ENGINEERING
MATHEMATICS
FENG 310
LECTURER:
Sallu Jaward
Phone:+23275727287
E-mail: sallujawardfat35096@gmail.com
jawards@coventry.ac.uk
Course Material:
ü Course material, including lecture notes, PowerPoint slides, sample exams,
Home work solutions, hints, etc … will be available to the students
throughout the semester.
ü Any student registered for the class should have access to all these materials.
2
q ADVANCED ENGINEERING MATHEMATICS by K.A. Stroud and Dexter J. Booth,
4th edition, McGraw-Hill Education, New York, NY. ISBN: 978-1-259-69653-4.
q ADVANCED ENGINEERING MATHEMATICS by Dennis G. Zill, 6 th edition, |
Burlington, MA : Jones & Bartlett Learning, [2017]. ISBN: 1284105903 (casebound).
q CALCULUS, EARLY TRANSCENDENTAL FUNCTIONS by Ron Larson, Robert
Hostetler and Bruce H. Edwards 4th edition, Houghton Mifflin , ISBN 13: 978-0-61873069-8 ISBN 10: 0-618-73069-9
q MULTIVARIABLE CALCULUS by Ron Larson 11th edition
Homework
ü A list of weekly problem sets along with due dates can be found throughout the
lecture slides. Although you may discuss homework problems with me, or your
classmates, you are expected to work out the problems independently
ü There is no specific format for the Home Work assignments. Please write legibly.
ü You are expected to turn in your assignments on the due date, typically by 12
noon of that day unless noted otherwise on the Assignment.
ü Late Home Work submissions will receive a zero grade unless permission is
obtained from the instructor before the due date.
4
Academic Honesty
ü You may discuss homework problems with others, but you must
not retain written notes from your conversations with other
students, or share data electronically (e.g. files, emails, etc …) to be
used in completing your homework.
ü Your written work must be completed without reference to such
notes, with the exception of class notes.
5
Classroom etiquette
üPlease try to arrive on time for each class because late arrivals are
distracting.
üIf you arrive late please enter quietly and take a seat near the back of
the room.
üPlease do not converse with your classmates during class.
üAll electronic devices (cell phones, iPods, iPads, laptops, …) must be
turned off
6
Exams
ü In addition to a 3-hour final exam, there are three tests during the
semester.
ü In case you miss a test, your score on that test will be zero without an
option for a make-up.
ü All tests and exams will be closed book.
ü The regular exams will be given at the end of the semester.
7
Grading
ü Each of the three tests will count as 15 points with the average of the scores
counting toward the final grade.
ü The final exam will count as 70 points, and the homework will count collectively
as 10 points, for a total of 95 points.
ü Regular attendance will count for 5 points.
ü Your final grade for the course will be based primarily on your total point score,
but other appropriate factors, such as participation and personal initiative, will
also be taken into account.
8
Derivatives of Functions
Derivatives of Functions
Integration Formulas
Integration Formulas
OUTLINE OF TOPICS TO BE COVERED
1. Introduction to Differential Equations
5. Applications of Differential Equations
ü Definition and classification (order, degree, linearity)
ü Mechanical vibrations (damped and undamped)
ü General and particular solutions
ü Electrical circuits (RLC circuits)
2. First-Order Differential Equations
ü Population models
ü Separable equations
6. Laplace Transforms
ü Homogeneous equations
ü Definition and basic properties
ü Exact equations and integrating factors
ü Inverse Laplace transform
ü Linear differential equations
ü Laplace transform of common functions
3. Higher-Order Differential Equations
ü Step functions and Dirac delta function
ü Second and higher-order linear differential equation
7. Applications of Laplace Transforms
ü Solutions using characteristic equations
ü Solving differential equations using Laplace transforms
ü Method of undetermined coefficients
ü Initial value problems
ü Variation of parameters
ü Systems of linear differential equations
4. Systems of Differential Equations
8. Introduction to Partial Differential Equations
ü Introduction to systems of first-order differential
ü Brief Introduction to Partial Differential Equations
equations
ü Method of Separation of Variables
LECTURE 1
Introduction to
Differential Equations
15
Definition and classification
Definition of Differential Equation: An equation containing the derivatives of one or more dependent
variables, with respect to one or more independent variables, is said to be a differential equation
(DE). To discuss these equations, we will categorize each differential equation by its type, order, and
linearity.
Classification by Type
Ordinary Differential Equations (ODEs): An equation containing only ordinary derivatives of
one or more functions with respect to a single independent variable.
Partial Differential Equations (PDEs): An equation involving only partial derivatives of one or
more functions with respect to two or more independent variables. Example:
The following example demonstrates both types of differential equations (ODEs and PDEs).
16
Examples: Types of Differential Equations
��
+ 6� = �−� ,
��
�2 �
��
+ − 12� = 0,
��2
��
are examples of ordinary differential equations
��
��
+ = 3x+2y
��
��
�2 � �2 �
= 0,
+
��2 ��2
��
��
=− ,
��
��
�2� �2� ��
=
−
,
��2 ��2 ��
are examples of partial differential equations. Notice that in the third equation, there are two
dependent variables and two independent variables.
This indicates that u and v must each be functions of two or more independent variables.
17
Classification by Order
The order of a differential equation (ODE or PDE) is the order of the highest derivative in the equation.
EXAMPLE: Order of a Differential Equation
�2�
�� 3
�4 �
�2 �
The differential equations 2 + 5
− 4� = �� ,
2 4 + 2 = 0 are examples of a second-order ordinary differential equation
��
��
��
��
and a fourth-order partial differential equation, respectively.
A first-order ordinary differential equation is sometimes written in the differential form M(�, �)�� + �(�, �)�� = 0
EXAMPLE 3 Differential Form of a First-Order ODE: Assuming y is the dependent variable in a first-order Ordinary Differential
Equation (ODE), we recall from calculus that the differential dy is defined as: �� = �’ ��
a)
By dividing both sides by the differential dx, an alternative form of the equation (� − �)�� + 4��� = 0 is given by: � − � +
��
��
4 = 0 �� 4 + � = �
��
��
b) By multiplying the differential equation 6��
��
��
+ �2 + � 2 = 0
An n-th order ordinary differential equation with one dependent variable can be generally expressed as:
� �, �, �,’ . . . . . , �� = 0
where F is a real-valued function of n+2 is �, �, �,’ . . . . . , ��
���
= �(�, �, �,’ . . . . . , �� )
�
��
Classification by Linearity
A D.E is is linear in y if we can write it in form
�� (�)�(�) + ��−1 (�)�(�−1) + . . . . + �1 (�)�’ + �0 (�)� + �(�) = 0 or
���
� � − 1�
��
+ � (�)� = �(�)
�� (�) � + ��−1 (�) � − 1 + , . . . , �1 (�)
0
��
��
��
The dependent variable y and all its derivatives y, y, … , y(n) are of the first degree; that
is, the power of each term involving y is 1.
The coefficients a0, a1, … , an of y, y , … , y(n) depend at most on the independent
variable x
EXAMPLE: Linear and Nonlinear Differential Equations
1. (� − �)�� + 4��� = 0
2. �’’ − 2�’ + � = 0
3�
�
��
3
3. � ��3 + 3� �� − 5� = ��
�4�
4. ��4
�2�
+ �2 = 0
5. ��2 + ���� − 5� = 0
6. (1 − �)�’ + 2� = ��
7. �’’’ + 3��’ + � = ��
8.
��
��
’’
+ 3� = ����
9. � + sin � � = ��
�� 3
10. �� �� = �3�
Solution of an ODE
General Solution
Particular Solution
(Arbitary constant)
(Initial Values)
verify that the indicated function is an explicit solution of the given differential equation.
Practice problems set 1
Verify that the indicated function is an explicit
solution of the given differential equation.
1) 2�′ + � = 0,
2) �′′ − � = 0
�
�=� 2
� = ��
3) �′′ − 6�′ + 13� = 0; � = �3� ���2�
4) �′′ + � = tan � ;
� = −(cos � )ln (sec � + tan � )
5) (1 − 2� − �2 )�′′ + 2(1 + �)�′ − 2� = 0;
�=�+1
Find values of m so that the function � = ��� is a solution
of the given differential equation.
