DIFFERENTIAL EQUATIONS

Michelle T. Panganduyon, PhD

November 2019

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

introduced by Pierre Simmon Marquis De Laplace

(1749-1827), a French Mathematician.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

introduced by Pierre Simmon Marquis De Laplace

(1749-1827), a French Mathematician.
powerful technique; it replaces operations of calculus by
operations of Algebra.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

introduced by Pierre Simmon Marquis De Laplace

(1749-1827), a French Mathematician.
powerful technique; it replaces operations of calculus by
operations of Algebra.
an Ordinary (or) Partial Differential Equation together with
Initial conditions is reduced to a problem of solving an
Algebraic Equation.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Recall:

Standard method:
(i) Consider a given ODE in the function F(t);

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Recall:

Standard method:
(i) Consider a given ODE in the function F(t);
(ii) find complementary function yc and particular solution yp ;

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Recall:

Standard method:
(i) Consider a given ODE in the function F(t);
(ii) find complementary function yc and particular solution yp ;
(iii) general solution = yc + yp ;

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Recall:

Standard method:
(i) Consider a given ODE in the function F(t);
(ii) find complementary function yc and particular solution yp ;
(iii) general solution = yc + yp ;
(iv) apply initial conditions to determine unknown constants;

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Recall:

Standard method:
(i) Consider a given ODE in the function F(t);
(ii) find complementary function yc and particular solution yp ;
(iii) general solution = yc + yp ;
(iv) apply initial conditions to determine unknown constants;
(v) deduce solution.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace transform method:

(i) Consider a given ODE in the function F(t);

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace transform method:

(i) Consider a given ODE in the function F(t);
(ii) take Laplace transform (using tables), incorporating the
initial conditions, to give an algebraic equation in the
transform f (s);

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace transform method:

(i) Consider a given ODE in the function F(t);
(ii) take Laplace transform (using tables), incorporating the
initial conditions, to give an algebraic equation in the
transform f (s);
(iii) solve for f (s);

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace transform method:

(i) Consider a given ODE in the function F(t);
(ii) take Laplace transform (using tables), incorporating the
initial conditions, to give an algebraic equation in the
transform f (s);
(iii) solve for f (s);
(iv) rearrange this expression for f (s) to appropriate form for
inversion;

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace transform method:

(i) Consider a given ODE in the function F(t);
(ii) take Laplace transform (using tables), incorporating the
initial conditions, to give an algebraic equation in the
transform f (s);
(iii) solve for f (s);
(iv) rearrange this expression for f (s) to appropriate form for
inversion;
(v) take inverse Laplace transform (using tables) to give
solution.

M.T. Panganduyon Differential Equations

Laplace Transform

defn. (Laplace Transform)

Let F(t) be a given function defined for all t ≥ 0. The Laplace
Transformation of F(t) is defined as
Z ∞
L{F(t)} = e−st F(t)dt = f (s).
0

M.T. Panganduyon Differential Equations

Laplace Transform

defn. (Laplace Transform)

Let F(t) be a given function defined for all t ≥ 0. The Laplace
Transformation of F(t) is defined as
Z ∞
L{F(t)} = e−st F(t)dt = f (s).
0

L is called the Laplace Transform Operator.

M.T. Panganduyon Differential Equations

Laplace Transform

defn. (Laplace Transform)

Let F(t) be a given function defined for all t ≥ 0. The Laplace
Transformation of F(t) is defined as
Z ∞
L{F(t)} = e−st F(t)dt = f (s).
0

L is called the Laplace Transform Operator.

F(t) – determining function, depends on t.

M.T. Panganduyon Differential Equations

Laplace Transform

defn. (Laplace Transform)

Let F(t) be a given function defined for all t ≥ 0. The Laplace
Transformation of F(t) is defined as
Z ∞
L{F(t)} = e−st F(t)dt = f (s).
0

L is called the Laplace Transform Operator.

F(t) – determining function, depends on t.
f (s) – generating function, depends on s.

M.T. Panganduyon Differential Equations

Laplace Transform

defn. (Laplace Transform)

Let F(t) be a given function defined for all t ≥ 0. The Laplace
Transformation of F(t) is defined as
Z ∞
L{F(t)} = e−st F(t)dt = f (s).
0

L is called the Laplace Transform Operator.

F(t) – determining function, depends on t.
f (s) – generating function, depends on s.

