MATH 2Z03 - ENGINEERING
MATHEMATICS III
Lecture Notes
Carlos Cordoba Caycedo & James Hornick
Spring 2025
McMaster University
Engineering Mathematics III (v. June 1, 2025)
Carlos Cordoba & James Hornick
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Contents
1 First Order Ordinary Differential Equations
1.1 Formal definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Existence and Uniqueness theory . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Geometric approach: Direction fields . . . . . . . . . . . . . . . . . . . . . . .
1.4 Autonomous differential equations . . . . . . . . . . . . . . . . . . . . . . . .
1.5 First order separable differential equations . . . . . . . . . . . . . . . . . . . .
1.6 First order linear differential equations . . . . . . . . . . . . . . . . . . . . . .
1.7 Applications of first order differential equations . . . . . . . . . . . . . . . . .
1.7.1 Population Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.2 Newton’s Cooling Law . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.3 Mixing Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2 Linear Differential Equations
2.1 Homogeneous linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Solution of nth order linear equations with constant coefficients . . . . . . . .
2.3 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Finding a particular solution . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Undetermined coefficients . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Variation of parameters . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Cauchy-Euler equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Application: Deflection of a beam . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 Boundary Value Problems: Eigenvalues and Eigenfunctions . . . . . . . . . .
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3 Laplace Transform
3.1 Motivation: Spring/Mass systems . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Inverse Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Solving IVPs with the Laplace transform . . . . . . . . . . . . . . . . . . . .
3.5 The unit step function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Convolution and Volterra integral equations . . . . . . . . . . . . . . . . . . .
3.7 Dirac delta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Preface
A goal without a plan is just a wish.
Antoine de Saint-Exupéry
Since the dawn of math, we have always been interested in applying our mathematical
knowledge to understand the behavior of the world we live in. However, using arithmetic
operations (addition, multiplication, exponentiation, etc) and unknown variables, might not
be enough to model the dynamics of the universe.
As an example, when we want do describe the how fast an object is moving through
space, we have to study the ratio of change of the position with respect to time, in other
words, we are analyzing how quickly the object is changing its position. Note that this
notion of studying the ratio of change of a given quantity cannot be captured completely
by simple arithmetic operations.
Moreover, this is just a simple example in physical dynamics and nowadays there are
almost infinitely many different problems that require modeling the ratio of change of a
certain variable. In order to study the behavior of unknowns with respect to their ratio of
change, or derivatives, the theory of differential equations was developed. During this
course, we attempt to understand the basics of this type of equations including how to find
solutions, what do the solutions describe and how to perform a graphical approach to the
construction of the solutions.
As a personal note for any future students who are reading this or anyone who is interested in going through this document, just remember that hard work is always going to
beat talent! So have fun while you all chase your dreams and never forget that you will
achieve each and every single one of them. With that being said, let’s embark on this new
adventure!
Chapter 1
First Order Ordinary Differential
Equations
How many things have been denied one day, only to become realities the next!
Jules Verne, From the Earth to the Moon.
During the first part of this course, we will discuss the introductory ideas that led into the
study and analysis of the theory of differential equations. The first couple of sections will
serve as a starting point for the new students into the field and then further topics about
the general solution of linear differential equations (Including Bernoulli equations and linear
models) will be discussed in further sections.
Moreover, we will study the direction fields generated by the explicit form of the equation
including the asymptotic behavior of autonomous systems, basic numerical methods to solve
differential equations and some applications as population growth, Newton’s cooling law and
exponential decay.
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Formal definitions
In previous math courses, you learned how to find solutions of certain equations in one
variable or multiple variables. Now, instead of looking for numbers that satisfy given conditions, we want to characterize the behavior of an unknown function y(x) so we introduce
the following type of equations:
Definition 1 (Differential equation). A differential equation is an equation involving
an unknown target function y(x) (most of the times this function is called the dependent
variable of the equation), independents variable x and the derivatives of y(x).
Before we start asking ourselves about the existence of solutions for differential equations, we are going to classify them according to three different properties. The first two
classifications serve as a general setting for differential equation, but we will write the third
classification focusing on equation that only have ordinary derivatives.
1. Classification by type: If a differential equation contains only ordinary derivatives
with respect to a single independent variable, then it is called an ordinary differential equation. For example,
dy
− 5y = 0
dx
d2 u du
+
− sin(x) = 1
dx2
dx
However, if the equation involves partial derivatives with respect to two or more independent variables, it is called a partial differential equation and some examples
are:
∂2u ∂2u
− 2 = 0 (Wave equation)
∂x2
∂y
1
∂V
∂V
∂2V
+ σ 2 S 2 2 + rS
− rV = 0 (Black-Scholes equation)
∂t
2
∂S
∂S
2. Classification by the order: We define the order of a differential equation to
be the highest derivative that appears in the equation. For example, the differential
equation
t2
dy
d3 y
−t
= et
3
dt
dt
is a third-order ordinary differential equation, whereas the equation
xy
∂2u
∂u
∂u
−
+y
=0
∂x∂y ∂x
∂y
is a second-order partial differential equation.
Lecture Notes (version June 1, 2025)
7
3. Classification by linearity: We call a differential equation linear if it has the
following form
an (x)
dn−1 y
dy
dn y
+
a
(x)
+ · · · + a1 (x)
+ a0 (x)y = g(x)
n−1
n
n−1
dx
dx
dx
Note that each term containing the dependent function y(x) and its derivatives are elevated to the power of 1 and each coefficient depends only on the independent variable.
For example, the equations
d2 y
− sin(x)y = x3
dx2
dy
x3
x2
= x
dx
e +1
x
are examples of a second order linear differential equation and a first order linear
differential equation respectively. Whereas the equations
3 2
d y
− 2y = x2
dx3
and
d4 y
dy
+ sin(y)
=0
dt4
dt
are nonlinear third-, and fourth order ordinary differential equations.
With this classification, we denote that an nth order ordinary differential equation can be
represented as
dn y
dy
F x, y, , · · · , n = 0
dx
dx
which is called the implicit representation of the differential equation. Furthermore, if
we can isolate the highest derivative of our dependent variable, then we get the explicit
representation which has the general form:
dn y
dy
dn−1 y
= f x, y, , · · · , n−1
dxn
dx
dx
Now, we turn back our attention to our main goal of this course which is finding solutions
of differential equations.
Definition 2 (Solution of a differential equation). Given a differential equation in implicit
form, we say that a function f (x) is a solution of a differential equation on an interval I
if the following equality holds
dn f
df
F x, f, , · · · , n = 0
dx
dx
Note that we can always check that a function is a solution of a differential equation by
computing the derivatives of the result from our computation as we observe in the following
examples.
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Example 1.1.1. The function y(x) = cos(x) is a solution of the differential equation
y ′′ + y = 0
Since y ′ (x) = − sin(x) and y ′′ (x) = − cos(x).
Example 1.1.2. The function y(x) = e2x is a solution of the differential equation
y ′ − 2y = 0
Since y ′ (x) = 2e2x .
It is important to mention that in Examples 1.1.1 and 1.1.2, if we multiply any of the
two solutions by a real constant it is still going to be a solution. Therefore, when asking if a
differential equation has an unique solution, we need to impose certain conditions that our
solutions must satisfy.
Definition 3 (Initial Value Problem). An Initial value problem (denoted by IVP) is
given by a differential equation and numerical conditions that our solution must satisfy. The
numerical conditions are imposed on solution of the differential equation and its derivatives
and this conditions are called initial conditions.
Normally, we will require to have as many conditions as the order of the differential equation since we are trying to determine a unique solution but some applications and examples
prevent us from getting all the information that we need from the equation. Nevertheless,
consider the following example where we add a condition to Example 1.1.2.
Example 1.1.3. The function y(x) = e2x is the unique solution of the initial value problem
given by
(
y ′ − 2y = 0
y(0) = 1
Since y ′ (x) = 2e2x and y(0) = 1. We will discuss the reason for uniqueness in further
sections.
Observation 1.1.4. When solving initial value problems, our solution f (x) must be continuous and differentiable as many times is required by the equation in order to be considered
a solution, on an interval I that contains all the points where the initial conditions are
imposed.
Example 1.1.5. The function y(x) = x2 is the unique solution for the IVP given by
(
y ′ = − xy
y(1) = 2
Note that as presented in Observation 1.1.4, the domain of the function y(x) is given by
(−∞, 0) ∪ (0, ∞). However, since we want our solution to be continuous and differentiable
on an interval that contains the initial condition (in this case is x0 = 1), then the largest
interval where this solution exists is (0, ∞).
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9
Definition 4 (Maximum interval of existence). Given that f (x) is a solution of an IVP,
then the Maximum interval of existence of a solution, I, is the largest sub interval of
the domain of the function where the function is continuous, differentiable and I contains
all points where the initial conditions are evaluated.
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Existence and Uniqueness theory
Now that the notation is established and the basics were already introduced, we are interesting in answering the following questions:
1. Does an initial value problem have a solution?
2. If it does have a solution, is that solution unique?
The following Theorem due to Charles Picard gives a sufficient condition to answer both
questions.
Theorem 1.2.1 (Picard’s existence and uniqueness theorem). Given an initial value problem
y ′ = f (x, y), y(x0 ) = y0
where x0 and y0 are real numbers. Let R be the rectangular region on the real plane defined
by
R = {(x, y) : a < x < b and c < y < d}
and (x0 , y0 ) ∈ R as shown in Figure 1.1.
If f (x, y) is continuous on the rectangle, then there is a solution y(x) for an interval
∂f
is continuous in R, then the solution is
I ⊆ (a, b) that contains x0 . Also, if in addition
∂y
unique.
y
d
R
(x0 , y0 )
c
a
I
b
x
Figure 1.1: Example of a rectangular region R where we can find
a solution of the differential equation (blue line) on a certain
interval I represented by the dashed blue lines.
Lecture Notes (version June 1, 2025)
11
Observation 1.2.2. Note that Theorem 1.2.1 is only saying that if we can prove certain
behavior of f (x, y), we can conclude that there is a solution of the differential equation.
However, it is not stating that if one of the conditions of continuity is not achieved, then we
can conclude the opposite of the theorem. (Non-solvable or non-uniqueness)
Example 1.2.3. Consider the following initial value problem
(
y ′ = y 1/3
y(0) = 0
Note that the function f (x, y) = y 1/3 is continuous at any point. But when we take the
partial derivative with respect to y, we get:
∂f
∂ 1/3 1 1
=
y
=
∂y
∂y
3 y 2/3
which is not continuous at any point with y-coordinate equal to 0. Moreover, since we want
our initial condition to lie inside our rectangle, we get that the partial derivative is not
continuous in any rectangle to which we would apply Theorem 1.2.1.
However, since f (x, y) is continuous, by Theorem 1.2.1 there is a solution of the differential equation but since the partial derivative is not continuous, we cannot conclude from
the theorem whether it is unique or not.
In particular, we can check that the constant function y(x) = 0 and the piece-wise
function:
(
3/2
2
, x≥0
3x
f (x) =
0,
x<0
are both solutions of the IVP.
Example 1.2.4. Consider the IVP given by
(
y ′ = x2 + y 2
y(0) = 4
Note that in this case we have that f (x, y) = x2 + y 2 is a continuous function for all real
numbers and if we take the partial derivative with respect to y, we get that:
∂
∂f
=
x2 + y 2 = 2y
∂y
∂y
is also a continuous function for all real numbers. Hence, there is a unique solution of the
differential equation but there is no simple explicit form for the solution.
All the readers should be aware that concluding that a solution exists from a given
differential equation is far different from exhibiting an explicit function that solves the
equation. Nevertheless, we will be working on equations where we can compute an explicit
solution but we will give an introduction on how to approximate solutions by series in future
chapters.
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Geometric approach: Direction fields
After studying our first type of differential equations, we were able to find an explicit or
implicit solution to a differential equation, but most of the times it is difficult to even
rearrange the equation into a separable form. In this chapter we introduce a geometrical
approach to understand the behavior of a solution of an initial value problem by constructing
the direction field of the equation.
Definition 5 (Direction field). Given an ODE of the form y ′ = f (x, y) and fixing a rectangular region R, the graph that contains lines representing the slope of the solution of the
ODE at the points of the rectangular grid is called the direction field of the equation.
Note that this approach uses the fact that having an equation of the form y ′ = f (x, y)
implies that evaluating the function f (x, y) at a point (a, b) is giving as a result the slop of
the solution y(x) at the point (a, b). Then, we can approximate the behavior of the solution
by "connecting" the slopes given by the method. Let’s observe how it works in the following
example:
Example 1.3.1. Consider the following IVP