1) 2�′ + � = 0,
3. 2�′′ + 9�′ − 5� = 0
2. �′′ − 5�′ + 6� = 0
4. 3�′ = 4�
find values of m so that the function y= �� is a solution
of the given differential equation.
1.
0
�2 �′′ − 7��′ + 15� = 0
2. x�′′ + 2�′ =
LECTURE 2
First-Order Differential Equations
vSeparable equations
vHomogeneous equations
v Exact equations and integrating factors
vLinear differential equations
Separable equations
A first-order differential equation is called separable if all terms involving � and y can be
grouped with dx and dy , respectively. In other words, a differential equation of the form
is said to have separable variables.
��
= g(x). h(y)
��
For example, the differential equations
��
= �2 �4 �5�−3�
��
are separable and nonseparable, respectively
and
��
= � + cos �
��
Separable variable is a foundational method for solving certain differential equations
Classwork set
2) Solve (1 + �)�� − ��� = 0
3) Solve the initial value problem
4) Solve
��
6) Solve
��
��
= �2 − 4.
( 2�
5) Solve cos � �
��
=
2�
�
.
− �)
��
��
��
��
�
= , �(4) =− 3.
�
= �� sin 2�, if y(0)=0.
7) Solve sin 3� �� + 2�cos3 3� �� = 0.
8) Solve csc � �� + sec2 � ��=0
Practice problems set 2
2.
3.
solve the given differential equation by separation of variables.
��
��
= P − �2
−� ) ��
� +�
= �2 ,
��
2 ��
( �
4. �
��
= � − ��, �(−1) =− 1.
2)
��
4)
��
3)
��
��
��
��
��
= 4(�2 + 1), x( � 4)=1
=
=
��+3�−�−3
��−2�+4�−8
��+2�−�−2
��−3��+�−3
�2 −1
.
.
5) �� = �2−1, y(2)=2
a) (1 + �� )2 �−� �� + (1 + �� )3 �−� �� = 0
Homogeneous equations
A differential equation is said to be a homogeneous DE if every term has
the same degree of
Homogeneous DE can be written as: M(x,y)dx + N(x,y)dy=0
If M is simpler use: x=vy, dx=vdy+ydv
If N is simpler use: y= vx, dy=vdx+xdv
Homogeneous DE
��
��
��
��
= �(�, �)
=�
�
�
�
, let v= , y=vx
�
Classwork set
Solve the following DE
Section A
1. (�� + �� + �� )�� − �� �� = �
2. �
� ��
��
= �� + �� �
��
�
�
3. ��� �� = �� + ���
��
4. � �� = � + � ��
��
5. (� − �) �� = � + �
Section B
��
1. (� + �) �� = � − �
2. ��
��
��
= �� + ���
��
3. ��� �� = �� + ���
��
4. �� �� = �� + � ��� + ��
��
5. � �� + � = �� + ��
Exact Equations
Exact equations are a special type of first-order differential equation where the left-hand side of the
equation is the exact differential of some function f(x,y). This allows us to integrate directly to find a
solution
Recognizing an Exact Equation
A first-order DE of the form:
�(�, �)�� + �(�, �)�� = �
is exact if there exists a function f(x,y) such that:
��
��
= �(�, �) and = �(�, �)
��
��
Criterion for Exactness
For M(x,y) and N(x,y) with continuous first partial derivatives, the equation is exact if:
�� ��
=
��
��
This is a necessary and sufficient condition for the equation to be exact. If this condition holds, the
equation represents the total differential of a function f(x,y)
Exact Equations: Method of Solution
1. Verify Exactness
��
��
Check whether the condition
= is satisfied.
��
��
2. Determine f(x,y)
Ø Integrate M(x,y) with respect to x, treating y as a constant, to find f(x,y) (up to a function of y, g(y)):
�(�, �) =
�(�, �)�� + �(�)
Ø Differentiate f(x,y) with respect to y and equate it to N(x,y) to find g′ (y)
��
= �′ (�) + ( ����� ���� �)
��
Ø Integrate g′ (y) to find g(y)
3. General Solution
Combine results to express the solution implicitly as
�(�, �) = �
where C is an arbitrary constant.
Practice problems set 3
Determine whether the given differential equation is exact. If it is exact, solve it.
1) (2� − 1)�� + (3� + 7)��� = 0
2) (2� − �)�� − (� + 6�)�� = 0
3) (���� − �����)�� + (���� + ����� − �)�� = 0
4) (2��2 − 3)�� + (2�2 � + 4)�� = 0
1
��
�
5) 2� − + ���3�
+ 2 − 4�2 + 3����3� = 0
�
�
��
�
6) (1 + ln � + � )�� = (1 − ln � )�� = 0
3
��
3
7) 1 − � + � �� + � = � − 1
8)
1
�
1
+ 2− 2
�
a) Solve
�
� +�2
�
1+�2
�
�� + (�� + 2
+ ���� − 2��
1
� +�2
��
��
)�� = 0
= �(� + ����), y(0)=1
Integrating Factors for Nonexact Equations
Integrating factors are used to transform a nonexact differential equation into an exact one, enabling us
to solve it using the techniques for exact equations.
Given a differential equation of the form:
�(�, �)�� + �(�, �)�� = �
the equation is nonexact if the criterion for exactness does not hold:
�� ��
≠
��
��
To make it exact, we multiply through by an integrating factor μ(x,y), such that:
μ(�, �)�(�, �)�� + �(�, �)�(�, �)�� = �
becomes exact
The modified equation is exact if
�(��) �(��)
=
��
��
To find μ(x,y), the following procedure is used:
1. Criterion for Exactness After Multiplication:
Expand
�(��)
�(��)
= 0 using the product rule. After simplification
�� ��
�� � − �� � = (
−
)�
�� ��
This is a partial differential equation for �(�, �)
��
−
Integrating Factors for
Nonexact Equations
��
2. Simplifying Assumptions
If �(�, �) is assumed to depend only on � or �, the equation simplifies:
• � = �(�):
The equation becomes:
• � = �(�):
The equation becomes:
��
�� ��−�� ��
=�
��
�
��
�� ��−�� ��
=�
��
�
3. Conditions for Use:
v if (�� ��−�� ��) � depends only on � ,use �(�)
v If (�� ��−�� ��) � depends only on �, use �(�)
Practice problems set 4
Verify that the given differential equation is not exact, apply the integrating factor, and solve the resulting
exact equation.
1. (−������ + 2�����)�� + 2������� = 0,
�(�, �) = ��
2. (�2 − 2�� − �2 )�� + (�2 + 2�� − �2 )�� = 0 �(�, �) = (� + �)−2
Solve the given differential equation by finding an appropriate integrating factor
1.
2.
3.
4.
(2�2 + 3�)�� + 2���� = 0
�(� + � + 1)�� + (� + 2�)�� = 0
6���� + (4� + 9�2 )�� = 0
2
������ + (1 + )������ = 0
�
−3�
5. (10 − 6� + � )�� − 2�� = 0
6. (�2 + ��2 )�� + (5�2 − �� + �3 ����)�� = 0
find an integrating factor of the form �� �� , then solve
1. (2�2 − 6��)�� + (3�� − 4�2 )�� = 0
2. (12 + 15��)�� + (6��−1 + 3�2 )�� = 0
First-Order Linear Differential Equation
The most common case is the first-order linear differential equation:
��
+ P(�)� = Q(�)
��
where P(x) and Q(x) are known
General Solution
1. Identify P(x) and Q(x)
2. Find the integrating factor �(�):
3.
4.
5.
�(�) = � �(�)�� .
Multiply through by �(�) to make the left-hand side exact:
��
�(�)
+ �(�)P(�)� = �(�)Q(�)
��
Recognize the left-hand side as a derivative
�
[�(�)�] = �(�)Q(�)
��
Integrate both sides:
�(�)� =
�(�)� ��
Practice problems set 5
LECTURE 3
Higher-Order Differential Equations
üSecond and higher-order linear differential equation
üSolutions using characteristic equations
üMethod of undetermined coefficients
üVariation of parameters
Reduction of Order
Reduction of order is a technique for finding a second, linearly independent solution �� (�) of a second-order
linear homogeneous differential equation when one solution �� (�) is already known. This method involves
reducing the original second-order equation into a first-order differential equation.