M.T. Panganduyon Differential Equations

Laplace Transform

defn. (Laplace Transform)

Let F(t) be a given function defined for all t ≥ 0. The Laplace
Transformation of F(t) is defined as
Z ∞
L{F(t)} = e−st F(t)dt = f (s).
0

L is called the Laplace Transform Operator.

F(t) – determining function, depends on t.
f (s) – generating function, depends on s.

Question will be in t and Answer will be in s.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace Transformation is useful since

Particular solution is obtained without first determining
the general solution.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace Transformation is useful since

Particular solution is obtained without first determining
the general solution.
Nonhomogeneous Equations are solved without obtaining
the complementary function

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace Transformation is useful since

Particular solution is obtained without first determining
the general solution.
Nonhomogeneous Equations are solved without obtaining
the complementary function
The Laplace Transformation is a very powerful technique,
that it replaces operations of calculus by operations of
algebra.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace Transformation is useful since

Particular solution is obtained without first determining
the general solution.
Nonhomogeneous Equations are solved without obtaining
the complementary function
The Laplace Transformation is a very powerful technique,
that it replaces operations of calculus by operations of
algebra.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Laplace Transformation is useful since

Particular solution is obtained without first determining
the general solution.
Nonhomogeneous Equations are solved without obtaining
the complementary function
The Laplace Transformation is a very powerful technique,
that it replaces operations of calculus by operations of
algebra. For e.g. With the application of L.T to an Initial
value problem, consisting of an Ordinary( or Partial )
differential equation (O.D.E) together with Initial
conditions is reduced to a problem of solving an algebraic
equation (with any given Initial conditions automatically
taken care of).

M.T. Panganduyon Differential Equations

Laplace Transforms

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Z ∞
L{F(t)} = e−st F(t)dt = f (s)
0
The symbol ‘L’ denotes the L.T operator, when it operated on a
function F(t), it transforms into a function f (s) of complex
variable s.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Z ∞
L{F(t)} = e−st F(t)dt = f (s)
0
The symbol ‘L’ denotes the L.T operator, when it operated on a
function F(t), it transforms into a function f (s) of complex
variable s.

We say the operator transforms the function F(t) in the ‘t’

domain (usually called time domain) into the function f (s) in the
‘s’ domain (usually called complex frequency domain or simply
the frequency domain).

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Because the upper limit in the integral is infinite, the domain of

integration is infinite.

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Because the upper limit in the integral is infinite, the domain of

integration is infinite. Thus the integral is an example of an
Improper Integral:
Z ∞
L{F(t)} = e−st F(t)dt
0

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Because the upper limit in the integral is infinite, the domain of

integration is infinite. Thus the integral is an example of an
Improper Integral:
Z ∞ Z T
L{F(t)} = e−st F(t)dt = lim e−st F(t)dt
0 T→∞ 0

M.T. Panganduyon Differential Equations

Laplace Transforms
Introduction

Because the upper limit in the integral is infinite, the domain of

integration is infinite. Thus the integral is an example of an
Improper Integral:
Z ∞ Z T
L{F(t)} = e−st F(t)dt = lim e−st F(t)dt
0 T→∞ 0

R ∞ Laplace Transformation of F(t) is said to exist if the integral

The
0
e−st F(t)dt converges for some values of s, otherwise it does
not exist.

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Elementary Functions

(Transforms of certain exponential, trigonometric functions and

of polynomials)
k
1 L{k} = , k any constant
s

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Elementary Functions

(Transforms of certain exponential, trigonometric functions and

of polynomials)
k
1 L{k} = , k any constant
s
1
2 L{e } =
kt
s−k

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Elementary Functions

(Transforms of certain exponential, trigonometric functions and

of polynomials)
k
1 L{k} = , k any constant
s
1
2 L{e } =
kt
s−k
n!
3 L{tn } = n+1 , s > 0, n a positive integer.
s

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Elementary Functions

(Transforms of certain exponential, trigonometric functions and

of polynomials)
k
1 L{k} = , k any constant
s
1
2 L{e } =
kt
s−k
n!
3 L{tn } = n+1 , s > 0, n a positive integer.
s
k
4 L{sin kt} = 2 ,s>0
s + k2

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Elementary Functions

(Transforms of certain exponential, trigonometric functions and

of polynomials)
k
1 L{k} = , k any constant
s
1
2 L{e } =
kt
s−k
n!
3 L{tn } = n+1 , s > 0, n a positive integer.
s
k
4 L{sin kt} = 2 ,s>0
s + k2
s
5 L{cos kt} = 2 ,s>0
s + k2

M.T. Panganduyon Differential Equations

Laplace Transforms

Remark:
. The Laplace transform of F(t) will exist even if the object
function F(t) is discontinuous, provided that the integral in
the definition of L{F(t)} exists.

