Big Picture in Focus
ULOa. Define and classify the different types of Differential Equation.
Metalanguage
In this section, the essential terms and symbols relevant in the study of
Differential Equations will be discussed to demonstrate ULOa. Please refer to these
definitions in case you will encounter difficulty in understanding educational concepts.
1. Differential Equations. Equations containing derivatives or differentials.
2. Ordinary Differential Equations (ODEs). Differential equations that depend on a single
variable.
3. Partial Differential Equations (PDEs). Differential equations that depend on a single
variable.
4. Order. Refers to the highest - ordered derivative in the given differential equation.
5. Degree. The exponent of the highest - ordered derivative.
6. Dependent Variable.
7. Independent Variable.
8. Parameters.
9. First Order ODE. An ODE that contains the first derivative only of the unknown function.
Essential Knowledge
To perform the aforesaid big picture (unit learning outcomes) for the first three
(3) weeks of the course, you need to review the fundamental concepts in Calculus that
will be laid down in the succeeding pages. Please note that you are not limited
to exclusively refer to these resources. Thus, you are expected to utilize other books,
research articles and other resources that are available in the university’s library
e.g. ebrary, search.proquest.com etc.
Differential Equation
A differential equation is any equation which contains derivatives, either ordinary
derivatives or partial derivatives.
There is one differential equation that everybody probably knows, that is Newton’s
Second Law of Motion. If an object of mass m is moving with acceleration a and being
acted on with force F then Newton’s Second Law tells us.
Example 1
𝐅 = 𝐦𝐚
Rewriting this equation to prove that it a differential equation,
𝐝𝐯 𝐝𝟐 𝐬
𝐚= = (𝐝𝐭𝟐 )
𝐝𝐭
At this point v is the velocity and s is the position of the particle at any time t. In addition,
force F may also be a function of time, velocity, and/or position. So, having this the
Newtons Second Law of Motion can now be written as a differential equation in terms of
𝐝𝐯/𝐝𝐭 and 𝐝𝐬/𝐝𝐭
𝐝𝐯
𝐅(𝐭, 𝐯) = 𝐦
𝐝𝐭
𝐝𝐬 𝐝𝟐 𝐬
𝐅(𝐭, 𝐬 )=𝐦 𝟐
𝐝𝐭 𝐝𝐭
This type of equation will be used in the succeeding topics.
Here are some other differential equation.
𝐚𝐲" + 𝐛𝐲′ + 𝐜𝐲 = 𝐠(𝐭)
𝐝𝟐 𝐲 𝐝𝐲
𝐬𝐢𝐧 𝐲 𝟐 = (𝟏 − 𝐲) + 𝐲 𝟐 𝐞−𝟓𝐲
𝐝𝐱 𝐝𝐱
𝐲 (𝟒) + 𝟏𝟎𝐲 ′′′ − 𝟒𝐲 ′ + 𝟐𝐲 = 𝐜𝐨𝐬 𝐭
𝛛𝟐 𝐮 𝛛𝐮
∝ =
𝛛𝟐 𝐱 𝛛𝐱
𝐚𝟐 𝐮𝐱𝐱 = 𝐮𝐭𝐭
𝟐
𝛛𝟑 𝐮 𝛛𝐮
( 𝟐 ) =𝟏+
𝛛 𝐱𝛛𝐭 𝛛𝐭
Definition of Integration.
Integration is the inverse operation to differentiation. In differentiation, we solve
for the differential of a given function whereas in integration, we solve for the
function corresponding to a given differential. The resulting function is called the
integral of the differential.
Indefinite Integral
The collection of all the possible antiderivatives of a given function is called the
indefinite integral. The indefinite integral comprises of the antiderivative and the
constant of integration (𝑪). The presence of this constant of integration (in
indefinite integration) introduces a family of functions which have the same
derivative in all points in their domain.
If 𝑭(𝒙) is a function whose derivative 𝑭′ (𝒙) = 𝒇(𝒙) on a certain interval of the x
axis, then 𝑭(𝒙) is called an antiderivative or indefinite integral of 𝒇(𝒙). The
indefinite integral of a given function is not unique; for example, 𝒙𝟐 , 𝒙𝟐 − 𝟔,
𝒙𝟐 + 𝟑 are all indefinite integrals of 𝒇(𝒙) = 𝟐𝒙. All indefinite integrals of 𝒇(𝒙) =
𝟐𝒙 are then included in 𝑭(𝒙) = 𝒙𝟐 + 𝑪.