The general second-order linear homogeneous differential equation is:
�2 (�)�′′ + �1 (�)�′ + �0 (�)� = 0
it is rewritten in standard form
�′′ + �(�)�′ + �(�)� = 0
If �� (�) is a known solution of the equation, we use the substitution
�2 = �(�)�1 (�)
where �(�) an unknown function to be determined. This substitution reduces the second-order equation
into a first-order equation.
�2 = �(�)�1 (�)
�2′ = �′ �1 + ��1 ′
�2′′ = �′′ �1 + �′ �1 ′ + �′ �1 ′ + ��1 ′′
Classwork and Practice problems set 6
The indicated function �� (�) is a solution of the given equation. Use reduction of order
or formula, to find a second solution �� (�).
1)
2)
3)
4)
5)
6)
7)
�′′ − 4�′ + 4� = 0
�′′ − � = 0
�′′ + 2�′ + � = 0
�′′ + 16� = 0
�′′ + 9� = 0
�′′ − � = 0
5. (1 − 2� − �2 )�′′ + 2(1 + �)�′ − 2� = 0
�1 = �2�
�1 = ��
�1 = ��−�
�1 = cos 4�
�1 = sin 3�
�1 = cosh �
�1 = � + 1
Superposition Principle for Homogeneous
Equations
The Superposition Principle of homogeneous equations states that if �� (�), �� (�), . . . , �� (�) are solutions to a
homogeneous linear differential equation
�(�) = �� �(�) + ��−1 �(�−1) + . . . . + �1 �’ + �0 � = 0
then any linear combination of these solutions
�� (�) = �� �� (�) + �� �� (�) + , . . . , + �� �� (�)
is also a solution of the homogeneous equation �(�) = �. The solutions to the homogeneous equation can be
combined linearly to form the general solution to the homogeneous equation.
Superposition Principle for Nonhomogeneous
Equations
The Superposition Principle for nonhomogeneous equations states that if ��� (�), ��� (�), . . . , ��� (�) are
particular solutions of k different nonhomogeneous linear differential equations, corresponding to different
forcing functions �� (�), �� (�), . . . , �� (�)), then the linear combination of these particular solutions:
�� (�) = ��� (�) + ��� (�) + . . . + ��� (�)
thus, the general solution can be expressed as
�(�) = �� (�) + �� (�)
where �� (�)is the general solution of the corresponding homogeneous equation �(�) = �.
Homogeneous Linear Equations with Constant
Coefficients
Auxiliary Equation
By assuming a solution of the form , we substitute into
constant coefficients, all solutions can be expressed
the differential equation, which leads to
��2 ��� + ����� + ���� = 0
as exponential functions or combinations of
or equivalently
exponential functions.
��� (��2 + �� + �) = 0
Since ��� ≠ �, the term in parentheses must vanish
General Form:
��2 + �� + �) = 0
The homogeneous linear equation with constant
This quadratic equation is called the auxiliary equation.
Solving it provides the roots, �� and �� , which
coefficients is written as:
determine the form of the general solution.
(�)
(�−1)
’
�� � + ��−1 �
+ . . . . + �1 � + �0 � = 0
For linear higher-order differential equations with
where �� are constants, �� ≠ �
For a second-order equation
��′′ + ��′ + �� = 0
Three Cases of Roots
1. Distinct Real Roots �� − ��� > � :
If the roots �� and �� are real and distinct, the general solution is
� = �1 ��1� + �2 ��2�
2. Repeated Real Roots �� − ��� = � :
If the roots �� = �� = � are real and equal, the general solution is
� = (�1 + �2 �)���
3. Complex Conjugate Roots �� − ��� < � :
If the roots are complex �� = � + �� and �� = � − ��, the general solution is expressed in terms of real
functions:
� = ��� (�1 cos �� + �2 sin ��)
Here, � represents the exponential decay/growth, while β governs the oscillatory
behavior.
Classwork and Practice problems set 7
find the general solution or the particular solution of the given second-order differential equation
1.
2.
�′′ − 6�′ + 9� = 0
�′′ + 5�′ + 6� = 0
3. 4�′′ + �′ = 0
4. �′′ + 9� = 0
5.
2�′′ − 5�′ − 3� = 0
6. �′′ + 4�′ + 7� = 0
7. �′′ − 10�′ + 25� = 0
8. �′′ + 8�′ + 25� = 0
9. �′′′ + 3�′′ − 4� = 0
10. 4�′′ − 5�′ + 6� = 0
i.
�′′ + 4� = 0,
�(0) = 4, �′ (0) = 6,
ii. �′′ − � = 0,
�(0) = 1, �′ (1) = 0
iii. �′′ + 2�′ + 2� = 0, �(0) = 2, �′ (0) = 1
iv. 4�′′ + 4�′ + 17� = 0,
v.
vi.
9
�2 �
+ 24
2
�� +
�4 �
�2 � +
��
�4 �
��4
vii. 16
a)
+2
��4
��
�2 � +
��2
+ 24
16� = 0
�(0) =− 1, �′ (0) = 2
�=0
��2
9� = 0
Two roots of a cubic auxiliary equation with real coefficients are �1 =−
1
and �2 = 3 + �. What is the corresponding homogeneous linear
2
differential equation?
LECTURE 4
Method of undetermined coefficients
The method of undetermined coefficients is a practical technique for solving nonhomogeneous linear differential
equations of the form
where:
�� �(�) + ��−1 �(�−1) + . . . . + �1 �’ + �0 � = �(�)
v The coefficients �� are constants
v g(x) is a simple function such as a polynomial, exponential, sine, or cosine, or their combinations.
The general solution of such an equation is given by
where:
� = �� + ��
�� : The complementary function, obtained by solving the associated homogeneous equation
�� �(�) + ��−1 �(�−1) + . . . . + �1 �’ + �0 � = 0
​ �� :A particular solution to the nonhomogeneous equation, determined using an educated guess based on �(�)
Method of undetermined coefficients
Guess the Form of ��
The form of depends on the nature of �(�).
For example
If �(�) is a polynomial of degree n
guess �� = ��� + ���−� + . . . , + �
If �(�) = ��� guess �� = ����
If �(�) = sin (��) �� cos (��),
guess �� = ���� (��) + ���� (��)
Account for Overlap with �� :
If any term in your guess for �� appears in �� , multiply ��
by �� , where k is the smallest integer making �� independent
of ��
Substitute �� into the Original Equation
Ø Differentiate �� as needed.
Ø Substitute �� and its derivatives into the given equation.
Ø Match coefficients of like terms in
�, ��� , ��� (��), ��� (��) to solve for the unknown
coefficients.
Write the general solution as
For combinations like �� ���
� = �� + ��
guess �� = (��� + ���−� + . . . , + �)���
Classwork and Practice problems set 8
solve the given differential equation by undetermined coefficients.
1. �′′ + 2�′ − 24� = 16 − (� + 2)�4�
2. �′′ + 5�′ + 6� = �2
3. �′′ + 9� = �2�
4. �′′ + 3�′ + 2� = cos �
5. �′′ + � = cos �
6. �′′ − 9� = ��� + sin 2�
7. �′′ + 4�′ − 2� = 2�2 − 3� + 6
8. �′′ − 3�′ + 2� = �� sin t
9. �′′ − 2�′ − 3� = 3�2�
10. �′′ − 2�′ − 3� = 3�−�
1. �′′ + � = 1 + �2
�(0) = 5, �(1) = 0,
1. �′′ − 4� = (�2 − 3)sin 2�
2. �′′ − 3�′ + 3� = � − 4��
3. 2�′′ + 3�′ − 2� = 14�2 − 4� − 11
�(0) = 0, �′ (0) = 0
Cont. Practice problems set 8
1.
a)
Consider the differential equation ��′′ + ��′ + �� = ��� , where a, b, c, and k are constants. The auxiliary
equation of the associated homogeneous equation is ��2 + �� + � = 0
If k is not a root of the auxiliary equation, show that we can find a particular solution of the form �� = ���� ,
where � = 1 (��2 +��+�).
b) If k is a root of the auxiliary equation of multiplicity one, show that we can find a particular solution of the
c)
form �� = ����� , where � = 1 (2�� + �). Explain how we know that k ≠ −� (2�).