M.T. Panganduyon Differential Equations

Laplace Transforms

Remark:
. The Laplace transform of F(t) will exist even if the object
function F(t) is discontinuous, provided that the integral in
the definition of L{F(t)} exists.

M.T. Panganduyon Differential Equations

Laplace Transforms

Remark:
. The Laplace transform of F(t) will exist even if the object
function F(t) is discontinuous, provided that the integral in
the definition of L{F(t)} exists.

Example: Find the Laplace transform of H(t), where

H(t) = t, 0 < t < 4,

= 5, t > 4.

M.T. Panganduyon Differential Equations

Laplace Transforms

Remark:
. The Laplace transform of F(t) will exist even if the object
function F(t) is discontinuous, provided that the integral in
the definition of L{F(t)} exists.

Example: Find the Laplace transform of H(t), where

H(t) = t, 0 < t < 4,

= 5, t > 4.

Answer:
1 e−4s e−4s
L{H(t)} = + − 2
s2 s s

M.T. Panganduyon Differential Equations

Laplace Transforms

Question: Can we obtain the Laplace transforms of ALL

functions F(t)?

M.T. Panganduyon Differential Equations

Laplace Transforms

Question: Can we obtain the Laplace transforms of ALL

functions F(t)?
Answer: NO.

M.T. Panganduyon Differential Equations

Laplace Transforms

Question: Can we obtain the Laplace transforms of ALL

functions F(t)?
Answer: NO.
No, but it can be shown that Laplace transforms exist (for some
s) for all “reasonable” functions F(t).

M.T. Panganduyon Differential Equations

Laplace Transforms

Question: Can we obtain the Laplace transforms of ALL

functions F(t)?
Answer: NO.
No, but it can be shown that Laplace transforms exist (for some
s) for all “reasonable” functions F(t).
. The functions must be (at least) piecewise continuous,
and not go to infinity too rapidly as t → ∞.

M.T. Panganduyon Differential Equations

Laplace Transforms

Question: Can we obtain the Laplace transforms of ALL

functions F(t)?
Answer: NO.
No, but it can be shown that Laplace transforms exist (for some
s) for all “reasonable” functions F(t).
. The functions must be (at least) piecewise continuous,
and not go to infinity too rapidly as t → ∞.

♥ READING ASSIGNMENT (Functions of Class A) ♥

M.T. Panganduyon Differential Equations

Laplace Transforms

Question: Can we obtain the Laplace transforms of ALL

functions F(t)?
Answer: NO.
No, but it can be shown that Laplace transforms exist (for some
s) for all “reasonable” functions F(t).
. The functions must be (at least) piecewise continuous,
and not go to infinity too rapidly as t → ∞.

♥ READING ASSIGNMENT (Functions of Class A) ♥

Sectionally Continuous Functions

M.T. Panganduyon Differential Equations

Laplace Transforms

Question: Can we obtain the Laplace transforms of ALL

functions F(t)?
Answer: NO.
No, but it can be shown that Laplace transforms exist (for some
s) for all “reasonable” functions F(t).
. The functions must be (at least) piecewise continuous,
and not go to infinity too rapidly as t → ∞.

♥ READING ASSIGNMENT (Functions of Class A) ♥

Sectionally Continuous Functions
Functions of Exponential Order

M.T. Panganduyon Differential Equations

Laplace Transforms
Functions of Class A

We use the term “a function of class A” for any function that is:
(a) Sectionally continuous over every finite interval in the
range t ≥ 0

M.T. Panganduyon Differential Equations

Laplace Transforms
Functions of Class A

We use the term “a function of class A” for any function that is:
(a) Sectionally continuous over every finite interval in the
range t ≥ 0
(b) Of exponential order as t → ∞

M.T. Panganduyon Differential Equations

Laplace Transforms
Functions of Class A

We use the term “a function of class A” for any function that is:
(a) Sectionally continuous over every finite interval in the
range t ≥ 0
(b) Of exponential order as t → ∞

Theorem
If F(t) is a function of class A, L{F(t)} exists.