2
y ′ = x(y − 1)
y

y(1) = 2
We would like to study this differential equation using a geometrical approach by constructing the direction field. (Clearly it is a separable equation and one of the exercises of the
previous section deals with the solution of the IVP)
The method that we use to construct the direction field is called the Method of isoclines, because we want to study first the values of (x, y) that generate constant values on
the slope of the solution. First, we will define our grid to be the following:
y
2
(1, 2)
1
-2
-1
1
2
x
-1
-2
Observe that the point in green represents the point where our solution should go through.
Now, we will develop our method in three key steps:
Lecture Notes (version June 1, 2025)
13
1. First, we will try to find if there are points where we get y ′ = ±∞. If there are any
points that satisfy that, we will get that our possible solutions cannot pass through
those points. In our example, observe that for any point with y = 0 (i.e. any point
over the x-axis), our function f (x, y) goes to infinity. With that in mind, we will draw
a line that cannot be crossed on our direction field in red as follows:
y
2
1
-2
-1
1
2
x
-1
-2
2. Secondly, we will try to find the points where y ′ = 0 since slope zero implies that our
solutions are horizontal. For our given function, observe that whenever y = ±1 our
function vanishes. A similar behavior is obtained when x = 0 but only if y ̸= 0.
Therefore, we will draw the lines x = 0 and y = ±1 as lines that our solutions must
cross horizontally in blue.
y
2
1
-2
-1
1
2
x
-1
-2
3. Notice that the red and cyan lines are dividing the plane into regions where the function
f (x, y) is neither zero nor infinite. Hence, the sign of the derivative is constant on
these regions, i.e. the direction of the derivative is the same. In our case, we label the
regions separated by the red and cyan lines as follows
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y
(1, 2)
2
I
II
1
III
-2
IV
-1
1
V
x
2
VI
-1
VII
VIII
-2
Since we are interested in our solution passing through the point (1, 2), we will study regions
I and II only for this example. To do that, we just need to evaluate our function f (x, y) at
one point lying inside of each region to determine how our solution behaves:
(−1) · (2)2 − 1
3
For I, we pick the point (−1, 2) to obtain: f (−1, 2) =
=−
2 2
(1) · (2)2 − 1
3
=
For II, we pick the point (1, 2) to obtain: f (1, 2) =
2
2
Hence, we get that in the region I the solution has negative slope meaning that it is decreasing, whereas in region II the solution has positive slope and so it is increasing in that
region.
Then, focusing at regions near our initial condition and following the rules stated during
this procedure, we get the approximate solution given by the black line presented below
y
(1, 2)
1
-2
-1
1
-1
-2
2
x
Lecture Notes (version June 1, 2025)
1.4
15
Autonomous differential equations
Definition 6 (Autonomous equations). An autonomous ordinary differential equation is an ODE where there is no explicit dependence on the independent variable in the
equation, meaning that the equation has the form:
y ′ = f (y)
Note that since only the dependent variable appears on the right hand side, the constant
solutions y(x) = c where f (c) = 0 are solutions of the differential equation. This type of
solutions are called singular solutions of the equation and they determine the asymptotic
behavior of non-constant solutions of the equation depending on the sign of f (y) between
those points.
Example 1.4.1. Consider the ODE y ′ = y 2 (4 − y 2 ). Note that this is an autonomous
equation since there is no explicit dependence on the independent variable and in this
example we can find the critical solution by solving the following equality:
0 = y 2 (4 − y 2 ) = y 2 (2 − y)(2 + y)
Note that this equality holds if y(x) = 0, y(x) = 2 or y(x) = −2 which are our singular
solutions to the ODE and looking at the direction field only considering the information
given by the singular solutions we get the following
y
4
y(x) = 2
2
-4
-2
2
4
y(x) = 0
y(x) = −2
-2
-4
Observe that these singular solutions are splitting the plane into regions where all the other
non-constant solutions must lie in, then we just have to classify the growth of any solution
near the singular solutions of the ODE by calculating the sign of f (y) to understand how
the solutions behave:
−2
Functions
0
2
y
y2
+
+
+
+
2−y
+
+
+
−
2+y
−
+
+
+
f (y)
−
+
+
−
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The graph above shows that f (y) is positive for all y ∈ (−2, 0) ∪ (0, 2) meaning that the
solutions are increasing on that region, and f (y) is negative for y ∈ (−∞, −2) ∪ (2, ∞)
meaning that solutions are decreasing in those regions, which we can visualize as follows
y(x) = 2, Attractor
y(x) = 0, Semi-stable
y(x) = −2, Repeller
Lecture Notes (version June 1, 2025)
1.5
17
First order separable differential equations
Note. The main tool that we will use for this section include integration by parts, partial
fraction decomposition and substitution techniques. Any reader should be familiar with
these techniques or recall them before the start of this chapter. Now we are going to study
a particular type of ODEs which can be solved by separating both sides of the equality and
integrating.
dy
= f (x, y) is
Definition 7 (Separable equations). A first order ODE in explicit form dx
called separable if it can be arranged so that we get:
dy
= h(x)g(y)
dx
Solution Technique [Separable equations] The standard procedure to solve a
separable equation is described below:
1. Rearrange the differential equation to make it of the form:
dy
= h(x)g(y)
dx
2. If g(y) ̸= 0, then we can rewrite the equation as
dy
= h(x) dx
g(y)
3. Finally, since both sides of the equality depend on a single variable, we integrate both sides and set a constant value out of the indefinite integral as
follows:
Z
Z
1
dy = h(x) dx + C
g(y)
Warning! This integral may not yield an explicit definition of y(x). However,
if we have an implicit definition of our target function we can use implicit
differentiation to make sure that we got the correct solution.
Example 1.5.1. Consider the differential equation y ′ ln |y| + x2 y = 0. First, we will rearrange the equation as follows
ln |y|
dy
= −x2 y
dx
Now, observe that if we divide by y both sides of the equality, we get the equality of the
integrals
Z
Z
ln |y|
x3
dy = −x2 dx + C = − + C
y
3
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Finally, we only need to solve the left integral by using the substitution and letting u = ln |y|
Z
Z
2
ln |y|
u2
(ln |y|)
dy = u du =
=
y
2
2
Therefore, we get the implicit solution of the differential equation given by
2
(ln |y|)
x3
=− +C
2
3
Observe that the constant C cannot be determined unless we impose initial conditions into
the differential equation so we must include it into our solution.
Example 1.5.2. Consider the following IVP:
y ′ = ex−y ,
y(1) = 1
Even though the previous equation doesn’t seem to be separable, we can separate it using
the following property of the exponential function:
y ′ = ex−y = ex e−y ⇐⇒
dy
= ex dx ⇐⇒ ey dy = ex dx
e−y
After integrating both sides of the equality we can find the family of solutions given by
y(x) = ln (ex + C). Finally, we can evaluate the initial condition of y(1) = 1 to find the
value of C that solves the equation.
1 = y(1) = ln e1 + C = ln (e + C) =⇒ e = e + C =⇒ C = 0
Hence the solution is given by
y(x) = ln (ex + 0) = ln (ex ) = x
Example 1.5.3. Find the solution for the IVP
y′ =
xy
(x2 + 1)(x + 1)
,
y(0) = e2
Note that this is a separable equation and we can separate both variables in the following
form:
x
dy
dx
= 2
y
(x + 1)(x + 1)
The left side of the equation can be easily integrated to obtain ln(y) whereas the right side
seems to be more complicated. However, we can use the partial fraction decomposition
procedure as follows:
x
(x2 + 1)(x + 1)
=
Ax + B
C
+
⇐⇒ x = (Ax + B) (x + 1) + (C) (x2 + 1)
2
x +1
x+1
where the values of A, B and C must be determined. In order to find these values, we simply
evaluate the equality at different values of x instead of solving a system of equations:
Lecture Notes (version June 1, 2025)
19
• At x = −1, we obtain the equality:
−1 = (A(−1) + B) (−1 + 1) + (C) ((−1)2 + 1) = 2C =⇒ C = −
1
2
• At x = 0, we obtain the equality:
0 = (A(0) + B) (0 + 1) + (C) (02 + 1) = B + C =⇒ B =
1
2
• At x = 1, we obtain the equality:
1 = (A(1) + B) (1 + 1) + (C) (12 + 1) = 2A + 2B + 2C =⇒ A =
1
2
Hence, the integral that we need to solve is given by:
Z Z
Z
Z
1 x
1 1
1 1
x
1
1
1
1
1
+
−
dx +
dx −
dx
dx =
2
2
2
2
2x +1 2x +1 2x+1
2
x +1
2
x +1
2
x+1
1
1
1
= ln(x2 + 1) + arctan(x) − ln(x + 1) + C
4
2
2
Getting the implicit description of the solution:
ln(y) =
1
1
1
ln(x2 + 1) + arctan(x) − ln(x + 1) + C
4
2
2
Now, we can find the value of C using the condition that y(0) = e2 .
ln(e2 ) =
1
1
1
1
1
1
ln(0 + 1) + arctan(0) − ln(0 + 1) + C ⇐⇒ 2 = · 0 + · 0 − · 0 + C = C
4
2
2
4
2
2
Concluding that the solution of the ODE is given by the implicit formula:
ln(y) =
1
1
1
ln(x2 + 1) + arctan(x) − ln(x + 1) + 2
4
2
2
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20
First order linear differential equations
Recall that a first order linear differential equation has the general form
a1 (x)
dy
+ a0 (x)y = g(x),
dx
y(x0 ) = y0
where a1 (x), a0 (x) and g(x) are continuous functions and a1 (x) is nonzero on an interval
containing x0 . Then we can manipulate the equation to obtain the form:
dy
+ P (x)y = Q(x)
dx
Now, we want to reduce the problem of solving a linear equation to a problem that we
already know how to solve. In this case, we will multiply the differential equation by an
integrating factor µ(x) so that we can transform our linear equation into a separable one as
it is outlined in the following technique.
Solution Technique [First order linear equations] Given a first order equation
of the form
a1 (x)
dy
+ a0 (x)y = g(x),
dx
y(x0 ) = y0
with the properties listed before for the functions a1 (x), a0 and g(x), we proceed
as follows:
1. Rearrange the equation to change it into the form:
dy
+ P (x)y = Q(x)
dx
2. Compute the integrating factor µ(x) given by
R
µ(x) = e P (s) ds
3. Compute the solution of the equation given by:
Z
1
y(x) =
µ(s)Q(s) ds + C
µ(x)
4. If an initial condition is given, use it to find the value of the constant C.
Example 1.6.1. Consider the IVP
(x + 2)y ′ + 4y =
1 + 2x2
,
x(x + 2)3
y(−1) = 2
2
1+2x
In this case, a1 (x) = x + 2, a0 (x) = 4 and b(x) = x(x+2)
3 . Since all of then are continuous
and a1 is non-zero on an interval containing −1, there is a solution of the IVP that we can
Lecture Notes (version June 1, 2025)
21
find using our integrating factor procedure. First, we isolate y ′ to get:
y′ +
4
x+2
y=
1 + 2x2
x(x + 2)4
Hence, we first compute the integrating factor µ(x) by the following integral
µ(x) = e
R
4
s+2 ds
= e4 ln(x+2) = (x + 2)4
Then, the solution has the form:
Z
1
y(x) =
µ(s)Q(s) ds + C
µ(x)
Z 1 + 2s2
1
4
(s + 2) ·
ds + C
=
(x + 2)4
s(s + 2)4
Z
1
1 + 2s2
=
ds
+
C
(x + 2)4
s
1
=
ln(|x|) + x2 + C
(x + 2)4
Finally, we can find the value of C by using the fact that y(−1) = 2:
1
ln(1) + (−1)2 + C
4
(−1 + 2)
=1+C ⇒C =1
2 = y(−1) =
Concluding that the solution to this equation is:
y(x) =
1
ln(|x|) + x2 + 1
4
(x + 2)
Example 1.6.2. Find the solution of the IVP:
sin(x)y ′ + 2 cos(x)y =
e3x
,
sin(x)
y
π
2
=0
Note that this is a linear equation but we need to isolate the function y ′ by diving the
whole ODE by sin(x) to obtain:
y′ + 2
cos(x)
e3x
y=
sin(x)
sin2 (x)
Note that this is a linear equation, so we start by computing the integrating factor
R
cos(t)
µ(x) = e 2 sin(t) dt = e2 ln(sin(x)) = sin2 (x)
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Then we solve for y(x) as follows:
Z
3x e
1
2
sin
(x)
dx
+
C
sin2 (x)
sin2 (x)
Z
1
=
e3x dx + C
2
sin (x)
1
1 3x
=
e
+
C
sin2 (x) 3
y(x) =
Now we can find the value of the constant C using the initial condition:
π
1
1 3π
1 3π
1 3π
2 + C
=
0=y
e
= e 2 + C ⇐⇒ C = − e 2
2
π
2
3
3
3
sin 2
Therefore, the solution of the ODE is given by:
1 3x 1 3π
1
2
e
−
e
y(x) =
3
sin2 (x) 3
22
Lecture Notes (version June 1, 2025)
1.7
23
Applications of first order differential equations
After understanding how first order differential equations work and introducing distinct
solution methods, we can start to develop some ground work on basic applications of these
type of equations. The applications that we will discuss are: Population dynamics, Newton’s
cooling law and Mixing problems.
1.7.1
Population Growth
One of the most important applications of differential equation arise when modeling living
behavior of certain organism. For this section, we introduce two distinct models that explain
how a population behaves.
First, we assume that a given population or system grows/decays proportional to its
population and so it can be represented by the following differential equation
dP
= aP,
dt
P (0) = P0 > 0
Observe that it is another separable equation and the general solution of this problem is
given by P (t) = P0 eat . Observe that this behavior means that the population grows unbounded in time and so it is called exponential growth
Another possible scenario is that the total population P (t) grows until it reaches a certain
capacity and so it satisfies the differential equation
dP
= P (a − bP ),
dt
P (0) = P0
with a, b > 0 and P0 > 0. Note that this is a separable ODE and we can use our method of
solving separable equations to get the following solution:
P (t) =
aP0
bP0 + (a − bP0 )e−at
This behavior is called a logistic growth and its relation with the exponential growth can
be observed in Figure 1.2.
