) e-
*'
-
=
dt
-
o
⇐
-
f- 9
, mummy
Laplace Transform and Di↵erential Equations
f. and gave
continuous
September 20, 2020
)
31 / 55
Definition
Let F (s) is a given function. If there exists a function f (t) such that
L[f ](s) exists and F (s) = L[f ](s), we say Fofis inverse Laplace transform
of f and denote, f (t) = L 1 [F (s)](t) or simply L 1 [F ] = f . .
¥
#
Example:
Itt
←
f-
Font
As L[Ha (t)] = e as /s, implies L 1 [e as /s] = Ha (t).
y
Linearity of Inverse L.T.
-
sa
-
-
,
of discontinuity
Note that as L is linear operator, it follows immediately that L 1 is also
linear i.e.
L 1 [aF (s) + bG (s)] = aL 1 [F (s)] + bL 1 [G ].
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Also note the following results:
I
L 1 [F (s
I
L
1

↵)] = e ↵t f (t)
F (s)
=
s
Z ⌧
f (⌧ )d⌧
0
I
L 1 [e ↵s F (s)] = H↵ (t)f (t
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Laplace Transform and Di↵erential Equations
a)
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Some useful Inverse Laplace Transform:
S. No.
1
2
3
L 1 [F ]
1
1
↵t cos(↵t))
(s 2 +↵2 )2✓ 2↵2 (sin(↵t)
→
s
t
sin(↵t)
2
2
2
(s +↵ )
2↵
F (s)
¥-00
←
(
-
-
s2
(s 2 +↵2 )2
1
2↵ (sin(↵t) + ↵t cos(↵t))
Table 2: Inverse Laplace transform of some functions
partial
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traction
Laplace Transform and Di↵erential Equations
Exercise
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When solving for linear ODEs we will encounter several functions of the
form P(s)/Q(s) where P and Q are polynomial such that degree of P is
less then degree of Q. We need to take their inverse Laplace transform.
This can be done via the following results:
O
Theorem
Let P(s) and Q(s) be polynomial of degree n and m respectively such
that n < m. If Q has m simple roots 1 , · · · m , then
L 1
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
m
X P( i )
P(s)
=
e it
0( )
Q(s)
Q
i
1
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Convolution
Definition
The convolution of f (t) and g (t), denoted by f ⇤ g , is the function define
as,
Z
-
-
-
-
t
f ⇤ g (t) =
f (t ✓)g (✓)d✓
O=
0
I
Some properties of convolution are given here:
1.
f ⇤g =g ⇤f
-
-
2.
f ⇤ (g1 + g2 ) = f ⇤ g1 + f ⇤ g2
-
-
-
3.
(f
⇤ g ) ⇤ h = f ⇤ (g ⇤ h)
←
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Convolution (cont.)
4.
f ⇤0=0⇤f =0
-
-
-
5.
1 ⇤ f = f ⇤ 1 6= f
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Convolution and L.T.
Theorem
Let f , g be piecewise continuous function of exponential order and
F (s) = L[f ] and G (s) = L[g ], then
=
=
L[f ⇤ g (t)](s) = F (s) · G (s)
-
and this implies,
-
I
L 1 [F (s)G (s)] = (f ⇤ g )(t).
=
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If
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Convolution and L.T. (cont.)
Proof:
F (s)G (s) =
Z 1
f ( )e
-
0
As
s
d
Z 1
0
f (⌧ )e
9
-
s⌧
d⌧
and ⌧ are independent variables, we can rewrite F (s)G (s) as,
Z 1 Z 1
F (s)G (s) =
f ( )e s( +⌧ ) d g (⌧ )d⌧.
0
0
We fix ⌧ and introduce the change of variable for internal integral as
t = + ⌧ , then dt = d and we get,
Z 1 Z 1
F (s)G (s) =
f (t ⌧ )e st dt g (⌧ )d⌧
0
=
Z 1 Z 1
⌧
f (t
⌧ )g (⌧ )e st dt d⌧
0
⌧
a
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Convolution and L.T. (cont.)
We reverse the order of integration to get,
-
t
÷÷t
z
F (s)G (s) =
=
Z 1 Z t
0
-
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Z 1 Z t
0
f (t
⌧ )g (⌧ )e st d⌧ dt
0
A.
0
f (t
-
⌧ )g (⌧ )d⌧ e st dt = L[f ⇤ g ]
-
Laplace Transform and Di↵erential Equations
-
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Consider a IVP with constant coefficients,
x 00 (t) + ax 0 (t) + bx(t) = g (t),
x(0) = x0 .
x 0 (0) = x1 .
Let X (s) = L[x(t)](s), then
s 2 X (s)
sx0
x1 + a(sX (s)
x0 ) + bX (s) = G (s).
Solving it for X , we get,
X (s) =
sx0 + x1 + ax0 G (s)
.
s 2 + as + b
Solution x(t) is,
x(t) = L
1
[X (s)] = L
-
Harish Kumar
1