M.T. Panganduyon Differential Equations

Laplace Transforms
Functions of Class A

We use the term “a function of class A” for any function that is:
(a) Sectionally continuous over every finite interval in the
range t ≥ 0
(b) Of exponential order as t → ∞

Theorem
If F(t) is a function of class A, L{F(t)} exists.

Theorem
If F(t) is a function of class A and if L{F(t)} = f (s),

lim f (s) = 0.
s→∞

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(a) Linearity: If α and β are constants then

L{αF(t) + βG(t)} = αL{F(t)} + βL{G(t)}.

Example: Find L{2 + 5e3t }.

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(a) Linearity: If α and β are constants then

L{αF(t) + βG(t)} = αL{F(t)} + βL{G(t)}.

Example: Find L{2 + 5e3t }.

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(a) Linearity: If α and β are constants then

L{αF(t) + βG(t)} = αL{F(t)} + βL{G(t)}.

Example: Find L{2 + 5e3t }.

Answer:
2 5
L{2 + 5e3t } = + , Re(s) > 3.
s s−3

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms
(b) First Shifting Property or First Translation Property:
If L{F(t)} = f (s), for Re(s) > b, then L{ekt F(t)} = f (s − k),
provided Re(s) > b + Re(k).

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms
(b) First Shifting Property or First Translation Property:
If L{F(t)} = f (s), for Re(s) > b, then L{ekt F(t)} = f (s − k),
provided Re(s) > b + Re(k).

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms
(b) First Shifting Property or First Translation Property:
If L{F(t)} = f (s), for Re(s) > b, then L{ekt F(t)} = f (s − k),
provided Re(s) > b + Re(k).
This property may also be expressed as

L{ekt F(t)} = f (s)s→s−k .

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms
(b) First Shifting Property or First Translation Property:
If L{F(t)} = f (s), for Re(s) > b, then L{ekt F(t)} = f (s − k),
provided Re(s) > b + Re(k).
This property may also be expressed as

L{ekt F(t)} = f (s)s→s−k .

Multiplication by ekt therefore produces a shift in the

variable, but leaves the functional form of the transform
unaltered.

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms
(b) First Shifting Property or First Translation Property:
If L{F(t)} = f (s), for Re(s) > b, then L{ekt F(t)} = f (s − k),
provided Re(s) > b + Re(k).
This property may also be expressed as

L{ekt F(t)} = f (s)s→s−k .

Multiplication by ekt therefore produces a shift in the

variable, but leaves the functional form of the transform
unaltered.
Example: Find L{te2t }.

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms
(b) First Shifting Property or First Translation Property:
If L{F(t)} = f (s), for Re(s) > b, then L{ekt F(t)} = f (s − k),
provided Re(s) > b + Re(k).
This property may also be expressed as

L{ekt F(t)} = f (s)s→s−k .

Multiplication by ekt therefore produces a shift in the

variable, but leaves the functional form of the transform
unaltered.
Example: Find L{te2t }.
Answer:

1  1
L{te } = 2 
2t
= , provided Re(s) > 0 + 2 = 2.
s s→s−2 (s − 2)2

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(c) Change of Scale Property: If L{F(t)} = f (s), then

1 s
 
L{F(kt)} = f .
k k
Example:

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(c) Change of Scale Property: If L{F(t)} = f (s), then

1 s
 
L{F(kt)} = f .
k k
Example:
1 Find L{cos 4t}.

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(c) Change of Scale Property: If L{F(t)} = f (s), then

1 s
 
L{F(kt)} = f .
k k
Example:
1 Find L{cos 4t}.
sin t 1 sin 3t
     
2 Given that L = arctan , find L .
t s t

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(c) Change of Scale Property: If L{F(t)} = f (s), then

1 s
 
L{F(kt)} = f .
k k
Example:
1 Find L{cos 4t}.
sin t 1 sin 3t
     
2 Given that L = arctan , find L .
t s t

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(c) Change of Scale Property: If L{F(t)} = f (s), then

1 s
 
L{F(kt)} = f .
k k
Example:
1 Find L{cos 4t}.
sin t 1 sin 3t
     
2 Given that L = arctan , find L .
t s t
Answer:
s
1 L{cos 4t} = , s>0
s2 + 16

M.T. Panganduyon Differential Equations

Laplace Transforms
Properties of Laplace Transforms

(c) Change of Scale Property: If L{F(t)} = f (s), then

1 s
 
L{F(kt)} = f .
k k
Example:
1 Find L{cos 4t}.
sin t 1 sin 3t
     
2 Given that L = arctan , find L .
t s t
Answer:
s
1 L{cos 4t} = 2 , s>0
s + 16  
sin 3t 3
 
2 L = arctan .
t s

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Any function of class A has a Laplace transform, but the

derivative of such a function may or may not be of class A.