If k is a root of the auxiliary equation of multiplicity two, show that we can find a particular solution of the
form �� = ��2 ��� , where A = 1 (2�).
LECTURE 5
Variation of Parameters
The general solution is
�(�) = �� (�) + �� (�)
of a nonhomogeneous linear differential equation.
where �� (�) is the complementary solution and �� (�) is the
particular solution
General form
�2 (�)�′′ + �1 (�)�′ + �0 (�)� = �(�)
or in standard form
�′′ + �(�)�′ + �(�)� = �(�)
Assumes �(�), �(�) ��� �(�) are continuous on an interval �
Solve the associated homogeneous equation
′′
′
� + �(�)� + �(�)� = 0
The complementary solution is:
�� = �1 �1 (�) + �2 �2 (�)
where �1 (�) ��� �2 (�)are linearly independent solutions.
Particular Solution ��
Assume in the form:
�� (�) = �1 (�)�1 (�) + �2 (�)�2 (�)
where �� (�) �nd �� (�) are functions to be determined
Using the product rule to differentiate �� twice
Key Results:
−�(�)�2
′
�1 = (
� �1 , �2 )
where �(�� , �� ) is the Wronskian
�1
W= �′
1
�(�)�1
′
�� = (
� �1 , �2 )
�2
�′ 2
Integrate to find �� (�) �nd �� (�)then substitute back to find ��
The general solution is
�(�) = �� (�) + �� (�)
Classwork and Practice problems set 9
solve each differential equation by variation of parameters.
SECTION A
1. �′′ − 4�′ + 4� = (� + 1)�2�
2. 4�′′ + 36� = csc 3�
3. �′′ + � = tan �
4. �′′ + � = sec �
5. �′′ + � = sec �tan �
6. �′′ + � = cos2 �
7.
�′′ + � = sec2 �
8. �′′ − � = cosh �
9. �′′ − � = sinh 2�
SECTION B
2
1. �′′ + � = ��
2. �′′ + �′ − 2� = ln x
3. �′′ − 2�′ + � = �� tan−1 �
4. 2�′′ + �′ = 6�
5.
��
� − 2� + � =
1+�2
�′′ + 2�′ + � = �−� ln �
′′
′
6.
7. �′′ + 3�′ + 2� = sin ��
8.
�′′ + 3�′ + 2� =
1
1+��
Cauchy–Euler Equations
Cauchy–Euler equations are a type of linear differential equation with variable coefficients.
Form:
��
� − 1�
�
�
��
�
�−1
+
,
.
.
.
,
+ � � = �(�),
�� �
+
�
�
�
�
�−1
1
0
�
�
−
1
��
��
��
where �� , ��−� , �� , . . . , �� are constants
The degree � of �� matches the order � of differentiation.
Homogeneous second-order equation
��2�′′ + ���′ + �� = 0
Solution interval: (0,∞) or (−∞,0) by substitution � =−� .
Assume
Cauchy–Euler Equations
Solution Assumption
Assume � = ��
Substitute � = �� into the equation
��2�′′ + ���′ + �� = 0
Results in
��(� − 1)�� + ���� + ��� = 0
Simplifies to the auxiliary equation
��2 + (� − �)� + � = 0
Cases of Roots
Case 1: Distinct Real Roots
Roots �� and �� where �1 ≠ �2
�(�) = �1 ��1 + �2 ��2
Case 2: Repeated Roots
Single root �� = ��
�(�) = �1 ��1 + �2 ��2 ln �
Case 3: Complex Roots
Roots r=α±iβ:
�(�) = �� [�1 cos( �ln �) + �2 sin (�ln � )
Classwork and Practice problems set 10
SECTION A
SECTION B
Use the substitution y=xr to solve the given
equations
Use the substitution� = (� − �0 )�
to solve the given equation
2
2 � � − 2� �� − 4� = 0
1. � ��2
2.
��
2 � � + 8� �� + � = 0
4� 2
��
��
2
3. 4�2 �′′ + 17� = 0, �(1) =− 1, �′ (1) =
1
−
2
1. (� + 3)2 �′′ − 8(� + 3)�′ +
14� = 0
2. (� + 2)2 �′′ + (� + 2)�′ + � =
0
3. (� − 4)2 �′′ − 5(� − 4)�′ +
9� = 0
CONT. Practice problems set 10
SECTION C
use the substitution � = �� to transform
the given Cauchy–Euler equation to a
differential equation with constant
coefficients. Solve the original equation
by solving the new equation
1. �2 �′′ + 9��′ − 20� = 0
2. �2 �′′ + 10��′ + 8� = �2
3. �2 �′′ − 4��′ + 6� = ln �2
SECTION D
find a homogeneous Cauchy–Euler differential
equation whose general solution is given.
1) � = �1 �4 + �2 �−2
2) � = �1 + �2 �5
3) � = �1 �−3 + �2 �−3 ln �
4) � = �1 cos( ln �) + �2 sin (ln � )
1
5) � = �1 �
2 cos(
1
1
1
ln �) + �2 � 2 sin ( ln � )
2
2
LECTURE 6
Differential Equation: Power Series Method
LECTURE 7
Applications of Differential Equations: Linear Models
ü Growth and Decay
ü Newton’s Law of Cooling
ü Mixture of two Solutions
ü Series Circuits
Applications of Differential Equations: Linear Models
This section introduces linear models for solving first-order differential equations, focusing on real-world
applications like population growth, radioactive decay, Newton’s Law of Cooling, and mixing problems
Growth and Decay
The differential equation
models growth or decay
��
= ��, �(�0 ) = �0
��
Figure 1: Shroud image in
Problem 5
Source: Zill (2017)
Class Work and Practice problems set 3
1.
2.
3.
4.
5.
3
A culture initially has P0 number of bacteria. At � = 1ℎ the number of bacteria is measured to be �� . If
2
the rate of growth is proportional to the number of bacteria P(t) present at time t, determine the time
necessary for the number of bacteria to triple.
A breeder reactor converts relatively stable uranium-238 into the isotope plutonium-239. After 15 years
it is determined that 0.043% of the initial amount A0 of the plutonium has disinte grated. Find the halflife of this isotope if the rate of disintegration is proportional to the amount remaining
A fossilized bone is found to contain 0.1% of its original amount of C-14. Determine the age of the
fossil
The population of a town grows at a rate proportional to the population present at time t. The initial
population of 500 increases by 15% in 10 years. What will the population be in 30 years? How fast is
the population growing at t =30?
The Shroud of Turin, which shows the negative image of the body of a man who appears to have been
crucified, is believed by many to be the burial shroud of Jesus of Nazareth. In 1988 the Vatican granted
permission to have the shroud carbon dated. Three independent scientific laboratories analyzed the
cloth and concluded that the shroud was approxi mately 660 years old, an age consistent with it
historical appearance. Using this age, determine what percentage of the original amount of C-14
remained in the cloth as of 1988.
Newton’s Law of Cooling
Governs the cooling or warming of an object relative to ambient temperature:
•
�� : Ambient temperature
��
= �(� − �� )
��
Class Work and Practice problems set 3
1.
2.
3.
4.
5.
When a cake is removed from an oven, its temperature is measured at 3000F. Three minutes later its
temperature is 2000F. How long will it take for the cake to cool off to a room tem perature of 700F. ?
A thermometer is removed from a room where the temperature is 700F and is taken outside, where the air
temperature is 10F. After one-half minute the thermometer reads 500F. What is the reading of the
thermometer at t 1 min? How long will it take for the thermometer to reach 150F?
A thermometer is taken from an inside room to the outside, where the air temperature is 50F.. After 1
minute the thermometer reads 550F, and after 5 minutes it reads 300F. What is the initial temperature of
the inside room?