I
sx0 + x1 + ax0 G (s)
.
s 2 + as + b
Laplace Transform and Di↵erential Equations
September 20, 2020
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Example: Consider the IVP,
x 00 (t) + x(t) = g (t),
-
-
We get,
-5¥
X (s) =
So,
x(t) = L 1
-

x(0) = 0,
x 0 (0) = k.
-
k + G (s)
.
s2 + 1
I


→
8M
k
G (s)
G (s)
+L 1 2
= k sin(t) + L 1 2
.
s2 + 1
sI
+1
s +1
-
.
*
-
-
Inverse Laplace transform of F (s) = 1/(s 2 + 1) is sin(t). Using the
convolution property,

Z t
G (s)
1
GH Asin H
L
=
sin(t ✓)g (✓)d✓,
s2 + 1
0
#
which gives,
x(t) = k sin(t) +
Z t
sin(t
¥4
,
✓)g (✓)d✓.
0
Harish Kumar
Laplace Transform and Di↵erential Equations
September 20, 2020
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Higher Order ODEs with Constant Coefficients
x (n) +a1 x (n 1) · · ·+an 1 x 0 +an x = g (t),
x(0) = x0 , · · · x (n 1) (0)xn 1 .
Taking Laplace transform of the ODE we get,
s n X (s)
s n 1 x0
s n 2 x1
···
xn 1 + · · · an X (s) = G (s),
which can be written as,
P(s)X (s)
Q(s) = G (s),
:-#
where
P(s) = s n + a1 s n 1 + · · · + an 1 s + an
=
and
=
Q(s) = s n 1 x0 + s n 2 x1 + · · · + an 2 (sx0 + x1 ) + an 1 x0 .
Harish Kumar
Laplace Transform and Di↵erential Equations
September 20, 2020
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Solving for X (s) gives,
X (s) =
G (s) + Q(s)
,
P(s)
-
-
9¥
.
I
.
and so the solution will be,
x(t) = L
-
1
[X (s)] = L
-
1
 →

Q(s)
G (s)
+L 1
.
P(s)
P(s)
f-
@tttpartialfratvntE.a
.
.
Harish Kumar
Laplace Transform and Di↵erential Equations
September 20, 2020
45 / 55
Linear ODE with variable coefficients:
Note that,
d
L[tx 0 (t)] =
[sX x0 ] =
ds
Similarly,
-
÷÷÷
X
s
dX
.
ds
dX
.
ds
In fact, if the coefficient are of the form at + b, taking Laplace transform
results in first order ODE of X (s). Solving this ODE and taking inverse
Laplace transform results in the solution.
L(tx 00 ) =
2sX
s2
Consider the IVP,
x + tx = 0,
±
00
x(0) = 0,
x 0 (0) = b.
Taking Laplace transform we get,
s 2 X (s)
Harish Kumar
b + L[tx(t)] = s 2 X
b
Laplace Transform and Di↵erential Equations
X 0 (s) = 0
September 20, 2020
46 / 55
So, we get,
X 0 (s)
s 2 X (s) =
b.
T
This a linear ODE which can be easily solved by method of integrating
factor. Then taking the inverse Laplace transform results in the
solution.
Harish Kumar
Laplace Transform and Di↵erential Equations
September 20, 2020
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Generalized Solutions
I In all the theory discussed so far for ODE we have always assumed
that coefficient and the forcing terms are atleast continuous.
I However, in several practical problems we have to consider the
forcing which are not continuous or not even function (e.g. Dirac
delta). These equations are still need to be solved.
-
-
I The method of Laplace transform is powerful enough to handle such
situations.
✓
I However, the solution we get may not be of desired smoothness. For
example solution of second order may not be C 2 . These solutions
are called generalized solutions.
-
Here we will try to find ”Generalized Solutions”
✓
v
e
if
¥bIde
,
Ho
Harish Kumar
CK
discontinuous
emt
Laplace Transform and Di↵erential Equations
September 20, 2020
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Generalized Solutions (cont.)
Example 1:
Consider the IVP,
T.EE
x 00 + 3x 0 + 2x = H1 (t)
H2 (t),
¥7
x(0) = 0,
x 0 (0) = 0.
.
Taking Laplace transform
s 2 X + 3sX + 2X =
-
which gives,
1
(e s e 2s )
s =
"
E e÷÷
X (s) =
e s e 2s
.
s(s 2 + 3s + 2)
Now using partial fractions,
1
1
=
s(s 2 + 3s + 2)
2s
Harish Kumar
1
1
+
.
s + 1 2(s + 2)
Laplace Transform and Di↵erential Equations
September 20, 2020
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Generalized Solutions (cont.)
So,
f (t) = L
①
1