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Any function of class A has a Laplace transform, but the

derivative of such a function may or may not be of class A.

Theorem
If F(t) is continuous for t ≥ 0 and is also of exponential order as
t → ∞, and F0 (t) is of class A, then

L{F0 (t)} = sL{F(t)} − F(0).

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Any function of class A has a Laplace transform, but the

derivative of such a function may or may not be of class A.

Theorem
If F(t) is continuous for t ≥ 0 and is also of exponential order as
t → ∞, and F0 (t) is of class A, then

L{F0 (t)} = sL{F(t)} − F(0).

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Any function of class A has a Laplace transform, but the

derivative of such a function may or may not be of class A.

Theorem
If F(t) is continuous for t ≥ 0 and is also of exponential order as
t → ∞, and F0 (t) is of class A, then

L{F0 (t)} = sL{F(t)} − F(0).

Note: The integral of a sectionally continuous function is

continuous.

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

From the theorem, we obtain, if F, F0 , F00 are suitably restricted,

L{F00 (t)} = sL{F0 (t)} − F0 (0),

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

From the theorem, we obtain, if F, F0 , F00 are suitably restricted,

L{F00 (t)} = sL{F0 (t)} − F0 (0),

or
L{F00 (t)} = s2 f (s) − sF(0) − F0 (0),

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

From the theorem, we obtain, if F, F0 , F00 are suitably restricted,

L{F00 (t)} = sL{F0 (t)} − F0 (0),

or
L{F00 (t)} = s2 f (s) − sF(0) − F0 (0),

and the process can be repeated as many times as we wish.

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives
Theorem
If F(t), F0 (t), . . . , F(n−1) (t) are continuous for t ≥ 0 and of
exponential order as t → ∞, and F(n) (t) is of class A, then from

L{F(t)} = f (s)

it follows that
n−1
X
L{F (t)} = s f (s) −
(n) n
sn−1−k F(k) (0).
k=0

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives
Theorem
If F(t), F0 (t), . . . , F(n−1) (t) are continuous for t ≥ 0 and of
exponential order as t → ∞, and F(n) (t) is of class A, then from

L{F(t)} = f (s)

it follows that
n−1
X
L{F (t)} = s f (s) −
(n) n
sn−1−k F(k) (0).
k=0

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives
Theorem
If F(t), F0 (t), . . . , F(n−1) (t) are continuous for t ≥ 0 and of
exponential order as t → ∞, and F(n) (t) is of class A, then from

L{F(t)} = f (s)

it follows that
n−1
X
L{F (t)} = s f (s) −
(n) n
sn−1−k F(k) (0).
k=0

Thus
L{F(3) (t)} = s3 f (s) − s2 F(0) − sF0 (0) − F00 (0),

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives
Theorem
If F(t), F0 (t), . . . , F(n−1) (t) are continuous for t ≥ 0 and of
exponential order as t → ∞, and F(n) (t) is of class A, then from

L{F(t)} = f (s)

it follows that
n−1
X
L{F (t)} = s f (s) −
(n) n
sn−1−k F(k) (0).
k=0

Thus
L{F(3) (t)} = s3 f (s) − s2 F(0) − sF0 (0) − F00 (0),

L{F(4) (t)} = s4 f (s) − s3 F(0) − s2 F0 (0) − sF00 (0) − F(3) (0), etc.
M.T. Panganduyon Differential Equations
Laplace Transforms
Transforms of Derivatives

Example: Using transform of derivatives, find the laplace

transform of the following.
1 F(t) = t3

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Example: Using transform of derivatives, find the laplace

transform of the following.
1 F(t) = t3
2 F(t) = t sin t

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Example: Using transform of derivatives, find the laplace

transform of the following.
1 F(t) = t3
2 F(t) = t sin t

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Example: Using transform of derivatives, find the laplace

transform of the following.
1 F(t) = t3
2 F(t) = t sin t

Answer:
6
1 L{t3 } = , s>0
s4

M.T. Panganduyon Differential Equations

Laplace Transforms
Transforms of Derivatives

Example: Using transform of derivatives, find the laplace

transform of the following.
1 F(t) = t3
2 F(t) = t sin t

Answer:
6
1 L{t3 } = , s>0
s4
2s
2 L {t sin t} = .
(s2 + 1)2

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

For functions of class A, the theorems of advanced calculus

show that it is legitimate to differentiate the Laplace transform
integral.