A small metal bar, whose initial temperature was 200C, is dropped into a large container of boiling water.
How long will it take the bar to reach 900C if it is known that its temperature increased 2 in 1 second?
How long will it take the bar to reach 980C?
Two large containers A and B of the same size are filled with different fluids. The fluids in containers A
and B are maintained at 00C and 1000C, respectively. A small metal bar, whose ini tial temperature is
1000C, is lowered into container A. After 1 minute the temperature of the bar is 900C. After 2 minutes the
bar is removed and instantly transferred to the other container. After 1 minute in container B the
temperature of the bar rises 10. How long, measured from the start of the entire process, will it take the
bar to reach 99.90C?
Mixture of two Solutions
Net rate of change of salt
1.
2.
3.
��
= input rate − output rate
��
A tank contains 200 liters of fluid in which 30 grams of salt is dissolved. Brine containing 1 gram of salt per
liter is then pumped into the tank at a rate of 4 L /min; the well-mixed solution is pumped out at the same
rate. Find the number A(t) of grams of salt in the tank at time t.
Solve Problem 1 assuming that pure water is pumped into the tank.
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon
is pumped into the tank at a rate of 5 gal/min. The well-mixed solution is pumped out at the same rate. Find
the number A(t) of pounds of salt in the tank at time t
Series Circuits:Resistor-Inductor (R-L)
The sum of voltage drops in a series circuit equals the impressed voltage
�
��
+ �� = �(�)
��
L: Inductance (Henrys), R: Resistance (Ohms)
LR-series circuit
1
�
�� + � = �(�)
�
R: Resistance (Ohms), C: Capacitance (Farads)
��
1
+ � = �(�)
��
�
�=
��
��
RC-series circuit
Class Work and Practice problems set 3
A 100-volt electromotive force is applied to an RC-series circuit in which the resistance is 200 ohms and the
capacitance is 10–4 farad. Find the charge q(t) on the capacitor if q(0) =0. Find the current i(t).
A 200-volt electromotive force is applied to an RC-series cir cuit in which the resistance is 1000 ohms and
the capacitance is 5 × 10−6 farad. Find the charge q(t) on the capacitor if i(0) =0.4. Determine the charge and
current at � = 0.005� . Determine the charge as t → ∞
An electromotive force
120,
�(�) =
0,
0 ≤ � ≥ 20
� < 20
is applied to an LR-series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the
current i(t) if i(0) = 0.
An LR-series circuit has a variable inductor with the inductance defined by
1
0 ≤ � ≥ 10
�(�) = 1 − 10 �,
� < 10
0,
Find the current i(t) if the resistance is 0.2 ohm, the impressed voltage is E(t) =4, and i(0)=0
Applications of Differential Equations: Nonlinear
Models
Population Dynamics and Logistic Growth
Exponential Growth Model
Governing Equation:
Assumption: Growth rate k is constant.
Growth limited by environmental factors
Reflects realistic population dynamics.
��
= ��
��
��
= ��(�)
��
Logistic Growth Model
Incorporates carrying capacity (K) to reflect resource limits
��
= �(� − ��)
��
Suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. If it is assumed
that the rate at which the virus spreads is proportional not only to the number x of infected students but also to the
number of students not infected, determine the number of infected students after 6 days if it is further observed
that after 4 days x(4) =50.
The number N(t) of people in a community who are exposed to a particular advertisement is governed by the
logistic equation. Initially N(0)=500, and it is observed that N(1)=1000. Solve for N(t) if it is predicted that the
limiting number of people in the community who will see the advertisement is 50,000.
LECTURE 7
Systems of Linear Differential Equations
v Theory of Linear Systems
v Homogeneous Linear Systems
Ø Distinct Real Eigenvalues
Ø Repeated Eigenvalues
Ø Complex Eigenvalues
Theory of Linear Systems
Systems of n linear differential equations in n unknowns c unknowns can be represented as:
�11 (�)�1 + �12 (�)�2 + . . . + �1� (�)�� = �1 (�)
�21 (�)�1 + �22 (�)�2 + . . . + �2� (�)�� = �2 (�)
:
:
.
.
�21 (�)�1 + �22 (�)�2 + . . . + �2� (�)�� = �2 (�)
�
Here ��� (�) are polynomials in the differential operator � =
��
Restricting the study to first-order systems, the equations take the form:
��1
= �1 (�, �1 , �2 , . . . , �� )
��
��2
= �2 (�, �1 , �2 , . . . , �� )
��
:
:
.
.
���
= �� (�, �1 , �2 , . . . , �� )
��
Theory of Linear Systems
Definition
A linear system arises when each �� (�, �1 , �2 , . . . , �� ) is linear in the dependent variables �1 , �2 , . . . , �� . This
results in:
��1
= �11 (�)�1 + �12 (�)�2 + . . . + �1� (�)�� + �1 (�)
��
��2
= �11 (�)�1 + �22 (�)�2 + . . . + �2� (�)�� + �2 (�)
��
:
:
.
.
���
= ��1 (�)�1 + ��2 (�)�2 + . . . + ��� (�)�� + �� (�)
��
Homogeneous: If �� (�) = 0 for all � = 1,2, . . . , �.
Non-Homogeneous: If �� ≠ 0 for any �
Cont. Theory of Linear Systems
Matrix Representation of Linear Systems
Using matrices, the system can be compactly written as:
�′ = �(�)� + �(�)
Where:
�11 (�) �12 (�) … �1� (�)
�� (�)
�� (�)
�� (�)
�� (�)
�21 (�) �22 (�) … �2� (�) ,
�=
,
�(t)
=
�(�)
=
⋮
⋮
⋮
⋮
⋮
�� (�)
�� (�)
�31 (�) �32 (�) �3� (�)
The matrix equation simplifies to:
Homogeneous Case
If F(t)=0, the system reduces to:
�′ = �� + �
�′ = ��
Theory of Linear Systems
Systems of n linear differential equations in n unknowns can be written in a normal form as first-order
differential equations. The equations are linear when the functions involved are linear in the dependent
variables.
Linear Systems and Notation
A first-order linear system is written as:
��
= �(�)� + �(�)
��
X′ = �� + �
Where:
• � is the column vector of unknown functions,
• �(�) is a matrix of coefficients,
• �(�) is a vector of nonhomogeneous terms.
For homogeneous systems, �(�) = � and the system simplifies to:
X′ = ��
Class Work and Practice problems set 3
1)
2)
1)
�
�
If � =
,or �
�
�
then write the matrix form of the homogeneous linear system:
��
��
= 3� + 4�, = 5� − 7�
��
��
��
= 6� + � + � + �
��
��
= 8� + 7� − � + 10�
��
��
= 2� + 9� − � + 6�
��
��
= 6� + � + � + �
��
��
= 8� + 7� − � + 10�
��
��
= 2� + 9� − � + 6�
��
A)
��
��
= 3� − 5�, = 4� + 8�
��
��
��
B). �� =− 3� + 4� − 9�
��
= 6� − �
��
C).
��
= 10� + 4� + 3�
��
��
=�−�+�+�−1
��
��
= 2� + � − � − 3�2
��
��
= � + � + � + �2 − � + 2
��
a.
b.
c.
��
��
= 4� − 7�, = 5�
��
��
��
=�−�
��
��
= � + 2�
��
��
=− � + �
��
��
=− 3� + 4� + �−� sin 2�
��
��
= 5� + 9� + 4�−� cos 2�
��
��
= � + 6� − �−�
��
Class Work and Practice problems set 3
Verify that the vector X is a solution of the given homogeneous linear system
1.
2.
��
= 3� − 4�
��
��
= 4� − 7�;
��
�=
��
=− 2� + 5�
��
��
=− 2� + 4�;
��
1
1
2
�=
�
5cos �
3cos �−sin �
3. �′ = −1 4 �; � =
1 −1
4.
3. �′ =
−3�
−1
2
��
�−3� 2
4.
5.