1
s(s 2 + 3s + 2)
=
-
Using the shift theorem,
⇥
⇤
x = L 1 [ e s L[f ] e 2s L[f ] = f (t
=
8
>
<0,
1
>2
:
e
+ 12 e 2(t 1)
(t 1)
+ 12 e 2(t 1) + e (t 2)
1
1
e t + e 2t .
2
2
a
1)H1 (t)
f (t
1
2(t 2)
2e
if
if
if
(t 1)
e
2)H2 (t)
0 < t < 1, +
.
1 < t < 2,
t > 2.
t &
.
=
Note that the solution is not C 2 (Check !).
Te
Harish Kumar
Laplace Transform and Di↵erential Equations
September 20, 2020
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Generalized Solutions (cont.)
Example 2:✓
Instead of Heaviside function on right if we take (t 1) then we
get,
x 00 + 3x 0 + 2x = (t 1),
x(0) = 0, x 0 (0) = 0.
ES
taking Laplace transform we get,
X =
e s
=e s
2
s + 3s + 2
Inverse Laplace transform gives,
(
0,
1
x(t) = L [X ] =
e (t 1)
=
✓
1
s +1
e
2(t 1)
1
s +2
.
=
=
,
Note that the solution x(t) is not even C 1 .
=
Harish Kumar
◆
Laplace Transform and Di↵erential Equations
if
if
0<t<1
t > 1.
¥:
S
s
September 20, 2020
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System of Equations
Consider a system of ODE,
¥8,8
x10 = a11 x1 + a12 x2 + g1 (t), x20 = a21 x1 + a22 x2 + g2 (t).
)
-
Let us define X1 = L[x1 ] and X2 = L[x2 ], then assume G1 = L[g1 ] and
G2 = L[g2 ] exists, we get,
=
a-
=
(a11
=
a
s)X1 + a12 X2 =
-
-
a21 X1 + (a22
[D
-
s)X2 =
-
x1 (0)
G1 (s),
x2 (0)
G2 (s).
I
=
which can be written as,
~ =
sI )X
I
(A
-
~ (0)
X
-
G~ .
-
This is a linear system for X1 and X2 . Solving this we get expression for
X1 and X2 is the form of s. Then inverse Laplace transform gives the
solution x1 (t) and x2 (t).
..
I
-
Harish Kumar
-
Laplace Transform and Di↵erential Equations
September 20, 2020
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System of Equations (cont.)
Example 1: Mixing problem involving two tanks
Consider the System:
8
2
x1 +
x2 + 6.
100
100
x10 (t) =
-
Similarly for tank T2 we have,
x20 (t) =
8
x1
100
8
x2 .
100
-
with initial conditions x1 (0) = 0 and x2 (0) = 150. Taking Laplace
transform of both the equations we get,
( 0.08
s)X1 + 0.02X2 =
(0.08)X1 + ( 0.08
Harish Kumar
s)X2 =
6
,
s
150.
Laplace Transform and Di↵erential Equations
September 20, 2020
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System of Equations (cont.)
Solving for X1 and X2 we get,
100
62.5
s
s + 0.12
100
125
X2 (s) =
+
s
s + 0.12
X1 (s) =
-
-
375.5
,
s + 0.04
75
.
s + 0.04
Taking inverse Laplace transformation we get,
I
c
x1 (t) = 100
62.5e 0.12t
x2 (t) = 100 + 125e 0.12t
Harish Kumar
37.5e 0.04t ,
75e 0.04t .
Laplace Transform and Di↵erential Equations
September 20, 2020
54 / 55