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

For functions of class A, the theorems of advanced calculus

show that it is legitimate to differentiate the Laplace transform
integral. That is, if F(t) is of class A, from
Z ∞
f (s) = e−st F(t)dt
0

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

For functions of class A, the theorems of advanced calculus

show that it is legitimate to differentiate the Laplace transform
integral. That is, if F(t) is of class A, from
Z ∞
f (s) = e−st F(t)dt
0

it follows that Z ∞
f (s) =
0
(−t)e−st F(t)dt
0

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

For functions of class A, the theorems of advanced calculus

show that it is legitimate to differentiate the Laplace transform
integral. That is, if F(t) is of class A, from
Z ∞
f (s) = e−st F(t)dt
0

it follows that Z ∞
f (s) =
0
(−t)e−st F(t)dt
0
The integral on the right is the transform of the function (−t)F(t).

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

Theorem
If F(t) is of class A, it follows from

L{F(t)} = f (s)

that
f 0 (s) = L{−tF(t)}.

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

Theorem
If F(t) is of class A, it follows from

L{F(t)} = f (s)

that
f 0 (s) = L{−tF(t)}.

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

Theorem
If F(t) is of class A, it follows from

L{F(t)} = f (s)

that
f 0 (s) = L{−tF(t)}.

When F(t) is of class A, (−t)k F(t) is also of class A for any

positive integer k.

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

Theorem
If F(t) is of class A, it follows from

L{F(t)} = f (s)

that for any positive integer n,

dn
f (s) = L{(−t)n F(t)}.
dsn

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

Theorem
If F(t) is of class A, it follows from

L{F(t)} = f (s)

that for any positive integer n,

dn
f (s) = L{(−t)n F(t)}.
dsn

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

Theorem
If F(t) is of class A, it follows from

L{F(t)} = f (s)

that for any positive integer n,

dn
f (s) = L{(−t)n F(t)}.
dsn
Example: Find L{tet } and L{t2 sin kt}.

M.T. Panganduyon Differential Equations

Laplace Transforms
Derivatives of Transforms

Theorem
If F(t) is of class A, it follows from

L{F(t)} = f (s)

that for any positive integer n,

dn
f (s) = L{(−t)n F(t)}.
dsn
Example: Find L{tet } and L{t2 sin kt}.
Answer:
1 2k(3s2 − k2 )
L{tet } = and L{t2 sin kt} =
(s − 2)2 (s2 + k2 )3

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

 simply the inverse operation to taking the Laplace

transform.

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

 simply the inverse operation to taking the Laplace

transform.
 we first find f (s) and then must obtain F(t) from f (s).

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

 simply the inverse operation to taking the Laplace

transform.
 we first find f (s) and then must obtain F(t) from f (s).

definition (Inverse Laplace Transform)

If L{F(t)} = f (s), we say that F(t) is an inverse Laplace transform, or
an inverse transform, of f (s) and we write

F(t) = L−1 {f (s)}.

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

 simply the inverse operation to taking the Laplace

transform.
 we first find f (s) and then must obtain F(t) from f (s).

definition (Inverse Laplace Transform)

If L{F(t)} = f (s), we say that F(t) is an inverse Laplace transform, or
an inverse transform, of f (s) and we write

F(t) = L−1 {f (s)}.

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

 simply the inverse operation to taking the Laplace

transform.
 we first find f (s) and then must obtain F(t) from f (s).

definition (Inverse Laplace Transform)

If L{F(t)} = f (s), we say that F(t) is an inverse Laplace transform, or
an inverse transform, of f (s) and we write

F(t) = L−1 {f (s)}.