2 1
�; � =
−1 0
1
3
�� +
1
2
1
�′ = 6 −1 0 �; � =
−1 −2 −1
1 0 1
�′ = 1 1 0 �; � =
−2 0 −1
4
4
���
1
6
−13
����
1
1
− sin �− cos �
2
2
−����+����
Homogeneous Linear Systems
A homogeneous linear system is of the form:
�′ = ��
where A is a square matrix of constants, X is a vector of unknowns, and �′ is its derivative with respect to time
t
The goal is to determine solutions for X(t).
�(t) = K�λ�
Where K is a constant vector and λ is a scalar (eigenvalue).
Eigenvalues and Eigenvectors
General Solution Form:
Assume a solution of the form:
� = K���
Where K is a constant vector (eigenvector) and λ is a scalar (eigenvalue).
Differentiating � and substituting it into
�′ = ��
gives;
K���� = �K���
K� = �K, or �K − K� = 0,
Cont. Homogeneous Linear Systems
Where I is � × � identity matrix
K = IK,
(� − ��)� = 0
(�11 − �)�1 +
�12 �2 + . . . + �1� �� = 0
�21 �1 + (�22 − �)�2 + . . . + �2� �� = 0
:
:
.
.
��1 �1 + ��2 �2 + . . . + (��� − �)�� = 0
Characteristic Equation
For a nontrivial solution (� ≠ 0)the determinant of (A−λI) must be zero:
���(� − ��)=0
The resulting equation (polynomial) is the characteristic equation, and its roots are the eigenvalues � of �
Each eigenvalue λ has a corresponding eigenvector K, satisfies:
(A−λI)K=0.
Case 1: Distinct Real Eigenvalues
When A has n distinct real eigenvalues (λ1 , λ2 , . . . , λ� ), the solution is:
�(t) = �1 �1 ��1� + �2 �2 ��2� + . . . + �� �� ��n�
Here (�1 , K2 , . . . , �� ) are linearly independent eigenvectors corresponding to (�1 , �2 , . . . , �� )
Example
Solve 1.
��
= 2� + 3�
��
��
��
3.
= 2� + �;
1 1 4
�′ = 0 2 0 �; �(�) =
1 1 1
2.
1
3
0
��
��
=− 4� + � + �
��
= � + 5� − �
��
��
=
��
� − 3�
Class Work and Practice problems set 3
Find the general solution of the given system
1.
��
= � + 2�
��
��
= 4� + 3�
��
2.
3.
��
5
=− � + 2�
��
2
��
3
= � − 3�
��
4
��
=− 4� + 2�
��
��
5
=− � + 2�
��
2
4.
��
��
��
��
5.
=�+�−�
= 2�
6.
1 0 4
�′ = 0 1 0 �
1 0 1
��
=�−�
��
7.
� =
��
8.
�′ =
��
= 2� − 7�
��
��
= 5� + 10� + 4�
��
= 5� + 2�
��
′
�
�
�
� −
� �,
�
−� �
�,
−� �
�(�) =
3
5
Case 2: Repeated Eigenvalues
Not all eigenvalues (�1 , �2 , . . . , �� ) of an � � matrix A are distinct; some may be repeated.
if m is a positive integer and (� − �� )� is a factor of the characteristic equation while (� − �� )�+1 is not a
factor, then �� is an eigenvalue with multiplicity �
1.
m linearly independent eigenvectors exist.
v The general solution is a linear combination of solutions of the form:
�(t) = �1 �1 ��1� + �2 �2 ��1� + . . . + �� �� ��1�
where K1 , �� , . . . , �� are eigenvectors.
2. Fewer than m linearly independent eigenvectors exist.
v Additional solutions can be constructed using generalized eigenvectors:
�� = K11 ��1�
�� = �21 ���1� + K22 ��1�
⋮
�−2
��−1
�
�� = ��1
��1� + ��2
��1� + ⋯ + ��� ��1�
(� − 1)!
(� − 2)!
where ��� are column vectors, can always be found.
Assume �1 is an eigenvalue of multiplicity 2 and only one eigenvector K can be associated with
it . Then, a second solution of �′ =AX will take the form
�� = ����1� + ���1�
Eigenvalue of Multiplicity Two
Assume �1 is an eigenvalue of multiplicity 2 and only one eigenvector K can be associated with it . Then,
a second solution of �′ =AX will take the form
Substituting into the system yields
�� = ����1� + ���1�
(�� − �1 �)���1� + (�� − �1 � − �)��1� = 0
this is an identity, which implies that
1 −2 2
Solve �′ = −2 1 −2 X
2 −2 1
(� − �1 �)� = 0
(� − �1 �)� = �
Eigenvalue of Multiplicity Three
Now, assume �1 is an eigenvalue of multiplicity 3 and only one eigenvector can be associated with it .
Using a similar argument as above, a second solution would take the same form as �� (given above)
and a third solution will take the form
Where
�� � �
�� = � � 1 + ����1� + ���1�
�
(� − �1 �)� = 0
(� − �1 �)� = �
4 1 0
Solve �′ = 0 4 1 X
0 0 4
(� − �1 �)� = �
Class Work and Practice problems set 3
Find the general solution of the given system
1.
��
= 3� − �
��
��
= 9� − 3�
��
2.
3.
��
=− 6� + 5�
��
��
=− 5� + 4�
��
4.
��
��
��
��
5.
= 3� − � − �
=�+�−�
��
=�−�+�
��
��
= 3� + 2� + 4�
��
��
��
= 2� + 2�
��
= 4� + 2� + 3�
��
6.
7.
8.
1 0 0
�′ = 2 2 −1 �
0 1 0
�′ =
�′ =
� �
�,
−� �
−� �
�,
−� ��
�(�) =
−1
6
Case3: Complex Eigenvalues in Systems
For a coefficient matrix A, if the eigenvalues �1 = � + �� and �2 = � − �� (where β ≠ �) are complex, their
corresponding eigenvectors generally have complex entries.
Solution Construction:
A solution for the system can be expressed using eigenvalues and eigenvectors:
� = �1 �1 ��1� + �2 �2 ��2�
where �1 and �2 are eigenvectors corresponding to �1 and �2 , respectively.
Euler's Formula:
Using Euler’s formula (���� = cos (��) + �sin (��)), the exponential terms with complex eigenvalues can be
expressed as:
��1� = ��� (cos (��) + �sin (��))
Real Solutions:
The general solution is rewritten in terms of real functions cos (��) ��� sin (��), ensuring linear independence
and real entries:
�1 = �1 cos (��) + �2 sin (��)
�1 = �1 cos (��) − �2 sin (��)
where �1 and �2 are derived from the real and imaginary parts of the eigenvector �1
Superposition Principle:
The overall solution is a linear combination of these real solutions:
with arbitrary constants �1 ��� �2
� = �1 �1 + �2 �2
Class Work and Practice problems set 3
find the general solution of the given system
1.
2.
3.
4.
5.
6.
�′ =
�
�
�, �(0) =
−� −�
� −�
�′ =
�,
� −�
�′ =
�′ =
�′ =
�′ =
2
−1
� −�
�,
� −�
� ��
�,
−� −�
−� �
�, �(0) =
−� −�
� �
�
−� �
0
3
6.
8.
��
=�
��
��
=− �
��
��
=�
��
��
= 2� + � + 2�
��
��
= 3� + 6�
��
��
=− 4� − 3�
��
9.
��
= 5� + �
��
��
=− 2� + 3�
��
��
10. �� = 4� + 5�
��
=− 2� + 6�
��
LECTURE 8
Nonhomogeneous Linear Systems
vUndetermined Coefficients
vVariation of Parameters
Nonhomogeneous Linear Systems
Overview:
v Methods of undetermined coefficients and variation of parameters apply to nonhomogeneous linear systems.
v Undetermined Coefficients: Quick for constant, polynomial, exponential, sine, cosine functions.
v Variation of Parameters: More powerful, applied when undetermined coefficients aren’t sufficient.
v General Form
General Solution:
�′ = �� + �(�)
� = �� + ��
�� : Complementary solution (from homogeneous system �� = ��)
�� : Particular solution (from nonhomogeneous system �� = �� + �(�)
Method 1 - Undetermined Coefficients
Apply only when F(t) consists of constants, polynomials, exponentials, sines, cosines.