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24
Figure 1.2: The blue line represents the logistic growth of a certain population whereas the red line shows the exponential behavior of the same
growth system. Notice that only the Logistic solution remains bounded
in time and asymptotically goes to the value ab .
1.7.2
Newton’s Cooling Law
According to Newton’s work on heat transfer,
"The rate of change of temperature of an object is proportional to the difference in
temperature between the object and its surroundings"
So the temperature of a body can be described by the following differential equation
dT
= k(T − Tm ),
dt
T (0) = T0
where Tm is the temperature of the ambient or the reservoir and k is a proportion constant.
Moreover, this is another separable equation and the general solution is given by:
T (t) = Tm + (T0 − Tm )ekt
Note that in this equation we can either get the value of k from the differential equation or
we could obtain an extra condition from the statement of the problem that is going to allow
us to obtain the value of k.
Lecture Notes (version June 1, 2025)
1.7.3
25
Mixing Problems
For this type of problems, we will consider a saltwater solution that is added at a specific
rate to a tank with a different concentration and from the same tank the solution is being
drained at another rate as in next figure.
Q(t)=?
rate in
Q(0) = Q0
rate out
Our goal is to determine the concentration in the tank as a function of time, Q(t). Our idea
is that the concentration must satisfy the following differential equation:
dQ
= rate in − rate out
dt
Example 1.7.1. Given that a tank initially contains 40 pounds of salt dissolved in 600
gallons of water. Starting at t = 0, water that contains 1/2 pounds of salt per gallon is
poured into the tank at a rate of 4 gal/min and the mixture is drained from the tank at the
same rate, we want to find Q(t) which is the amount of salt in the tank at a time t.
We know that Q(t) satisfies the differential equation
dQ
= rate in − rate out
dt
so it suffices to find the rate in of salt into the tank and the rate out as well.
• Rate in: Water that contains 1/2 pounds of water per gallon is poured at a rate of
4 gal/min. Then we get that the rate in is:
rate in = (0.5 pounds/gal)(4 gal/min)
= 2 pounds/min
• Rate out: Since the mixture is drained from the tank at 4 gal/min and the amount
of salt per gallon that is drained is equal to Q(t)/600 (where 600 is the total amount
of water in the tank.
rate out = (Q(t)/600 pounds/gal)(4 gal/min)
=
Q(t)
pounds/min
150
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26
Therefore, the differential equation is given by
dQ
Q(t)
dQ Q(t)
= rate in − rate out = 2 −
⇐⇒
+
=2
dt
150
dt
150
Note that this is a linear differential equation, so we start by computing the integrating
factor µ(t) which is given by the following integral:
µ(t) = e
R
1
150 ds
= et/150
Hence the solution Q(t) has the following form:
Z
1
2es/150 ds + C
Q(t) = t/150
e
i
1 h
= t/150 300et/150 + C
e
= 300 + Ce−t/150
Using that Q(0) = 40, we can obtain that C = −260 and so the solution is:
Q(t) = 300 − 260e−t/150
Chapter 2
Linear Differential Equations
"You know my methods, Watson.
There was not one of them which I did not apply to the inquiry."
Sherlock Holmes in "The Crooked Man".
Sir Arthur Conan Doyle, The Memoirs of Sherlock Holmes.
At this time, we already introduced the standard methods to solve a first order ODE and
most of them worked similarly: We started from a possibly complicated equation and after several transformations (changing the variables of differentiation or multiplying by an
integrating factor), we got a separable equation which we could easily solve to obtain our
desired solution.
Notice that for higher order equations, these methods must be generalized to a, possibly, loss of information from whole functions. In this chapter, we will start our discussion
with homogeneous linear equations which are different from the first order homogeneous
equations previously discussed, then we will state a general solution for n-th order linear
equations with only constant coefficients.
We conclude the chapter with three distinct methods to solve non-homogeneous equations: Undetermined coefficients and variation of parameters. Followed by the Cauchy-Euler
equations which are a straightforward application of the solution methods.
27
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Homogeneous linear equations
Now that we are moving one order in our solutions, we will first classify linear differential
equations depending on their homogeneity as presented in the following definition.
Definition 8 (Homogeneous linear equations). A linear differential equation is called homogeneous if it can be written in the form
y ′′ + p(x)y ′ + q(x)y = 0
Furthermore, if the equation has the form:
y ′′ + p(x)y ′ + q(x)y = g(x)
is called non-homogeneous.
Similar as in the previous chapter, we will first give a sufficient condition for the existence and uniqueness of solutions for a given initial value problem with a second order
homogeneous linear equation.
Theorem 2.1.1 (Existence and Uniqueness theorem for linear ODEs). Suppose that p(x)
and q(x) are continuous on an open interval (a, b), x0 ∈ (a, b) be any point. Then the initial
value problem
y ′′ + p(x)y ′ + q(x)y = 0,
y(x0 ) = y0 ,
y ′ (x0 ) = y1
has a unique solution on (a, b). Before we continue working towards finding solutions of
higher order equations, consider the following example:
Example 2.1.2. For the differential equation
y ′′ + y = 0
Both y(x) = sin(x) and y(x) = cos(x) are both solutions of the equation but given an initial
condition, we should be able to determine uniquely a solution for this equation. Hence, we
have our first problem, how can we possibly know what is the correct linear combination that
we should pick as solution of our equation? Or are both solutions that we found necessary
to describe all possible functions that solve the equation?.
The issue that we observed on the previous example is summarized in the following
theorem:
Theorem 2.1.3. If y1 and y2 are solutions of the homogeneous equation
y ′′ + p(x)y ′ + q(x)y = 0
on (a, b), then any linear combination
y = c1 y1 + c2 y2
is a solution of the homogeneous equation on (a, b). It seems that now the problem is being
reduced into finding a family of functions that are capable of describing all possible solutions
of a homogeneous equation, those functions deserve a specific name.
Lecture Notes (version June 1, 2025)
29
Definition 9 (Fundamental set of solutions). A family of functions {y1 , . . . , yn } is a fundamental set of solutions on (a, b) of an nth order homogeneous equation of the form:
y (n) + an−1 (x)y (n−1) + · · · + a1 (x)y ′ + a0 (x)y = 0
if every solution of the equation on (a, b) can be written as a linear combination of the functions on the family.
Also, the general linear combination
y(x) = c1 y1 (x) + · · · + cn yn (x)
is called the general solution of the equation on (a, b).
Now we want to start working towards finding a fundamental set of solutions of an
equation, so the first idea that we have to comprehend is the independence of functions on
an interval.
Definition 10 (Linear independence of functions on an interval). A family of functions
{y1 , . . . , yn } is called linearly independent on an interval I if the only selection of real
constants c1 , . . . , cn such that
c1 y1 (x) + · · · + cn yn (x) = 0,
for all x ∈ I
is the trivial selection, i.e. each ci = 0.
This definition generalizes the definition of linear independence of vectors in a certain
vector space. However, we need to find an easier way to determine whether two functions
are linearly independent or not, to do that we introduce the following operator:
Definition 11 (Wronskian). Given a set of functions {y1 , . . . , yn }, we define the Wronskian of the set as:
y1
y1′
W (y1 , . . . , yn ) = ..
.
(n)
y1
···
···
···
···
yn
yn′
..
.
(n)
yn
Lemma 2.1.4. Two functions y1 and y2 are linearly independent on any interval I where
the Wronskian W (y1 , y2 ) has no zeros.
Example 2.1.5. Consider the functions y1 (x) = cos(x) and y2 (x) = sin(x), we will prove
that both functions are linearly independent for all R by calculating their Wronskian:
W (sin(x), cos(x)) =
sin(x) cos(x)
= − sin2 (x) − cos2 (x) = −1
cos(x) − sin(x)
Since the Wronskian has no zeros, we conclude that both functions are linearly independent
on any interval of R.
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30
Example 2.1.6. Consider the functions y1 (x) = x and y2 (x) = ex , we will conclude that
these functions are linearly independent but not in all of R.
W (x, ex ) =
x
1
ex
= xex − ex = ex (x − 1)
ex
Notice that this Wronskian vanishes at x = 1. Therefore, both functions x and ex are
linearly independent in any interval of R that does not contain 1. Because if the interval
contained 1, we can show that
0 = e ∗ (y1 (1)) − 1 ∗ (y2 (1))
so there is a non-trivial linear combination of y1 and y2 that vanishes at x = 1.
Similar to linear algebra, the property that we want from our Wronskian is for it to not
be zero because that will give us an intuition of what is the linear dependence between our
family of functions.
Theorem 2.1.7 (Abel’s Formula). If y1 and y2 are solutions of the homogeneous equation
y ′′ + p(x)y ′ + q(x)y = 0
on (a, b). Let x0 be any point in (a, b). Then,
−
W (y1 , y2 )(x) = W (y1 , y2 )(x0 )e
Rx
p(t) dt
x0
for any x ∈ (a, b). So, either the Wronskian vanishes on (a, b), or it has no zeros in the
interval.
Notice that the first big assumption on this theorem is that the functions y1 and y2 are
solutions of the homogeneous equation so that we can get the desired behavior from the
Wronskian. Now, the following theorem represents the importance of using the Wronskian
to characterize independence of our solutions:
Theorem 2.1.8. Suppose that p(x) and q(x) are continuous on an open interval (a, b), let
y1 and y2 be two solutions of
y ′′ + p(x)y ′ + q(x)y = 0
on (a, b). Then y1 and y2 are linearly independent on (a, b) if and only if W (y1 , y2 ) is non
zero on (a, b).
Lecture Notes (version June 1, 2025)
2.2
31
Solution of nth order linear equations with constant
coefficients
In this section we will focus in finding fundamental sets of solutions for linear equations
with constant coefficients. Before we introduce our method to find a fundamental set, we
will start with the following example:
Example 2.2.1. Let m1 ̸= m2 be two complex numbers, then we will show that em1 x and
em2 x are linearly independent functions. In order to do that, we will compute the Wronskian
for both functions as follows:
W (em1 x , em2 x ) =
em 1 x
m1 em1 x
em 2 x
m2 em2 x
= m2 e(m1 +m2 )x − m1 e(m1 +m2 )x
= (m2 − m1 )e(m1 +m2 )x
which is never zero by our assumptions on m1 and m2 . So both exponential functions are
linearly independent functions on all R.
Hence, it seems like having two exponential functions to a distinct power of the form rx
is giving us a pair of linearly independent functions. Now, we will focus on solving a second
order linear equation with constant coefficients which has the general form
ay ′′ + by ′ + cy = 0
The way that we start solving this type of equations is by assuming that our solution is
given by the form erx since Example 2.2.1 shows that exponential functions of this form are
linearly independent. First, we will observe how this equation looks like if we assume that
the solution is given by the exponential form mentioned:
′′
′
0 = a (erx ) + b (erx ) + c (erx )
= ar2 (erx ) + br (erx ) + c (erx )
= (erx ) ar2 + br + c
Note that since erx is not vanishing, then the only way that an exponential of that form is
a solution is for our value of r to be a root of the polynomial on the right.
Definition 12 (Auxiliary Equation). Given an nth order linear equation with constant
coefficients:
an y (n) + an−1 y (n−1) + · · · + a1 y ′ + a0 y = 0
where a1 , . . . , an are real constants. Then the auxiliary equation for this ODE is given by
an rn + an−1 rn−1 + · · · + a1 r + a0 = 0
and the roots of this equations give you the exponential functions that appear as the fundamental set of solutions.
Engineering Mathematics III (v. June 1, 2025)
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32
With the previous definition in mind, we start introducing the following examples:
Example 2.2.2. We will find the general solution of the ODE
y ′′ + y ′ − 6y = 0
In order to do that, we will find the roots of the auxiliary equation which is given by the
polynomial
0 = r2 + r − 6 = (r + 3)(r − 2)
Hence, the values of r = −3 and r = 2 are two distinct real solutions of the equation.
Therefore, the exponential functions e−3x and e2x are linearly independent functions that
satisfy the ODE.
Which implies that both of them form a fundamental set of solutions of the equation
and so the general solution is given by
y(x) = c1 e−3x + c2 e2x
Example 2.2.3. Now we consider the differential equation
y ′′ − 2y ′ + y = 0
First, we find the roots of the auxiliary polynomial as follows:
0 = r2 − 2r + 1 = (r − 1)2
Observe that now we have only one root which is r = 1 and it has multiplicity 2, i.e. it is
repeated twice as a root. So the method says that ex is one solution of the equation and in
order to find another solution which still has an exponential of the form ex in it, we must
multiply by x as many times as the multiplicity minus one.
In this case with multiplicity 2, it means that two linearly independent functions will
have a factor of ex . One of them is ex and the second one is given by xex . Before we