1 1
 
For instance, since L{e−3t } = , then L−1 = e−3t .
s+3 s+3

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

The inverse transform, L−1 , is a linear operator and so satisfies

the linearity property:

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

The inverse transform, L−1 , is a linear operator and so satisfies

the linearity property:
If α and β are constants then

L−1 {αF(t) + βG(t)} = αL−1 {F(t)} + βL−1 {G(t)}.

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

The inverse transform, L−1 , is a linear operator and so satisfies

the linearity property:
If α and β are constants then

L−1 {αF(t) + βG(t)} = αL−1 {F(t)} + βL−1 {G(t)}.

Theorem
L−1 {f (s)} = e−at L−1 {f (s − a)}

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1
s2 + 4

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1
s2 + 4

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1
 
2 L−1 + 2
s s +9

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1
 
2 L−1 + 2
s s +9

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
s s +9 3

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1
(s + 1)(s + 2)

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1
(s + 1)(s + 2)

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1 = −e−t + 2e−2t
(s + 1)(s + 2)

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1 = −e−t + 2e−2t
(s + 1)(s + 2)
15
 
4 L−1 2
s + 4s + 13

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1 = −e−t + 2e−2t
(s + 1)(s + 2)
15
 
4 L−1 2
s + 4s + 13

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1 = −e−t + 2e−2t
(s + 1)(s + 2)
15
 
4 L−1 2 = 5e−2t sin 3t
s + 4s + 13

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1 = −e−t + 2e−2t
(s + 1)(s + 2)
15
 
4 L−1 2 = 5e−2t sin 3t
s + 4s + 13
s+1
 
5 L−1 2
s + 6s + 25

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1 = −e−t + 2e−2t
(s + 1)(s + 2)
15
 
4 L−1 2 = 5e−2t sin 3t
s + 4s + 13
s+1
 
5 L−1 2
s + 6s + 25

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Example. Obtain the following:

2
 
1 L−1 = sin(2t)
s2 + 4
3 1 1
 
2 L−1 + 2 = 3 + sin(3t)
( s s + 9 ) 3
s
3 L−1 = −e−t + 2e−2t
(s + 1)(s + 2)
15
 
4 L−1 2 = 5e−2t sin 3t
s + 4s + 13
s+1
   
5 L−1 2 = e−3t cos 4t − 21 sin 4t
s + 6s + 25

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
1 L−1
s3 + 4s2 + 3s

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
1 L−1
s3 + 4s2 + 3s

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3
2 L−1
s3 (s + 1)2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3
2 L−1
s3 (s + 1)2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3 3
2 L−1 = 3 − t2 − 3e−t + 2te−t
s (s + 1)
3 2 2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3 3
2 L−1 = 3 − t2 − 3e−t + 2te−t
s (s + 1)
3 2 2
( )
16
3 L−1
s(s + 4)2
2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3 3
2 L−1 = 3 − t2 − 3e−t + 2te−t
s (s + 1)
3 2 2
( )
16
3 L−1
s(s + 4)2
2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3 3
2 L−1 = 3 − t2 − 3e−t + 2te−t
s (s + 1)
3 2 2
( )
16
3 L−1 = 1 − cos 2t − t sin 2t
s(s + 4)2
2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3 3
2 L−1 = 3 − t2 − 3e−t + 2te−t
s (s + 1)
3 2 2
( )
16
3 L−1 = 1 − cos 2t − t sin 2t
s(s + 4)2
2

s
 
4 L−1 2
s + 6s + 13

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3 3
2 L−1 = 3 − t2 − 3e−t + 2te−t
s (s + 1)
3 2 2
( )
16
3 L−1 = 1 − cos 2t − t sin 2t
s(s + 4)2
2

s
 
4 L−1 2
s + 6s + 13

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

Exercises: Obtain the following:

s2 − 6
( )
5 1
1 L−1 = −2 + e−t + e−3t
s + 4s + 3s
3 2 2 2
( 3 )
5s − 6s − 3 3
2 L−1 = 3 − t2 − 3e−t + 2te−t
s (s + 1)
3 2 2
( )
16
3 L−1 = 1 − cos 2t − t sin 2t
s(s + 4)2
2

s 3
 
4 L−1 2 = e−3t cos 2t − e−3t sin 2t
s + 6s + 13 2

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

M.T. Panganduyon Differential Equations

Laplace Transforms
Inverse Transforms

M.T. Panganduyon Differential Equations

M.T. Panganduyon Differential Equations