Steps:
1. Solve homogeneous system �� :
v Use characteristic equation of matrix A to find eigenvalues and eigenvectors.
2. Guess particular solution �� based on form of �(�)
3. Substitute guess into system to determine unknown constants.
Ex
1.
2.
�′ =
�′ =
−� �
�+
−� �
� �
�+
� �
−�
�
��
−���+�
Method 2 - Variation of Parameters
Source:Zill (2017)
Method 2 - Variation of Parameters
Source:Zill (2017)
Method 2 - Variation of Parameters
Source:Zill (2017)
Class Work and Practice problems set 3
In Problems 1–6,use variation of parameters and in 7-11 , , use the method of undetermined coefficients
to solve the given system
1.
2.
3.
4.
5.
�′ =
−� �
�+
� −�
� −�
′
� = �
�+
−�
�′ =
�
� −�
�+
� �
� �
�+
� −�
��
= 3� − 3� + 4
�′ =
��
��
= 2� − 2� − 1
��
��
�−�
�
−�
6.
� �
�
�����
������
��
��
�
���
7.
8.
��
= 2� − �
��
��
= 3� − 2� + 4�
��
��
= 2� + 3� − 7
��
��
=− � − 2� + 5
��
��
= 5� + 9� + 2
��
��
=− � + 11� + 6
��
9.
�′ =
−� �
�+
−� �
����
−�����
� � �
�
10. � = � � � � + −��
� � �
��
′
� � �
�
11. �′ = � � � � + −� ���
� � �
�
LECTURE 8
The Laplace Transforms
üDefinition and basic properties
üInverse Laplace transform
üLaplace transform of common functions
üStep functions and Dirac delta function
The Laplace Transforms
A specific integral transform that transforms a
function f(t), defined for � ≥ 0 t≥0, into a
function of S
Definition:
{�(�) =
∞
0
�−�� �(�) �� = lim
�→∞
{�(�)} = �(�)
provided the integral converges.
Key Examples
1. �{1}
2. �{��� }
3. �{�� }
4. �{sin �� }
5. �{cos �� }
�
0
�−�� �(�) ��
Properties of the Laplace Transform
Linearity:
For constants a,b,
{��(�) + ��(�)} = � {�(�)} + � {�(�)}
Examples
{1 + 5�}
{2�6� − 15sin 2� }
{10�−3� − 5� + 8}
}
Piecewise Continuity
A function f(t) is piecewise continuous on [0, ∞) if:
It is continuous on every subinterval[�� , ��+1 )of
[0, ∞), except possibly at a finite number of points
Properties of the Laplace Transform
Exponential Order
A function f(t) f(t) is of exponential order c if constants � > �, � > � , ��� � > �exist such that:
�(�) ≤ ���� , for all � > �
This means �(�)does not grow faster than ��� for larger t
Existence Theorem:
If f (t) is continuous and exponential order with constant c, then
�(s) = L{f(t)}
is defined for all � > �
Class Work and Practice problems set 3
Use the Definition to find {�(�))}
1) �(�) =
−1,
1,
3) �(�) =
2� + 1,
0,
2) �(�) =
4) �(�) =
5) �(�) =
6) �(�) =
4,
0,
0≤�<1
�≥1
0≤�<1
�≥1
0≤�<1
�≥1
−1,
1,
0≤�<1
�≥1
����,
0,
0≤�<� 2
�≥� 2
����,
0,
0≤�<�
�≥�
The Inverse Laplace Transforms
if L{f(t)}=F(s), then the inverse Laplace transform is denoted as:
�(�) = �−1 {�(�)}
This process determines the time-domain function f(t) corresponding to its Laplace transform F(S)
Common Inverse Laplace Transforms
1
1
1. �{��� } = �−� ↔ �−1 �−� = ���
1
1
2. �{1} = � ↔ �−1 � = 1
�
�!
3. �{� } = ��+1 ↔ �
�
−1
1
��
��−1
=
(�−1)!
�
4. �{cos �� } = �2+�2 ↔ �−1 �2+�2 = cos ��
�
1
1
5. �{sin �� } = �2+�2 ↔ �−1 �2+�2 = � sin ��
Linearity of the Inverse Laplace Transform
The transform is linear, meaning:
�−1 {��(�) + ��(�)} = ��−1 {�(�)} + ��−1 {�(�)}
find the given inverse transform.
2
−1 (�+2)
6. �
�3
−1 1
1. �
5
2.
3.
�
1
�−1 2
� +7
1
48
�−1 2 − 5
�
�
4. �−1
5.
2
1 2
− 3
�
�
3
−1 (�+1)
�
�4
7.
�−1
1
1
4
6
2 − +
�
�
Partial Fractions
1
�−2
1
8. �−1 � − �5 − �+8
1
9. �−1 4�+1
1
10. �−1 4�+1
11. �−1
1
5�−2
16. �−1 �2+3�
10�
18. �−1 �2+2�−3
5
12. �−1 �2+49
13. �−1 �2+16
2�−6
14 �−1 �2+9
�+1
15. �−1 �2+2
1
�+1
17. �−1 �2−4�
�
1
19. �−1 �2+�−20
�
20. �−1 (�−2)(�−3)(�−6)
Transforms of Derivatives
Laplace transform converts derivatives into algebraic terms.
′
�{� (�)} =
∞ −��
∞ −��
∞
−��
� �(�) �� = � �(�) + � 0 � �(�)��
0
0
∞
=− �(0) + ��{�(�)}
�{�′ (�)} = ��(�) − �(0)
∞
∞
�{�′′ (�)} = 0 �−�� �(�) �� = �−�� �′(�) + � 0 �−�� �′(�)��
0
=− �′(0) + ��{�′(�)}
= �[��(�) − �(0)] − �′(0)
�{�′′ (�)}
= �2 �(�) − ��(0) − �′(0)
Transform of n-th Derivative:
If f, �′ , . . . , ��+1 are continuous on [0, ∞) and are of exponential order and if �� (�) is piecewise
continuous on [0, ∞), then
�{�� (�)} = �� �(�) − ��−1 �(0) − ��−2 �′ (0),
where �(�) = �{�(�)}
Class Work and Practice problems set 3
In Problems, use the Laplace transform to solve the given initial-value problem
1.
��
+ 3� = 13���2�,
��
�(0) = 6
2. �′′ − 3�′ + 2� = �−4� , �(0) = 1, �′ (0) = 5
3.
��
− � = 1,
��
��
4. 2 �� + � = 0,
6. �′′ + 6�′ = �4� ,
7. �′′ − �′ = 2cos 5� ,
�(0) = 0
�(0) =− 3
�(0) = 2,
�(0) = 0,
8
�′′ − �′ = 2cos 5� ,
�(0) = 0,
9. �′′ + 5�′ + 4� = 0, �(0) = 1, �′ (0) = 0
10. �′′ − 4�′ = �3� − 3�−� ,
11. �′′ + �′ = 2sin 2� ,
12. �′′ − 9�′ = ��
�(0) = 1, �′ (0) =− 1
�(0) = 10, �′ (0) = 0
�(0) = 10, �′ (0) = 0
Translation on the s-Axis (First Translation Theorem)
The first translation theorem allows for easy evaluation of Laplace transforms for exponential multiples of
functions.
if �{�(�)} = �(�), then �{��� �(�)} = �(� − �)
where
��
�{�
∞ −�� ��
�(�)} = 0 � � �(�) ��
∞
= 0 �−(�−�)� �(�) �� == �(� − �)
The theorem shifts F(s) on the s-axis by a units:
• Right Shift: � > 0
• Left Shift: � < 0
Inverse Transformation:
To find �−1 {�(� − �)} identify F(s), take �−1 to find f(t), and multiply by ���
Class Work and Practice problems set 3
find either F (s) or f (t), as indicated
{��−10� }
8.