conclude with the general solution, we will show that the functions ex and xex are linearly
independent by computing the following Wronskian
W (ex , xex ) =
ex
ex
xex
= xe2x + e2x − xe2x = e2x ̸= 0
e + xex
x
Since they are linearly independent solutions of the equation, the general solution of the
ODE is given by
y(x) = c1 ex + c2 xex
Example 2.2.4. Finally, we consider the following differential equation
y ′′ − 2y ′ + 2y = 0
Lecture Notes (version June 1, 2025)
33
Following the same procedure as the before, we will find the roots of the auxiliary equation
which is given by r2 − 2r + 2. Note that our previous factorization method seems to fail on
this example, so we will use the quadratic formula to find the roots of the polynomial:
√
√
2± 4−4∗2
2 ± −4
2 ± 2i
=
=
=1±i
r=
2
2
2
This means that the roots of the equation are complex roots that happen to be conjugates
from one another. However, our method is still going to work because we can write down
our exponential as follows:
e(1±i)x = ex e±ix = ex (cos(±x) + i sin(±x)) = ex (cos(x) ± i sin(x))
So our complex roots are generating an exponential multiplied by cosine and sine terms.
Those are going to be our family of linear independent functions y1 (x) = ex cos(x) and
y2 (x) = ex sin(x). First, we will prove that those functions happen to be linearly independent
by computing the Wronskian:
ex cos(x)
ex sin(x)
x
x
e cos(x) − e sin(x) e sin(x) + ex cos(x)
= e2x sin(x) cos(x) + cos2 (x) − e2x sin(x) cos(x) − sin2 (x)
W (y1 , y2 ) =
x
= e2x (sin2 (x) + cos2 (x)) = e2x ̸= 0
So they are linearly independent functions on all R and the reader can check that they are
solutions of the equation. Therefore, they form a fundamental set and the general solution
is given by
y(x) = c1 ex cos(x) + c2 ex sin(x)
The previous computations can be summarized in the following theorem
Theorem 2.2.5. Let p(r) be the characteristic polynomial of the second order linear equation with constant coefficients:
ay ′′ + by ′ + cy = 0
Then we get the following behavior:
• If p(r) = 0 for two distinct real roots r1 , r2 , then the general solution of the equation
is given by:
y(x) = c1 er1 x + c2 er2 x
• If p(r) = 0 for a unique repeated real root r, then the general solution of the equation
is:
y(x) = c1 erx + c2 xerx
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34
• If p(r) = 0 for two conjugate complex roots of the form r = α + iβ, then the general
solution of the equation is:
y(x) = c1 eαx cos(βx) + c2 eαx sin(βx)
Furthermore, for higher order equations we just follow the same procedure as before but
now we take care of each one of the possible roots and their multiplicities one by one as in
the next example:
Example 2.2.6. Consider the following linear ODE with constant coefficients
y (v) − 2y (iv) + 2y ′′′ − 4y ′′ + y ′ − 2y = 0
Note that this is a fifth order equation, meaning that we need to find 5 different solutions
that are linearly independent. So, we start by computing the auxiliary equation and finding
the roots of the equation:
0 = r5 − 2r4 + 2r3 − 4r2 + r − 2
= r5 + 2r3 + r + −2r4 − 4r2 − 2
{z
} |
{z
}
|
odd powers
4
even powers
2
= r(r + 2r + 1) − 2(r4 + 2r2 + 1)
= (r − 2)(r2 + 1)2
So we have two distinct roots of the equation, and we just need to study each one separately:
• Starting with r = 2, this root appears only once so the exponential y1 (x) = e2x must
be part of the general solution.
• Now, the complex root r = i appears twice on the factorization. Hence, there must be
two functions that contain the information of the exponential function eix = cos(x) +
i sin(x). So, we are not only adding the solutions y2 (x) = cos(x) and y3 (x) = sin(x)
as part of this root, we also have to multiply each one of them by x to obtain the
information from the second order of the complex root, obtaining the remaining two
functions y4 (x) = x cos(x) and y5 (x) = x sin(x)
Therefore, the general solution of this fifth order equation is given by
y(x) = c1 y1 + c2 y2 + c3 y3 + c4 y4 + c5 y5
=
c1 e2x
| {z }
comes from r = 2
+ c2 cos(x) + c3 sin(x) + c4 x cos(x) + c5 x sin(x)
|
{z
}
comes from r = i
Lecture Notes (version June 1, 2025)
2.3
35
Linear Operators
On the previous section, we learned how to solve linear ODEs with constant coefficients
that were homogeneous. Now, we need to construct the basic facts and definitions for
non-homogeneous problems to obtain a solution that includes the information to achieve a
distinct function from zero in the differential equation. So we will consider a general linear
equation of order n given by
an (x)y (n) + · · · + a1 (x)y ′ + a0 y = g(x)
(2.1)
Definition 13 (Homogeneous solution). Given a differential equation as 2.1, a function
yh (x) is called a solution of the homogeneous problem if it satisfies
(n)
an (x)yh + · · · + a1 (x)yh′ + a0 yh = 0
Up to this point, we know how to find yh for ODEs with constant coefficients, for us to
find a solution that also takes into account the function g(x), we introduce the following
definition.
Definition 14 (Linear operator). A functional L that takes continuous functions as inputs
is called a linear operator if for any real numbers α and β, continuous functions y1 (x)
and y2 (x), the following holds:
L[αy1 + βy2 ] = αL[y1 ] + βL[y2 ]
In particular for an ODE as 2.1, we can define the following linear operator:
L[y] = an (x)y (n) + · · · + a1 (x)y ′ + a0 y
Now, we are interested in using the linearity of L[y] to find solutions for the general equation
similar to the following example:
Theorem 2.3.1 (Linear combinations applied to a linear operator). Given a linear operator
L[y] and continuous functions yi (x) for i = 1, . . . , n such that
L[yi ] = gi (x),
for i = 1, . . . , n.
Then we get that
L[α1 y1 + · · · + αn yn ] = α1 g1 + · · · + αn gn
Example 2.3.2. Consider a linear operator L[y], and functions y1 (x), y2 (x) such that
L[y1 ] = x2
L[y2 ] = ex cos(x)
Now, in case that we were looking for function y(x) such that L[y] = 2x2 − 3ex cos(x), we
use the linearity of the operator backwards as follows:
2x2 − 3ex cos(x) = 2 (L[y1 ]) − 3 (L[y2 ])
= L[2y1 ] + L[−3y2 ]
= L[2y1 − 3y2 ]
So we get that our solution y(x) = 2y1 − 3y2 .
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Now, it is time to start introducing the theory for the solution of non-homogeneous
problems. So first, we will define what type of functions we are trying to find.
Definition 15 (Particular solution). A function yp (x) is called a particular solution of
the differential equation if L[yp ] = g(x) or
an (x)yp(n) + · · · + a1 (x)yp′ + a0 yp = g(x)
Finally, we need to relate the homogeneous solution with the particular solution of the
equation which is obtained based on the following theorem.
Theorem 2.3.3 (General solution of a non-homogeneous ODE). Given a differential equation of the form
an (x)y (n) + · · · + a1 (x)y ′ + a0 y = g(x)
the general solution of the ODE is given by the sum of the homogeneous equation and the
particular equation
y(x) = yh (x) + yp (x)
Observation 2.3.4. The proof of this theorem is left as an exercise and the readers could
prove it by linearity of the operator assigned to the ODE.
Lecture Notes (version June 1, 2025)
2.4
37
Finding a particular solution
In this section we will develop three different methods to find a particular solution of a
second order differential equation of the form.
2.4.1
Undetermined coefficients
Given a second order ODE with constant coefficients as:
ay ′′ + by ′ + cy = g(x)
Our first method works as a guessing method, we will construct our particular solution yp
depending on the function g(x).
Warning: This method only works when the function g(x) is either:
• A polynomial with natural degree.
• An exponential of the form eαx .
• sin(βx) or cos(βx)
Or any sum or product of functions of this form.
To introduce our method, we will consider the following examples:
Example 2.4.1. Consider the ODE
y ′′ − 7y ′ + 12y = 4e2x
Recall that the solution of the equation has the form y(x) = yh (x) + yp (x) where yh solves
the homogeneous equation and yp is the particular solution of the ODE. So we first solve
the homogeneous equation which has the form
y ′′ − 7y ′ + 12y = 0
Observe that the auxiliary equation is r2 − 7r + 12 = 0 which has roots r = 3 and r = 4, so
the homogeneous solution is given by
yh (x) = c1 e3x + c2 e4x
Now, we have to find our particular solution yp which satisfies that
yp′′ − 7yp′ + 12yp = 4e2x
since the right hand side has an exponential of the form e2x , we will guess that our particular
solution yp (x) = Ae2x and then we will solve for our unknown coefficient A as follows:
′′
′
′′
4e2x = Ae2x − 7 Ae2x + 12 Ae2x
= 4Ae2x − 7 2Ae2x + 12 Ae2x
= 2Ae2x
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Then A = 2 which means that yp (x) = 2e2x is the particular solution of the ODE. Therefore,
the general solution is
y(x) = c1 e3x + c2 e4x + 2e2x
Example 2.4.2. Given the ODE
y ′′ − 7y ′ + 12y = 5e4x
which is exactly the same homogeneous problem as before but with a distinct function g(x),
so the homogeneous solution is yh (x) = c1 e3x + c2 e4x . Now, observe that if we tried to
guess that yp (x) = Ae4x , we would get something that was already part of the homogeneous
solution.
Thus, we need to change that to another function that keeps the information of the
exponential but is not annihilated by the derivatives, our choice is yp (x) = Axe4x which
now satisfies the following:
′′
′
5e4x = Axe4x − 7 Axe4x + 12 Axe4x
= 16Axe4x + 4Ae4x + 4Ae4x − 7 4Axe4x + Ae4x + 12 Axe4x
= 8Ae4x − 7Ae4x
= Ae4x
Getting that A = 5, and so the general solution of the equation is
y(x) = c1 e3x + c2 e4x + 5xe4x
Solution Technique For a differential equation
ay ′′ + by ′ + cy = eαx Q(x)
where Q(x) is a polynomial of degree n, then we can get a particular solution of
the form:
• If eαx is not part of the solution of the homogeneous problem, then yp (x) =
eαx q(x) where q(x) is a generic polynomial of the same degree as Q(x).
• If eαx is part of the solution of the homogeneous problem but xeαx is not,
then yp (x) = xeαx q(x) where q(x) is a generic polynomial of the same degree
as Q(x).
• If both eαx and xeαx are part of the solution of the homogeneous problem,
then yp (x) = x2 eαx q(x) where q(x) is a generic polynomial of the same degree
as Q(x).
Observe that the previous method takes care of sums and products between polynomials
and exponential functions, now the question is what to do with sin and cos functions. So,
Lecture Notes (version June 1, 2025)
39
we will study ODEs of the form:
ay ′′ + by ′ + cy = eαx (P (x) cos(βx) + Q(x) sin(βx))
Starting with the case of α = 0, we just have an ODE of the form:
ay ′′ + by ′ + cy = P (x) cos(βx) + Q(x) sin(βx)
Theorem 2.4.3. Suppose β is a real number and k is the highest degree between the polynomials Q(x) and P (x), then the ODE
ay ′′ + by ′ + cy = P (x) cos(βx) + Q(x) sin(βx)
has a particular solution of the form:
yp = A(x) cos(βx) + B(x) sin(βx)
where A(x) and B(x) are generic polynomials of degree k. However, if cos(βx) and sin(βx)
appear in the homogeneous solution of the ODE, then the particular solution has the form:
yp = xA(x) cos(βx) + xB(x) sin(βx)
Example 2.4.4. Consider the ODE
y ′′ + y = 8 cos(2x) + 12 sin(2x)
First we solve the homogeneous problem given by y ′′ +y = 0 which has as auxiliary equation
the polynomial r2 + 1 = 0 and yields the homogeneous solution yh (x) = c1 sin(x) + c2 cos(x).
By Theorem 2.4.3, we guess the particular solution yp (x) = A cos(2x) + B sin(2x) since
none of these terms is part of the homogeneous solution. Then, we can substitute our
solution into the equation to get:
yp′′ + yp = (−4A cos(2x) − 4B sin(2x)) + (A cos(2x) + B sin(2x)) = −3A cos(2x) − 3B sin(2x)
Which implies that A = − 38 and B = −4, and the solution of the ODE becomes:
y(x) = c1 sin(x) + c2 cos(x) −
8
cos(2x) − 4 sin(2x)
3
Furthermore, the case when α ̸= 0, yields the following theorem:
Theorem 2.4.5. Given an ODE of the form
ay ′′ + by ′ + cy = eαx (P (x) cos(βx) + Q(x) sin(βx))
the substitution y = ueλx generates a linear equation of constant coefficients on the variable
u of the form:
a∗ u′′ + b∗ u′ + c∗ u = P (x) cos(βx) + Q(x) sin(βx)
Engineering Mathematics III (v. June 1, 2025)
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If we let up be the particular solution for the differential equation of the variable u, then
the function
yp (x) = up eαx
is a particular solution of the original ODE on the variable y.
Example 2.4.6. Consider the following ODE
y ′′ − y ′ + y = ex (2 + x) sin(x)
As presented on Theorem 2.4.5, we do the change of variables y = uex to obtain a differential
equation for u:
′′
′
′′
ex (2 + x) sin(x) = (uex ) − (uex ) + (uex )
= (uex + 2u′ ex + u′′ ex ) − (uex + u′ ex ) + (uex )
= (u′′ + u′ + u) ex
Getting the ODE for u to be u′′ +u′ +u = (2+x) sin(x). Now, we need to find the particular
solution of the equation to conclude the solution that we were interested in. Therefore, we
will solve the homogeneous problem first to obtain the auxiliary equation r2 + r + 1 which
has complex roots of the form:
√
√
−1 ± 1 − 4
1
3
r=
=− ±
i
2
2
2
Since sin(x) is not part of the particular solution because i is not a root of the auxiliary
equation, we get that the particular solution up and its derivatives have the form
up = (Ax + B) cos(x) + (Cx + D) sin(x)
u′p = (Cx + D + A) cos(x) + (C − B − Ax) sin(x)
u′′p = (2C − B − Ax) cos(x) + (−2A − D − Cx) sin(x)
Therefore, we get the following equality after adding up all the derivatives of up
u′′p + u′p + up = (−Ax + (C − 2A − B)) sin(x) + (Cx + (2C + A + D)) cos(x) = (2 + x) sin(x)
Now, after setting equal the coefficients at both sides of the equation, we get the following
system:


−A = 1
A = −1






C − 2A − B = 2
B = 0
=⇒




C = 0
C = 0


2C + A + D = 0
D=1
Hence we obtain that up = −x cos(x) + sin(x) and so the particular solution of the equation
is:
yp (x) = ex (−x cos(x) + sin(x))
Lecture Notes (version June 1, 2025)
41
Example 2.4.7. We find the general solution of the following equation using the method
of undetermined coefficients.
y ′′ − 3y ′ + 2y = e3x x2 + 2x − 1 + 2e4x
First, we solve the homogeneous equation given by y ′′ − 3y ′ + 2y = 0 and its auxiliary
equation has roots at r = 2 and r = 1, so the homogeneous solution is
yh (x) = c1 e2x + c2 ex
Now, observe that the right side of the original equation can be split as follows:
e3x x2 + 2x − 1 + |{z}
2e4x
|
{z
}
yp2
yp1
So we can divide our problem between finding a particular solution for yp1 and then for yp2 .
First, we will find a particular solution for yp1 , since e3x is not part of the homogeneous
solution, using the method of undetermined coefficients, we will guess that our particular
solution has the form:
yp1 (x) = e3x Ax2 + Bx + C
After computing the derivatives and replacing yp1 into the original equation we get the
following equality:
x2 + 2x − 1 = x2 (2A) + x(2B + 6A) + (2A + 3B + 2C)
Solving each coefficient, we get A = 12 , B = − 21 and C = − 41 which implies that our
particular solution is:
yp1 (x) = e
3x
1
1 2 1
x − x−
2
2
4
On the other hand, for yp2 , since e4x is not part of the homogeneous solution we can guess
that the particular solution has the form yp2 (x) = Ae4x and after substituting it into the
ODE we obtain
2e4x = 6Ae4x ⇒ A =
1
3
which means that yp2 (x) = 13 e4x and we conclude that the general solution of the ODE is
given by
y(x) = c1 e2x + c2 ex + e3x
1 2 1
1
x − x−
2
2
4
1
+ e4x
3
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42
Variation of parameters
Consider an ODE of the form
a2 (x)y ′′ + a1 (x)y ′ + a0 (x)y = f (x)
Recall that a solution of the equation has the form y(x) = yh (x) + yp (x), where yh is the
homogeneous solution of the equation and since it is a second order ODE, yh can be written
as
yh (x) = c1 y1 + c2 y2
and the functions {y1 , y2 } form a fundamental set of solutions of the ODE. So the method
of variation of parameters assumes that the particular solution has the form yp (x) =
u1 (x)y1 (x) + u2 (x)y2 (x) where the functions u1 (x) and u2 (x) have to be determined.
Solution Technique First, we will put the ODE in standard form by diving the
equation by a2 (x) to get it in the form
y ′′ + P (x)y ′ + Q(x)y = g(x)
and we define the following quantities:
W (y1 , y2 ) =
y1
y1′
y2
,
y2′
W1 =
0
y2
,
g(x) y2′
W2 =
y1
y1′
0
g(x)
Now, we can find the functions u′1 and u′2 by the following quotients
u′1 =
W1
,
W (y1 , y2 )
u′2 =
W2
W (y1 , y2 )
After integrating, we get that the particular solution yp (x) can be written as:
yp (x) = u1 y1 + u2 y2
Note that in this situation we need to find all the homogeneous solution before applying
the method, but sometimes we cannot find all the solution and the next method solves that
issue.
Example 2.4.8. We will solve the differential equation given by
y ′′ − 3y ′ + 2y =
4
1 + e−x
using the method of Variation of parameters. First, we will solve the homogeneous equation
which is
y ′′ − 3y ′ + 2y = 0
Lecture Notes (version June 1, 2025)
43
giving us the auxiliary equation r2 − 3r + 2 = 0 which has r = 1 and r = 2 as solutions.
Therefore, the homogeneous solution of the problem is
yh (x) = c1 ex + c2 e2x
Now, denoting y1 (x) = ex and y2 (x) = e2x , we compute W , W1 and W2 to proceed with
the method of variation of parameters.
W (y1 , y2 ) =
y1
y1′
ex
y2
′ =
ex
y2
e2x
= 2e3x − e3x = e3x
2e2x
W1 =
4e2x
0
y2
′ = −y2 g(x) = −
g(x) y2
1 + e−x
W1 =
y1
y1′
4ex
0
= y1 g(x) =
g(x)
1 + e−x
Then, we get that u′1 is given by the following integral:
4e2t
dt
1 + e−t
Z
1
dt Using v = et + 1, dv = et dt = v dt
= −4
t
e +1
Z
1
= −4
dv
v(v − 1)
Z
u1 (x) = −
At this moment we need to use partial fractions to write down the integral as
Z 1
1
u1 = −4
− +
dv = −4 [− ln(v) + ln(v − 1)] = 4 ln(ex + 1) − 4x
v v−1
Moreover, the reader can solve for u2 (x) to find that it is
u2 (x) = −4x + 4e−x + 4 ln (ex + 1)
Concluding that the particular solution is
yp (x) = u1 y1 + u2 y2 = 4ex ln(ex + 1) − 4xex − 4xe2x + 4ex + 4e2x ln(ex + 1)
Example 2.4.9. We find the general solution of the following equation using the method
of variation of parameters.
y ′′ + 3y ′ + 2y =
1
1 + ex
First, we solve the homogeneous equation given by y ′′ + 3y ′ + 2y = 0 and its auxiliary
equation has roots at r = −2 and r = −1, so the homogeneous solution is
yh (x) = c1 e−2x + c2 e−x
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44
Before we apply our method of variation of parameters, we need to compute the following
three functions by letting y1 = e−2x and y2 = e−x .
W (y1 , y2 ) =
W1 =
W2 =
e−2x
−2e−2x
0
1
1+ex
−2x
e−x
= −e−3x + 2e−3x = e−3x
−e−x
e−x
e−x
−x = −
−e
1 + ex
e
−2e−2x
0
1
1+ex
=
e−2x
1 + ex
Now, by the method of variation of parameters we assume that the particular solution has
the form yp = u1 y1 + u2 y2 and we compute both u1 and u2 as follows:
Z
e2t
W1
dt = −
dt Let v = 1 + et , dv = et dt
t
W
1+e
Z
v−1
=−
dv = − (v − ln(v)) = − (1 + ex ) + ln(1 + ex ) + D1
v
Z
Z
W2
et
dt =
dt Let v = 1 + et , dv = et dt
u2 =
t
W
1+e
Z
1
=
dv = ln(v) = ln(1 + ex ) + D2
v
Z
u1 =
Therefore, the particular solution is:
yp (x) = u1 y1 + u2 y2
= (D1 − (1 + ex ) + ln(1 + ex ))e−2x + (D2 + ln(1 + ex ))e−x
= (D1 − 1)e−2x + (D2 − 1)e−x + ln(1 + ex ) e−2x + e−x
Notice that the functions in red are already part of the homogeneous solution of the equation.
Therefore the particular solution is only given by:
yp (x) = ln(1 + ex ) e−2x + e−x
Concluding that the general solution of the equation is
y(x) = c1 e−2x + c2 e−x + ln(1 + ex ) e−2x + e−x
Lecture Notes (version June 1, 2025)
2.5
45
Cauchy-Euler equations
One common type of differential equation is called the Cauchy-Euler equations which
can be solved by a transformation of variables and this section introduces the procedure to
solve this type of equations.
Definition 16 (Cauchy-Euler equation). A differential equation of the form
a2 x2 y ′′ + a1 xy ′ + a0 y = g(x)
is called a Cauchy-Euler equation of second order.
This type of equations satisfy that each derivative of the dependent variable is being
multiplied by the independent variable to elevated to the same power of differentiation.
Moreover, since x2 vanishes at x = 0, the existence and uniqueness theorem implies
that we can find a unique solution for either positive or negative values of x. For each
case, a change of independent variables will reduce the problem to an ODE with constant
coefficients that we solve with one of the previous methods.
Solution Technique Assuming that we are working on the positive branch, i.e.
x > 0, we change the independent variables by the transformation x = et . Now by
letting y be a function of t, the ODE becomes
a2 y ′′ + (a1 − a2 ) y ′ + a0 y = g et
which can be solved by as an equation with constant coefficients and the particular
solution can be obtained by any of the previous methods.
In case that x < 0, we use the transformation of variables x = −et and we obtain
the equation:
a2 y ′′ + (a1 − a2 ) y ′ + a0 y = g −et
Example 2.5.1. We will solve the following Cauchy-Euler equation
x2 y ′′ − 2xy ′ + 2y = 0,
x>0
Since x > 0, we perform the change of variables x = et to get the following differential
equation for y(t)
y ′′ + (−2 − 1)y ′ + 2y = 0 ⇐⇒ y ′′ − 3y ′ + 2y = 0
which has as auxiliary equation r2 − 3r + 2 = 0 and the roots are r = 1 and r = 2, so the
solution to the equation is given by
t → ln(x)
y(t) = c1 et + c2 e2t −−−−−−→ y(x) = c1 x + c2 x2
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46
Application: Deflection of a beam
After understanding how higher order differential equations work, we develop some ground
work on a particular application of higher order equations which is the deflection of a beam.
If we study the behavior of the deflection or distortion of a beam in a structure under a
certain weight or more external forces, after using the equation of momentum of a beam and
Young’s modulus, we can model the deflection y(x) as a function of the horizontal position
x using the following equation:
α
d4 y
= w(x)
dx4
where w(x) is the load per unit length and α is the flexural rigidity.
Note. We denote that a positive deflection is generated when it goes downwards.
More importantly, we need to have a procedure to determine multiple conditions to solve
this fourth order equation. These conditions arise from the different ways that the endpoints
of the beam are supported, we can summarize these condition in the following table:
Type of endpoint
Endpoint condition
Embedded
y = 0, y ′ = 0
Free
y ′′ = 0, y ′′′ = 0
Simply supported
y = 0, y ′′ = 0
Example 2.6.1. Consider a beam of length L embedded at both ends, we will find the
deflection if we are given that w(x) = w0 for 0 < x < L. First, notice that we are in the
following situation: Therefore, we can model the situation with the following differential
equation where the initial conditions are given on the boundary of the set-up:

(iv)

= w0
 αy
For x = 0, y(0) = y ′ (0) = 0


For x = L, y(L) = y ′ (L) = 0
First we solve the differential equation as y(x) = yh (x) + yp (x). Note that the auxiliary
equation has a unique root (r = 0), so the solution of the homogeneous problem is of the
form yh (x) = C1 + C2 x + C3 x2 + C4 x3 .
Furthermore, using the method of undetermined coefficients, we can guess a particular
solution of the form yp (x) = Ax4 where A is indeterminate. Then we use the equation to
obtain:
w0 = α
d4
d4
w0
4
Ax
=
Aα
x4 = 24Aα ⇒ A =
dx4
dx4
24α
Concluding that the general solution of the equation is given in the form:
y(x) = yh (x) + yp (x) = C1 + C2 x + C3 x2 + C4 x3 +
w0 4
x
24α
Lecture Notes (version June 1, 2025)
47
Due to both conditions at x = 0, we obtain that both C1 and C2 must vanish. Afterwards,
we can obtain the following 2 × 2 system of equations using the conditions at the right
endpoint of the beam:

w L4

0 = y(L) = C3 L2 + C4 L3 + 0


24α



0
= y ′ (L) = 2C3 L + 3C4 L2 +
w0 L3
6α
After solving for the constants C3 and C4 , we can simplify the solution of the ODE to
the function:
y(x) =
w0 2
2
x (x − L)
24α
where the constants w0 , α and L are given in the statement of the problem.
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48
Boundary Value Problems: Eigenvalues and Eigenfunctions
The previous application of higher order equations to is related to a model where the conditions are given on the boundary of the problem that we are trying to study. Equations that
have the previous behavior are called Boundary value problems. We will summarize the
study of these type of problems using the following example:
y ′′ + λy = 0,
y(0) = y(L) = 0
where L ̸= 0 and λ ∈ R. In order to study the possible solutions of the equation we will
perform an analysis by cases:
1. Case 1: λ = 0. Then the equation becomes y ′′ = 0 where the solutions are of the form
y(x) = C1 + C2 x. Now we use the initial conditions to get the system of equations:
(
For x = 0:
0 = y(0) = C1 + C2 ∗ 0 = C1
For x = L: 0 = y(L) = C1 + C2 ∗ L
From the first equation we get that C1 = 0 and after replacing it on the second
equation we can conclude that C2 must vanish as well, so the only solution of the
equation is y(x) = 0.
2. Case 2: λ < 0. We write λ = −a2 for a real number a, then the equation becomes
y ′′ − a2 y = 0 where the roots of the auxiliary equation are r = ±a, then solutions
are of the form y(x) = C1 eax + C2 e−ax . Now we use the initial conditions to get the
system of equations:
(
For x = 0:
0 = y(0) = C1 + C2
For x = L: 0 = y(L) = C1 eaL + C2 e−aL
From the first equation we get that C2 = −C1 and after replacing it on the second
equation we get the following equation:
0 = C1 eaL + C2 e−aL = C1 eaL − C1 e−aL = C1 eaL − e−aL
Since the factor eaL − e−aL is not zero, then we are forcing C1 = 0 and so C2 = 0
as well. Therefore we have a similar situation as before where the only solution is
y(x) = 0.
3. Case 3: λ > 0. We write λ = a2 for a real number a, then the equation becomes
y ′′ + a2 y = 0 where the roots of the auxiliary equation are r = ±ai, then solutions are
of the form y(x) = C1 cos(ax) + C2 sin(ax). Now we use the initial conditions to get
the system of equations:
(
For x = 0:
0 = y(0) = C1 cos(0) + C2 sin(0) = C1
For x = L: 0 = y(L) = C1 cos(aL) + C2 sin(aL)
Lecture Notes (version June 1, 2025)
49
The first equation gives us C1 = 0, so the second one is reduced to 0 = C2 sin(aL).
Notice that if sin(aL) ̸= 0, then the constant C2 must vanish and so the solution would
be trivial. However, in case that sin(aL) vanishes, we would obtain a solution given
by a constant times the sine function where a is such that sin(aL) = 0. Now we find
those values of a that do the trick for us:
sin(aL) = 0 ⇐⇒ aL = n · π ⇐⇒ a =
2
nπ
L
for n ∈ Z
2
Therefore, λn = a2n = nLπ2 are values of our parameter for which there is a non-zero
solution to the problem. Those parameters are called eigenvalues of the boundary
value problem and have a dependence on the value of n.
Furthermore, the non-zero solutions are called eigenfunctions of the boundary
value problem, in our case they are of the form:
nπ yn (x) = sin
x
L
Figure 2.1: This graph contains a comparison of the eigenfunctions obtained for the previous boundary value problem for
n = 0, 1, 2, 3 and taking L = 1. Note that the n-th eigenfunction has n different local critical points (either local maximum
or local minimum). Furthermore, note that only values of n > 1
produce an eigenfunction since n = 0 yields a zero solution for
the equation.
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Chapter 3
Laplace Transform
"The secret is to win going as slowly as possible."
Andreas Nikolaus "Niki" Lauda. (1984)
Three-time Formula One World Drivers’ Champion.
(1975 and 1977 with Ferrari, 1984 with McLaren)
Nowadays, advanced computational methods require techniques to simplify certain nonhomogeneous equations including piecewise functions and even distributions as part of the
function on the right hand side of the equation.
The basic technique that modern mathematicians use is transforming a difficult and
computationally expensive analytical problem into an algebraic problem that can be solved
easily. After solving the problem in the algebraic setting, we transform back our solution
into the analytical setting which yields a solution of the original problem.
In our setting of differential equations, this method is the Laplace transform and even
if for some examples this method could be more complex, most dynamical applications can
be solved efficiently using this transformation as discontinuous or instantaneous impulses.
Therefore this chapter begins with the analysis of the Spring/Mass system as a motivation for the technique. Starting from that point, we introduce both the transformation
and the inverse transformation before computing certain properties of the transform and
presenting practical techniques as the Heaviside function, the convolution and the Dirac
delta function.
In particular, we will emphasize that this chapter requires background in factorization
techniques and partial fraction decomposition.
51
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Motivation: Spring/Mass systems
As a motivation to use Laplace Transformation, we will study the movement of a mass
attached to a spring as in Figure 3.1.
k
m
Figure 3.1: Classic description of the Spring/Mass system.
Letting m be the mass and k be the spring constant, we describe the position of the
mass y(t) by the following differential equation
my ′′ + ky = F (t)
where F (t) is characterizing an external force to the system. Note that in absence of any
external force, this equation has the form
y ′′ + ω 2 y = 0 ⇒ y(t) = c1 cos(ωt) + c2 sin(ωt)
k
where ω 2 = m
, and the solution of the movement yields an oscillatory behavior with the
constants depending on the initial condition of the system.
Note that in the previous equation, we are assuming that the ambient space where all the
movement is happening, is not generating any resistance to the movement of the system.
But that is not always the case, so we have to include this resistance to our differential
equation by assuming that this opposition to the movement is proportional to the speed of
the system.
k
c
m
Figure 3.2: Classic description of the Spring/Mass system with
a damping factor.
Hence, we define c to be the damping constant of the environment and our ODE becomes:
my ′′ + cy ′ + ky = F (t)
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53
Note that in the absence of an external force, the differential equations turns into the
following equation with constant coefficients where the auxiliary equation is given by:
√
−c ± c2 − 4mk
2
mr + cr + k = 0 ⇒ r =
2m
Then, we get the following three scenarios depending on the roots of this equation.
√
1. Under-damped Motion In case that c < 4mk, then the square root gets an
imaginary number as a solution. Therefore, we will have complex roots of the form
α + iβ where α < 0. Hence, the solution of the system is given by
y(t) = eαt (c1 cos(βt) + c2 sin(βt))
which is an oscillatory movement that is exponentially decaying until it vanishes as in
Figure 3.3.
Figure 3.3: The green line represents the solution y(t) and the
dashed red line is the exponential decay of the solution.
√
2. Over-damped Motion In case that c > 4mk, then we get two distinct real roots
r1 , r2 which satisfy the following inequality r1 < r2 < 0. So the solution of the system
is given by
y(t) = c1 er1 t + c2 er2 t
which is not an oscillatory movement and does not have any zeros. However, it still
decays exponentially as in Figure 3.4.
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Figure 3.4: The green line represents the solution y(t).
√
3. Critically damped Motion If c = 4mk, then the term inside the square root
c
vanishes and so we have a repeated negative root r = − 2m
. Thus, we get the following
solution for the equation
y(t) = c1 ert + c2 tert = ert (c1 + c2 t)
meaning that it is not oscillatory but it vanishes at t = − cc21 . Furthermore, it vanishes
exponentially as it is presented in Figure 3.5
Figure 3.5: The green line represents the solution y(t).
Note.
Notice that the graphs might vary depending on the initial conditions of the
movement, but the analysis constructed before reflects an optimal scenario.
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55
Laplace transform
In the previous section, we were able to solve the homogeneous version of the Spring/Mass
system. However, when we start treating with a non-homogeneous version of the problem,
we normally deal with discontinuous functions which may not have an easy form to deal with.
Therefore, we introduce a transformation that helps to deal with discontinuous functions.
Definition 17 (Laplace Transform). Given a function f (t), we define the Laplace Transform (F (s) or L {f (t)}) by the following integral,
Z∞
L {f (t)} = F (s) =
e−st f (t) dt
0
which is defined for some s > s0 and s0 depends on the function f (t).
Now, we will compute some basic Laplace transforms:
Example 3.2.1. We compute L {1} as follows:
Z∞
L {1} =
e
−st
e−st
dt =
(−s)
0
∞
=
0
1
s
for s > 0. Note that if s ≤ 0, the integral does not converge and so we need to make sure
that the integral exists.
Example 3.2.2. Computing L {eat } for a constant a as follows:
L eat =
Z∞
e
Z∞
−st at
e dt =
0
e
−(s−a)t
e−(s−a)t
dt =
(−(s − a))
0
∞
=
0
1
s−a
for s > a. Note that we need that s ≤ a, or else the integral does not converge.
Example 3.2.3. Computing L {t}, we get the following integral:
Z∞
L {t} =
e−st t dt
u = t, du = dt and dv = e−st dt, v =
e−st
−s
0
=
te−st
−s
∞!
0
1
+
s
Z∞
e−st dt =
1
s2
0
for s > 0, or else the integral does not converge.
Note that our next step is to explore if there are more properties coming from this
integral transform, since integrals distribute over linear combinations, we get the following
theorem
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Theorem 3.2.4 (Linearity of the Laplace Transform). Let α, β be constants and f (t), g(t)
be functions of t that have Laplace Transform. Then we get the following equality:
L {αf (t) + βg(t)} = αL {f (t)} + βL {g(t)}
Proof. Starting with the set up in the theorem, we can compute as follows:
Z∞
L {αf (t) + βg(t)} =
e−st (αf (t) + βg(t)) dt
0
∞

∞

Z
Z
= α  e−st f (t) dt + β  e−st g(t) dt
0
0
= αL {f (t)} + βL {g(t)}
With the linearity of the transform, we can find L{sin(kt)} and L{cos(kt)} by using the
following complex formula
eikt = cos(kt) + i sin(kt)
Lemma 3.2.5. For a constant k, we get the following:
L{cos(kt)} =
s
s2 + k 2
,
L{sin(kt)} =
k
s2 + k 2
Proof. We start noticing that we have the following equality using the linearity of the Laplace
transform:
L{eikt } = L{cos(kt) + i sin(kt)} = L{cos(kt)} + iL{sin(kt)}
So we just need to find the real and imaginary parts of L{eikt } and each part describes the
transform of sine and cosine.
1
1
s + ik
=
·
s − ik
s − ik s + ik
s + ik
s + ik
=
= 2
(s − ik)(s + ik)
s + k2
s
k
= 2
+i 2
2
s +k
s + k2
| {z }
| {z }
L{eikt } =
L{cos(kt)}
L{sin(kt)}
Before we continue studying more properties of the Laplace transformation, we need
to describe sufficient conditions for the existence of a Laplace transformation for a given
function f (t)
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57
Definition 18 (Piecewise continuous functions). A function f (t) is piecewise continuous
on [0, T ] if f (0+) and f (T −) are finite and the function is continuous on the open interval
(0, T ) except at possibly finitely many points where it has jump discontinuities.
Furthermore, a function f (t) is piecewise continuous on all [0, ∞) if it is piecewise
continuous on [0, T ] for all T > 0.
Definition 19 (Exponential order). A function f (t) has exponential order s0 if there
exists M and t0 such that
|f (t)| ≤ M es0 t
for all t ≥ t0
Note. Be aware that there are functions that do not have exponential order because
2
they might never be bounded by a single exponential, for example the function f (t) = et
doesn’t have exponential order.
With both previous definitions, we get the following sufficient condition for the existence
of a Laplace transform of a function.
Theorem 3.2.6 (Existence of Laplace transform). If a function f (t) is piecewise continuous
and of exponential order s0 , then L{f (t)} exists for all s > s0 .
Example 3.2.7. Consider the piecewise function f (t) defined as
(
1,
0≤t<1
f (t) =
−3e−t , t ≥ 1
we will show that we can apply Theorem 3.2.6. First, observe that f (t) is piecewise and
since for any t ≥ 1 we are in an exponential decay region, the hypothesis for Theorem 3.2.6
are satisfied which means that we can compute the Laplace transform as follows:
Z∞
L{f (t)} =
e−st f (t) dt =
0
e−st dt +
0
Z1
=
Z1
e−st dt − 3
0
Z∞
Z∞
e−st −3e−t dt
1
e−(s+1)t dt
1
First, we will compute the later integral since it is the one that will describe the exponential
order of the transform
−(s+1)t ∞
Z∞
e
3e−(s+1)
−(s+1)t
dt = −3
=−
−3 e
−(s + 1)
s+1
1
1
so the tail of the integral exists whenever s > −1. Therefore, we have to compute the first
integral for all s > −1:
Z1
e
0
−st
e−st
dt =
−s
1
=−
0
e−s
1
+
s
s
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Note that this result is defined for s ̸= 0, so if we set s = 0 before integrating, the integral
simply yields 1. Therefore, we obtain the following Laplace transform:

3e−(s+1)
 1 e−s
−
−
, s > −1, s ̸= 0
L{f (t)} = s
s
s+1

s=0
1 − 3e ,
which is defined for s > −1.
Now, we will introduce two important properties of the Laplace transformation that can
be used to simplify computations of the transform:
Theorem 3.2.8 (Shifting Theorem). If L{f (t)} = F (s) for s > s0 , then for any constant
a, we get that:
L eat f (t) = F (s − a)
for s > s0 + a.
So if we are computing the Laplace transform of a function that is being multiplied by
an exponential function, we just have to find the transform of the function and evaluate it
at a value that is shifted.
Example 3.2.9. We will find L{e3t sin(t)}, note that we are going to apply Theorem 3.2.8
where the function f (t) = sin(t). Using the fact that
F (s) = L{sin(t)} =
1
s2 + 1
we get that following result:
L{e3t sin(t)} =
1
(s − 3)2 + 1
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59
Inverse Laplace transform
So far, we have been computing transformations of different functions. But at this point,
we will work towards the doing the opposite transformation.
We will start with a function in the variable s and then find the function f (t) such that
can be transformed into our starting function using the Laplace transformation. In order
to do that, we first have to define the inverse Laplace transform.
Definition 20 (Inverse Laplace transform). Given a function F (s), we define the inverse
Laplace transform, L−1 , as follows:
L−1 {F (s)} = f (t) ⇐⇒ L{f (t)} = F (s)
Example 3.3.1. Since L{1} = 1s , then
L
−1
1
=1
s
1
Example 3.3.2. As presented in a previous example, L{e3t sin(t)} = (s−3)
2 +1 , then
L−1
1
(s − 3)2 + 1
= e3t sin(t)
Moreover, we get the following linearity property of the inverse transform
Theorem 3.3.3 (Linearity of the inverse Laplace transform). Let c1 , . . . , cn be constants
and assume that F1 (s), . . . , Fn (s) are functions that are Laplace transforms, i.e. for all
i ∈ {1, . . . , n}, there is a function fi (t) such that Fi (s) = L{fi (t)}, then:
L−1 {c1 F1 (s) + · · · + cn Fn } = c1 L−1 {F1 (s)} + · · · + cn L−1 {Fn (s)} = c1 f1 (t) + · · · + cn fn (t)
During the rest of this chapter, we will compute different examples that explain how to
use the previous theorem:
Example 3.3.4. We will compute the following inverse transform
6
9
1
3
−1
−1
−1
+
= 6L
+ 3L
L
s + 5 s2 + 9
s+5
s2 + 9
since L{e−5t } =
1
3
and L{sin(3t)} = 2
, we get the following result:
s+5
s +9
6
9
L−1
+ 2
= 6e−5t + 3 sin(3t)
s+5 s +9
Example 3.3.5. We will find the inverse Laplace transform of the following quotient
8 + 3s
(s2 + 1)(s2 + 4)
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We will use partial fractions to describe the quotient as a linear combination of polynomial
fractions by assuming that the fraction has the form:
As + B
Cs + D
(As + B)(s2 + 4) + (Cs + D)(s2 + 1)
+
=
s2 + 1
s2 + 4
(s2 + 1)(s2 + 4)
3
As + 4As + Bs2 + 4B + Cs3 + Cs + Ds2 + D
=
(s2 + 1)(s2 + 4)
3
2
s (A + C) + s (B + D) + s(4A + C) + (4B + D)
=
(s2 + 1)(s2 + 4)
Hence we obtain the following system of equations:

Coefficient of s3 : A + C = 0



Coefficient of s2 : B + D = 0

Coefficient of s: 4A + C = 0



Coefficient of 1: 4B + D = 1
which has the solution given by
A = 0,
B=
1
,
3
C = 0,
D=−
1
3
Therefore, we get the following result from the quotient:
1 8 + 3s 8 + 3s
1
1
s
1
1
s
−
=
8
·
+
3
·
−
8
·
+
3
·
3 s2 + 1
s2 + 4
3
s2 + 1
s2 + 1
3
s2 + 4
s2 + 4
1
1
s
1
2
s
=
8· 2
+3· 2
+
4· 2
+3· 2
3
s +1
s +1
3
s +4
s +4
8
1
s
4
2
s
= · 2
+
− ·
−
3 s + 1 s2 + 1 3 s2 + 4 s2 + 4
Thus, we can take the inverse transform to obtain the following result:
s
4
s
8 + 3s
8
1
2
−1
−1
+
− ·
−
L
=L
·
(s2 + 1)(s2 + 4)
3 s2 + 1 s2 + 1 3 s2 + 4 s2 + 4
8
1
s
= L−1
+ L−1
2
2
3
s +1
s +1
4 −1
2
s
−1
−L
− L
3
s2 + 4
s2 + 4
8
4
= sin(t) + cos(t) − sin(2t) − cos(2t)
3
3
Example 3.3.6. In this example, we will compute the inverse transform of the quotient
2
(s − 3)2 − 9
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61
Note that this quotient could be solved by using another property of the transform, but in
our case we will solve factorize the denominator and use partial fractions after that:
2
2
2
2
= 2
= 2
=
(s − 3)2 − 9
s − 6s + 9 − 9
s − 6s
s(s − 6)
By using partial fractions, we can simplify the quotient as follows
A
B
A(s − 6) + Bs
s(A + B) − 6A
+
=
=
s
s−6
s(s − 6)
s(s − 6)
Hence we obtain the following system of equations:
(
Coefficient of s: A + B = 0
Coefficient of 1: −6A = 2
which yields the solution A = − 13 and B = 13 . So the quotient becomes:
2
1 1 1
1
=− · + ·
(s − 3)2 − 9
3 s 3 s−6
So taking the inverse transform and using the linearity we obtain:
2
1
1 −1 1
1 −1
1 1
−1
L
=− L
+ L
= − + e6t
2
(s − 3) − 9
3
s
3
s−6
3 3
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Solving IVPs with the Laplace transform
After being able to compute the Laplace transform of multiple types of functions and understanding the inverse of the transformation, we need to start working towards using the
Laplace transform to solve differential equation.
Our first step is to relate the Laplace transform to the derivative of a function.
Theorem 3.4.1 (Laplace transform of a derivative). Assuming that y(t) and y ′ (t) are
continuous on [0, ∞) and of exponential order s0 , and that y ′′ is piecewise continuous on
[0, ∞). Then y(t), y ′ (t) and y ′′ (t) have Laplace transforms for s > s0 andthey have the
form.
L {y ′ } = sL {y} − y(0)
L {y ′′ } = s2 L {y} − y ′ (0) − sy(0)
Proof. We will only prove the first one of these properties and then we will apply the formula
to conclude the second formula.
Z∞
′
L {y } = e−st y ′ dt
u = e−st , du = −se−st . dv = y ′ dt, v = y
0
= ye−st
∞
Z∞
+s
e−st y dt = −y(0) + sL {y}
0
0
Now, we can prove the second formula by iterating over using the first one:
n
o
′
L {y ′′ } = L (y ′ ) = sL {y ′ } − y ′ (0) = s [sL {y} − y(0)] − y ′ (0) = s2 L {y} − y ′ (0) − sy(0)
Note. When solving an ODE with the Laplace transform, we need to get the initial
conditions at t = 0 or at least at the same point t = a. Otherwise, our property for the
derivative of a function does not hold.
Example 3.4.2. Considering the differential equation
y ′′ + y = 4,
y(0) = 0, y ′ (0) = 1
First, we will take the Laplace transform of the left hand side of the equation using Theorem
3.4.1 as follows:
L {y ′′ + y} = L {y ′′ } + L {y}
= s2 Y (s) − y ′ (0) − sy(0) + Y (s)
= s2 Y (s) − 1 + Y (s)
= Y (s)(s2 + 1) − 1
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63
Moreover, we will compute the Laplace transform of the right hand side of the equation as
follows:
L {4} = 4L {1} =
4
s
Since we took the Laplace transform at both sides of the equality, we can set both transforms
equal and isolate the function Y (s) = L{y}.
4
s
4
Y (s)(s2 + 1) = 1 +
s
1
4
Y (s) = 2
+
s + 1 s(s2 + 1)
Y (s)(s2 + 1) − 1 =
⇐⇒
⇐⇒
As we want to take the inverse transform at both sides, we first have to express the quotient
4
as a linear combination of two quotients using partial fractions.
s(s2 + 1)
4
s(s2 + 1)
=
A Bs + C
A(s2 + 1) + (Bs + C)s
s2 (A + B) + s(C) + A
+ 2
=
=
2
s
s +1
s(s + 1)
s(s2 + 1)
Having the following system of equations:

2

Coefficient of s : A + B = 0
Coefficient of s: C = 0


Coefficient of 1: A = 4
Thus, B = −4 and the right hand side of our equation can be written as:
Y (s) =
1
s2 + 1
+
4
s
−4· 2
s
s +1
Finally, we take inverse transforms to get the following solution of the ODE:
1
4
s
1
1
s
−1
−1
−1
y(t) = L−1
+
−
4
·
=
L
+
4L
−
4L
s2 + 1 s
s2 + 1
s2 + 1
s
s2 + 1
Concluding that
y(t) = sin(t) + 4 − 4 cos(t)
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The unit step function
One of the most important application of the Laplace transform is the solution of nonhomogeneous differential equations with a piecewise continuous function as the target function. Using the Laplace transform on this type of equations allow us to simplify the computation by describing the piecewise function as a linear combination of special piecewise
functions called the Heaviside functions whose Laplace transform can be computed easily.
Definition 21 (Heaviside function). The Heaviside function denoted by u(t − a) is a
piecewise continuous function defined as:
(
0, t < a
u(t − a) =
1, t ≥ a
Example 3.5.1. Consider the function f (t) = u(t − π) sin(t) using the definition of the
Heaviside, we can write our function as
(
0,
t<π
f (t) =
sin(t), t ≥ π
In Figure 3.6 we observe what happens after multiplying a function by a Heaviside.
Figure 3.6: The dashed red line represents sin(t) and the dashed
blue line is the same function after being multiplied by the Heaviside at π.
The most important property of the Heaviside function is that we can describe piecewise continuous functions using combinations of distinct Heaviside as it is expressed in the
following theorem:
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65
Theorem 3.5.2 (Separation of piecewise continuous functions). Let f (t) be a piecewise
discontinuous function of the form,

g1 (t), 0 ≤ t < a1




g2 (t), a1 ≤ t < a2
f (t) =
..