{�5� �3 }
1
(�+2)3
1
9. �−1 (�−4)4
{�−2� cos 4� }
1
10. �−1 �2−6�+10
{�(�� + �2� )2 }
1
11. �−1 �2+2�+5
{�2� (� − 1)2 }
{�� sin 3� }
�−1
�
2
{�3� (9 − 4� + 10sin )}
�
12. �−1 �2+4�+5
2�+5
13. �−1 �2+6�+34
14. �−1
�
(�+1)2
2�−1
16. �−1 �2(�+1)3
5�
15. �−1 (�−2)2
17. �′′ − 6�′ + 9� = �2 �3� , �(0) = 2, �′ (0) = 17
18. �′′ + 4�′ + 6� = 1 + �−� , �(0) = 0, �′ (0) = 0
19. �′′ + 2�′ + � = 0, �(0) = 1, �′ (0) = 0
20. �′ − � = 1 + ��� , �(0) = 0,
21. �′′ − 2�′ + 5� = 1 + �, �(0) = 0, �′ (0) = 4
Unit Step Function
Unit Step Function
Translation on the t-Axis (Unit Step Function)
The unit step function �(� − �) is defined to be
�(� − �) = �(�) =
The function is used to "turn on" or "turn off" parts of
0,
1,
0≤�<�
�≥�
Second Translation Theorem
Second Translation Theorem
If F(s)=L{f(t)} and a> 0 then
{�(� − �)�(� − �)} = �−�� �(�)
{�(� − �)�(� − �)} = �−�� �(�)
Inverse Form of Theorem
If f (t) =�−1 {F(s)}, the inverse form of Theorem
�−1 {�−�� �(�)} = �(� − �)�(� − �)
Find the Laplace transform of the function f whose graph is given in Figure
In problem 1-12,ind either F(s) or f (t),
as indicated
{cos � �(� − �)}
{(� − 1)�(� − 1)}
{cos � �(� − �)}
{cos 2� �(� − �)}
{(3� + 1)�(� − 1)}
2−�
7.
{�
�(� − 1)}
−2�
−1 �
�
�3
8.
−��
−1 �
�
�2 +1
9. �
−1
�−2�
�2 (�−1)
−�� 2
−1 ��
10. �
�2 +4
−2� )2
−1 (1+�
11. �
�+2
−2�
−1 �
12. �
�−4
In Problems 13–16, use the Laplace transform to solve the
given initial-value problem.
13. �′ + � = �(�),
�(0) = 5, where
0,
0≤�<�
�(�) =
3cos � ,
�≥�
14. �′ + � = �(�),
�(0) = 0 �ℎ��
0,
0≤�<1
�(�) =
5,
�≥1
15. �′ + � = �(�),
�(0) = 0 �ℎ��
1,
0≤�<1
�(�) =
−1,
�≥1
16. �′′ + 4� = �(�),
�(0) = 0, �′(0) =− 1, �ℎ��
1,
0≤�<1
�(�) =
0,
�≥1
1. The differential equation for the instantaneous
charge q(t) on the capacitor in an LRC-series circuit is
�2�
��
1
� 2 + � + � = �(�).
��
��
�
Use the Laplace transform to find q(t) when � =
1ℎ, � = 20�ℎ��, � = 0.005�, �(�) = 150�, � >
0, �(0) = 0 ��� �(0) = 0. What is the current i(t)?
2. Use the Laplace transform to find the charge q(t)
in an RC-series when q(0) = 0 and E(t) = �0 �−�� �
> 0. Consider two cases: k ≠ 1/RC and k =1/RC.
write each function in terms of unit step functions.
Find the Laplace transform of the given function
a) �(�) =
2,
−2,
c) �(�) =
0,
�2 ,
1,
b) �(�) = 0,
1,
d) �(�) =
0≤�<3
�≥3
0≤�<4
4≤�<5
�≥5
0≤�<1
�≥1
0,
0 ≤ � < 3� 2
����,
� ≥ 3� 2
Graph for Problem g)
Graph for Problem h)
In Problems 49–54, match the given graph with
one of the given functions in (a)–(f ). The
graph of f (t) is given in FIGURE 4.3.11.
a)
b)
c)
d)
e)
f)
�(�) − �(�)�(� − �)
�(� − �)�(� − �)
�(�)�(� − �)
�(�) − �(�)�(� − �)
�(�)�(� − �) − �(�)�(� − �)
�(� − �)�(� − �) − �(� − �)�(� − �)
Derivatives of Transforms
Multiplying a Function by ��
The Laplace transform of �� �(�) can be computed using derivatives of the Laplace transform of �(�)
If �(�) = �{�(�)} and if we assume that interchanging of differentiation and integration is possible, then
that is
similarly
�
�
�(�) =
��
��
• General result for
∞
0
�−�� �(�)�� =
∞
� −��
[� �(�)�� =−
��
0
�
�{��(�)} =− �{�(�)}
��
∞
0
�−�� ��(�)�� =− �{��(�)};
�
�{� �(�)} = �{� ∙ ��(�)} =− �{��(�)}
��
2
�
=− 2 �{�(�)}
��
�
�{� �(�)}, If �(�) = �{�(�)} ��� � = 1,2,3. . . , then
2
�{�
�
�
�(�)} = (−1)� � �(�)
��
�
EXAMPLES
1. Evaluate �{������}.
2.
Solve . �′′ + 16� = ���4�, �(0) = 0, �′ (0) = 1
Transform of an Integral
Convolution
The convolution of two functions f(t) and g(t), denoted (� ∗ �)(�), is defined as
�
�(�)�(� − �) ��
0
(� ∗ �)(�) =
The variable � (Greek letter “tau”) is a dummy variable used for integration.
The result of the convolution is a new function of t, emphasizing its role in the definition.
Key Properties of Convolution
Commutative Property:
Linearity: The convolution of sums or scalar multiples follows the rules of linearity
(�� + ��) ∗ ℎ = �(� ∗ ℎ) + �(� ∗ ℎ),
Laplace Transform of Convolution (Convolution Theorem):
if f(t) and g(t) g(t) are piecewise continuous and of exponential order, then:
�{(� ∗ �)} = �{�(�)} + �{�(�)} = �(�)�(�)
Inverse Convolution Theorem
The Convolution Theorem is also useful for inverse Laplace transforms. If:
�−1 {�(�) ∙ �(�)} = � ∗ �
we can compute inverse transforms of products of Laplace transforms by using the convolution formula.
In Problems 1–7,find the
Laplace transform of f * g
using Convolution Theorem
1)
2)
3)
4)
5)
6)
7)
In Problems 8–12, evaluate the
given inverse transform.
1
8. �−1 2 2 2
(� + � )
�{�� ∗ ����}.
1
�{1 ∗ �3 }.
9. �−1 �3(�−1)
�{�2 ∗ ��� }.
�{�−� ∗ �� ����}.
1
−1
10.
�
�(�−1)
�{�2� ∗ ����}.
�
�{ 0 ���� ���(� − �) �� }.
1
�
11. �−1 �2(�−1)
�{ 0 ����� ��}.
In Problems 12–16, use Derivatives of
Transforms to evaluate the given
Laplace transform
12. �{��−10� }.
13. �{�3 �� }.
14. �{����2�}.
15. �{��2� ���6�}.
16. {��−3� ���6�}.
Transform of a Periodic Function
A function f(t) is periodic if �(� + �) = �(�) for all , where T is the period.
The Laplace transform of a periodic function is:
�{�(�)} =
1
1 − �−��
�
0
�−�� �(�)��
In Problems (P)1–5, find the Laplace transform of the given periodic function
Fig 1: Graph for P 1
Fig 2: Graph for P 2
Fig 3: Graph for P 3
Fig 4: Graph for P 4
Fig 6: Graph for P 6
Fig 5: Graph for P 5
the Dirac Delta Function
The Dirac Delta Function, δ(t−t ), is a mathematical construct introduced to model forces of large magnitude
acting over very short periods, such as a hammer strike or lightning on a wing.
It is defined as the limit of a family of functions,
Series Solutions of Linear Differential Equations
Solutions about Ordinary Points
Review of Power Series
Power Series Solutions
Solutions about Singular Points
Special Functions
Bessel Functions
Legendre Functions
Introduction to Partial Differential Equations
ü Brief Introduction to Partial Differential Equations
ü Method of Separation of Variables