.



gn (t), t ≥ an−1
where a1 , . . . , an−1 are constant values. Then we can write down f (t) as the following
addition of Heaviside functions with the gi functions:
f (t) = g1 + u(t − a1 )(g2 − g1 ) + u(t − a2 )(g3 − g2 ) + · · · + u(t − an−1 )(gn − gn−1 )
Example 3.5.3. Consider the piecewise continuous function
(
1,
0 ≤ t < π2
f (t) =
−1, t ≥ π2
Using Theorem 3.5.2, we can write f (t) as the following addition:
π
π
f (t) = 1 + u t −
(−1 − (1)) = 1 − 2u t −
2
2
Observe that Theorem 3.5.2 implies that we can rewrite piecewise continuous functions
into a sum of functions that are multiplied by a Heaviside. Hence, in order to compute the
Laplace transform of piecewise functions, we need to compute the Laplace transform of a
product of a Heaviside with another function as presented in the following Theorem:
Theorem 3.5.4 (Laplace transformation of the Heaviside). Let f (t) be a function that
admits a Laplace transform and a a constant value, then
L{u(t − a)f (t)} = e−as L{f (t + a)}
Example 3.5.5. We will compute the following transform using Theorem 3.5.4
2
2
2
2
2
−s
2
−s
−s
L{u(t − 1)(t + 1)} = e L (t + 1) + 1 = e L t + 2t + 2 = e
+ 2+
s3
s
s
Example 3.5.6. Consider the piecewise function
(
2t + 1, 0 ≤ t < 2
f (t) =
3t,
t≥2
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Before computing the Laplace transform of f (t), we will describe f (t) using Theorem 3.5.2
to get the following description:
f (t) = (2t + 1) + u(t − 2)(3t − (2t + 1)) = 2t + 1 + u(t − 2)(t − 1)
Now, we can take the Laplace of f (t) using the result from Theorem 3.5.4 and the linearity
of the transformation as follows:
L{f (t)} = L {2t + 1 + u(t − 2)(t − 1)}
= 2L {t} + L {1} + L {u(t − 2)(t − 1)}
1
2
= 2 + + e−2s L{(t + 2) − 1}
s
s
1
2
= 2 + + e−2s L{t + 1}
s
s
2
1
1
1
−2s
= 2 + +e
+
s
s
s2
s
Moreover, we need to understand how the inverse transform behaves with the Heaviside
in order to solve equations with piecewise functions.
Theorem 3.5.7 (Inverse transform with Heaviside functions). Let g(t) be a function such
that G(s) = L{g(t)}, then
L−1 e−as G(s) = u(t − a)g(t − a)
Example 3.5.8. For this example we will compute the following inverse transform:
−2s e
−1
L
= L−1 e−2s L{t} = u(t − 2)(t − 2)
s2
In this case, our function g(t) = t and we used the result from Theorem 3.5.7.
Now, we introduce an example about an application of this properties into an IVP
Example 3.5.9. Consider the following IVP
y ′′ + y = f (t),
y(0) = 2,
y ′ (0) = −1
where the function f (t) is the following discontinuous function.
(
1,
0 ≤ t < π2
f (t) =
−1, t ≥ π2
Using the same strategy of taking Laplace transform at both sides of the equality, we will
begin by taking Laplace of the left hand side which yields the following:
L{y ′′ + y} = L{y ′′ } + L{y}
= s2 Y (s) − y ′ (0) − sy(0) + Y (s)
= s2 Y (s) + 1 − 2s + Y (s)
= Y (s)(s2 + 1) + 1 − 2s
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67
Now, we will use our computation from Example 3.5.3 to compute the Laplace transform of
the right hand side as:
n
π o 1
1
= − 2e−πs/2 ·
L{f (t)} = L 1 − 2u t −
2
s
s
Substituting it into the equality and isolating Y (s), we get the following:
1
1
− 2e−πs/2 ·
s
s
1
2
−πs/2 1
Y (s)(s + 1) = −1 + 2s + − 2e
·
s
s
s
1
1
1
−πs/2
+2· 2
+
− 2e
Y (s) = − 2
·
s +1
s + 1 s(s2 + 1)
s(s2 + 1)
Y (s)(s2 + 1) + 1 − 2s =
⇐⇒
⇐⇒
Note that the only quotient that we can’t solve with our previous transformations is the
factor G(s) = s(s21+1) , so we use partial fractions to deal with this function as follows:
1
s(s2 + 1)
=
A Bs + C
A(s2 + 1) + (Bs + C)s
s2 (A + B) + s(C) + A
+ 2
=
=
2
s
s +1
s(s + 1)
s(s2 + 1)
Hence, we get the system of equations

2

Coefficient of s : A + B = 0
Coefficient of s: C = 0


Coefficient of 1: A = 1
Then B = −1 and so our function can be written as:
G(s) =
1
s
− 2
⇒ g(t) = L−1 {G(s)} = L−1
s s +1
s
1
− 2
s s +1
= 1 − cos(t)
Thus, we can describe the transform of our solution y(t) as
1
s
Y (s) = − 2
+2· 2
+ G(s) − 2e−πs/2 · G(s)
s +1
s +1
Finally, by taking the inverse transform at both sides we get:
π
π y(t) = − sin(t) + 2 cos(t) + (1 − cos(t)) − 2u t −
1 − cos t −
2
2
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Carlos Cordoba & James Hornick
68
Convolution and Volterra integral equations
One possible issue with the Laplace transform is that it does not behave nicely with multiplication of functions as we observe in the following example.
Example 3.6.1. Let f (t) = 2 and g(t) = 21 , computing the Laplace transform of the
product of both functions we get
1
1
L {f (t) · g(t)} = L 2 ·
= L {1} =
2
s
However, when we compute the product of the Laplace transforms of f (t) and g(t), we get
the following result:
1
2
1
1
L {f (t)} · L {g(t)} = L {2} · L
=
·
= 2
2
s
2s
s
Therefore, we observe that L {f (t) · g(t)} =
̸ L {f (t)} · L {g(t)}.
In general, the negation above holds for most functions f (t) and g(t). So, we will define
a different way of computing a multiplication of functions which is called the convolution.
Definition 22 (Convolution product). Let f (t) and g(t) be two functions, we define their
convolution denoted f ∗ g, as the following integral product:
Zt
(f ∗ g)(t) =
f (τ )g(t − τ ) dτ
0
Two important properties of the convolution are the following linearity properties which
imply that the behavior of the product is similar to the usual multiplication of functions.
Lemma 3.6.2 (Properties of the Convolution). Let f (t), g(t) and h(t) be functions and α,
β constant values. Then, we have the following properties of the convolution:
1. Commutative property: f ∗ g = g ∗ f .
2. Linearity of the transformation: Similar as the Laplace transform, the convolution satisfies the following linearity:
[αf (t) + βh(t)] ∗ g(t) = α (f ∗ g) + β (h ∗ g)
Going back to the application of the Laplace transform, we have the following theorem
that shows why this type of products behaves nicely with the transformation.
Theorem 3.6.3 (Laplace of a convolution). Let f (t) and g(t) be two functions that admit a
Laplace transformation, then f ∗ g admits a Laplace transform. Moreover, we can compute
it as:
L {f ∗ g} = L {f } · L {g}
Lecture Notes (version June 1, 2025)
69
Example 3.6.4. In this example we will compute the convolution of two functions and show
that its Laplace transform is the same as the product of the transforms of each function.
Let a ̸= b and consider f (t) = eat and g(t) = ebt , then we will compute their convolution
using the definition as
Zt
(f ∗ g)(t) =
Zt
f (τ )g(t − τ ) dτ =
0
eaτ eb(t−τ ) dτ
0
= ebt
Zt
e(a−b)τ dτ =
ebt (a−b)t
e
−1
a−b
0
1
=
eat − ebt
a−b
Now we compute the Laplace transform of the convolution:
1
1
1
1
1
eat − ebt
−
L {(f ∗ g)(t)} = L
=
=
a−b
a−b s−a s−b
(s − a)(s − b)
where the later term is equal to the product of the transforms of f (t) and g(t).
The use of convolutions also yields important methods to solve certain integral equations
similar to the ones presented in the following example.
Example 3.6.5. We will find the function h(t) such that
Zt
h(t) =
(t − τ )5 τ 7 dτ
0
Note that if we let f (t) = t and g(t) = t5 , then the right hand side of the equation is
f ∗ g. So, our approach for the solution of this problem is taking Laplace at both sides of
the equality and then taking inverse Laplace to find the solution.
7
7! 5!
7! · 5!
·
= 14
s8 s6
s
Finally, taking the inverse transform at both sides we get the following answer:
7! · 5!
7! · 5! −1 13!
7! · 5! 13
−1
h(t) = L
=
L
=
t
14
14
s
13!
s
13!
H(s) = L{f ∗ g} = L{f (t)} · L{g(t)} = L{t7 } · L{t5 } =
Another important type of integral equations that can be solved using the Laplace transformation are the Volterra integral equations.
Definition 23 (Volterra Integral equation). Let f (t) and k(t) be known functions that admit
a Laplace transformation. Then a Volterra integral equation
Zt
k(t − τ )y(τ ) dτ
y(t) = f (t) +
0
where the function y(t) is an unknown function that must be determined.
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70
Note. A general version of a Volterra integral equation can involve derivatives of the
unknown function y(t). In that case, we would need initial conditions to compute the
Laplace transform but they can be solved using the same approach as a normal Volterra
integral equation.
Example 3.6.6. We will solve the following Volterra integral equation
Zt
y(t) = sin(t) − 2
cos(t − τ )y(τ ) dτ
0
Taking Laplace at both sides of the equality, we do the following computation
⇐⇒
⇐⇒
⇐⇒
⇐⇒
Y (s) = L {sin(t)} − 2L {cos(t) ∗ y(t)}
s
1
−2 2
· Y (s)
Y (s) = 2
s +1
s +1
2s
1
Y (s) 1 + 2
= 2
s +1
s +1
2
s + 2s + 1
1
Y (s)
= 2
s2 + 1
s +1
1
Y (s) = 2
s + 2s + 1
Finally, we take the inverse transform at both sides to obtain the solution of the equation
1
1
−1
−1
y(t) = L
=L
= te−t
s2 + 2s + 1
(s + 1)2
Lecture Notes (version June 1, 2025)
3.7
71
Dirac delta function
Definition 24 (Dirac delta function). For some constant value t0 , we define the Dirac
delta function at t0 denoted by δ(t − t0 ) as a generalized function with the properties:
Z∞
Z∞
δ(t − t0 ) dt = 1,
f (t)δ(t − t0 ) dt = f (t0 ),
and
−∞
−∞
where f (t) is a function on the variable t.
A possible way to think about the Dirac delta function is as the following piecewise
function
(
+∞, t = t0
δ(t − t0 ) ≃
0,
t ̸= t0
Moreover, we can compute the Laplace transform of the Dirac delta function which is
described in the following Theorem:
Theorem 3.7.1 (Laplace transform of the Dirac delta function). Let t0 be a constant value,
then δ(t − t0 ) admits a Laplace transform. Moreover, the Laplace transform of the Dirac
delta function is
L{δ(t − t0 )} = e−st0
Note. The Laplace transform of the Dirac delta function is giving us an exponential
function depending on s, meaning that most problems involving the Dirac delta function
will generate a Heaviside function as part of the solution of the differential equation.
Example 3.7.2. Consider the following IVP
y ′′ + 4y = δ(t) + δ(t − π),
y(0) = y ′ (0) = 0
We are going to solve it using the Laplace transform and so we start computing the transform
of the left hand side of the equation:
L{y ′′ + 4y} = s2 Y (s) − y ′ (0) − sy(0) + 4Y (s) = Y (s)(s2 + 4)
Now, for the right hand side we will use the linearity of the transform and Theorem 3.7.1
L{δ(t) + δ(t − π)} = L{δ(t)} + L{δ(t − π)} = e−s·0 + e−sπ = 1 + e−sπ
Now, we can isolate Y (s) just dividing by the polynomial s2 + 4 which yields the following:
1
1
−1
−sπ
y(t) = L
+e
s2 + 4
s2 + 4
1 −1
2
1 −1 −sπ
2
= L
+ L
e
2
s2 + 4
2
s2 + 4
1
1
= sin(2t) + u(t − π) sin (2(t − π))
2
2