The Notation
The symbol ∫ 𝒇(𝒙)𝒅𝒙 is used to indicate the indefinite integral of 𝒇(𝒙). Thus, we
write ∫ 𝟐𝒙 𝒅𝒙 = 𝒙𝟐 + 𝑪. In the expression ∫ 𝒇(𝒙)𝒅𝒙, the function 𝒇(𝒙) is called
the integrand and is mathematically expressed as:
The
Concept of Integration
For example, find the indefinite integral for 𝒇(𝒙) = 𝒙𝟓 .
Base on the definition if 𝑭(𝒙) is such a function we should have:
′
(𝑭(𝒙)) = 𝒙𝟓
𝒙𝟔
Obviously, the function satisfies the above equation but other functions such
𝟔
𝒙𝟔 𝒙𝟔
as 𝟔 + 𝟏, 𝟔 − 𝟑, and etc. can be considered as an answer so we can write
the answer generally as:
𝒙𝟓
∫ 𝒙𝟒 𝒅𝒙 = +𝑪
𝟓
Using the definition of the indefinite integral, we can find the integral of a
simple functions directly:
o ∫ 𝟎 𝒅𝒙 = 𝑪
o ∫ 𝟏 𝒅𝒙 = 𝒙 + 𝑪
𝒙𝒏+𝟏
o ∫ 𝒙𝒏 𝒅𝒙 = + 𝑪 (𝒏 ≠ −𝟏)
𝒏+𝟏
𝟏
o ∫ 𝒙−𝟏 𝒅𝒙 = ∫ 𝒙 𝒅𝒙 = 𝒍𝒏 |𝒙| + 𝑪 ( 𝒙 ≠ 𝟎)
𝒂𝒙
o ∫ 𝒂𝒙 𝒅𝒙 = ∫ 𝒍𝒏 𝒂 + 𝑪 (𝒂 ≠ 𝟏, 𝒂 > 𝟎)
o ∫ 𝒄𝒐𝒔 𝒙 𝒅𝒙 = 𝒔𝒊𝒏 𝒙 + 𝑪
o ∫ 𝒔𝒊𝒏 𝒙 𝒅𝒙 = −𝒄𝒐𝒔 𝒙 + 𝑪
Rules of Integration
1. The derivative of the indefinite integral is the integrand:
′
(∫ 𝒇(𝒙)𝒅𝒙) = 𝒇(𝒙)
2. The differential of the indefinite integral is equal to the element of the integration:
𝒅 (∫ 𝒇(𝒙)𝒅𝒙) = 𝒇(𝒙)𝒅𝒙
3. The indefinite integral of a differential of a function is equal to that function plus a
constant:
∫ 𝒅(𝑭(𝒙)) = 𝑭(𝒙) + 𝑪
4. If 𝒂 ≠ 𝟎 and is a constant, then:
∫ 𝒂𝒇(𝒙)𝒅𝒙 = 𝒂 ∫ 𝒇(𝒙)𝒅𝒙
A constant coefficient goes in and comes out of the integral sign
5. The indefinite integral of the sum or difference of two integrable functions is
equal to the sum or difference of their individual indefinite integral:
∫[𝒇(𝒙) ± 𝒈(𝒙)]𝒅𝒙 = ∫ 𝒇(𝒙)𝒅𝒙 ± ∫ 𝒈(𝒙)𝒅𝒙
6. If ∫ 𝒇(𝒙)𝒅𝒙 = 𝑭(𝒙) + 𝑪, then:
𝟏
∫ 𝒇(𝒂𝒙 + 𝒃)𝒅𝒙 = 𝑭(𝒂𝒙 + 𝒃 ) + 𝑪
𝒂
And if 𝒃 = 𝟎, then
𝟏
∫ 𝒇(𝒂𝒙 )𝒅𝒙 = 𝑭(𝒂𝒙 ) + 𝑪
𝒂
o Using the last rule, we can easily calculate some integrals without applying a
specific method:
Example:
𝟏
a. ∫ 𝒆𝒂𝒙 𝒅𝒙 = 𝒆𝒂𝒙 + 𝑪
𝒂
𝒅𝒙
b. ∫ 𝒙−𝒂 = 𝒍𝒏 |𝒙 − 𝒂| + 𝑪
−𝒄𝒐𝒔 (𝒂𝒙 )
c. ∫ 𝒔𝒊𝒏(𝒂𝒙 )𝒅𝒙 = +𝑪
𝒂
The Substitution Method of Integration
o If the integrand is in the form of 𝒇(𝒈(𝒙))𝒈′(𝒙) 𝒅𝒙, and substituting 𝒖 = 𝒈(𝒙),
then we will have
∫ 𝒇(𝒈(𝒙))𝒈′(𝒙) 𝒅𝒙 = ∫ 𝒇(𝒖)𝒖′ 𝒅𝒙 = ∫ 𝒇(𝒖) 𝒅𝒖
o And if ∫ 𝒇(𝒖) 𝒅𝒖 = 𝑭(𝒖) + 𝑪, then:
∫ 𝒇(𝒈(𝒙))𝒈′(𝒙) 𝒅𝒙 = 𝑭(𝒈(𝒙)) + 𝑪
Example:
𝒙
1. Find the indefinite integral ∫ 𝒅𝒙.
𝟏+𝒙𝟐
Solution:
Let 𝒖 = 𝟏 + 𝒙𝟐 , then 𝒅𝒖 = 𝟐𝒙𝒅𝒙, and we will have:
𝟏
∫
𝒙
𝒅𝒙 = ∫ 𝟐 𝒅𝒖 = 𝟏 ∫ 𝒅𝒖 = 𝟏 𝒍𝒏 |𝒖| + 𝑪
𝟏 + 𝒙𝟐 𝒖 𝟐 𝒖 𝟐
since 𝒖 = 𝟏 + 𝒙𝟐 , thus,
𝟏 𝟏
𝒍𝒏 |𝒖| + 𝑪 = 𝒍𝒏 |𝟏 + 𝒙𝟐 | + 𝑪
𝟐 𝟐
𝟐
2. Find ∫ 𝒙𝒆𝒙 𝒅𝒙.
Solution:
Let 𝒖 = 𝒙𝟐 , then 𝒅𝒖 = 𝟐𝒙𝒅𝒙, and:
𝟐 𝟏 𝟏 𝟏 𝟐
∫ 𝒙𝒆𝒙 𝒅𝒙 = ∫ 𝒆𝒖 (𝟐 𝒅𝒖) = 𝟐
𝒆𝒖 + 𝑪 = 𝟐
𝒆𝒙 + 𝑪
𝒅𝒙
3. Find ∫ , ( 𝒙 > 𝟎)
𝒙𝒍𝒏 𝒙
Solution:
𝒅𝒙
Let 𝒖 = 𝒍𝒏 𝒙, then 𝒅𝒖 = , and:
𝒙
𝒅𝒙 𝒙𝒅𝒖 𝒅𝒖
∫ 𝒙𝒍𝒏 𝒙 = ∫ 𝒙 𝒖 = ∫ = 𝒍𝒏 |𝒖| + 𝑪 = 𝒍𝒏 |𝒍𝒏 (𝒙)| + 𝑪
𝒖
Note that, having success with this method requires finding a relevant substitution,
which comes after lots of practice.
ORDER
It is the largest derivative present in the differential equation. As per listed above a is a
first order DE, while b, c, f, and g are second order DE, h is a third order DE, and e is a
fourth order DE.
DEGREE
It is the represented by the power of the highest order derivative in the given differential
equation and which the differential coefficients are free from radicals and fractions. From
the examples from c to g it is considered as first degree while h is considered as a second
degree.
TYPE OF DIFFERENTIAL EQUATION
As stated earlier at the equations that exists is either an ordinary differential equation
(ODE) or a partial differential equation (PDE). ODE are those equations that contains
ordinary differential signs e.g. 𝐝𝐲/𝐝𝐱, 𝐲 ′ , 𝐮𝐱𝐲 , while in the other hand PDE has partial
𝛛𝟐 𝐮
differential signs like e.g. . From the examples c to e are ODE and f to h are PDE.
𝛛𝟐 𝐱
LINEAR DIFFERENTIAL EQUATIONS
A linear differential equation is any differential equation that can be written in the following
form.
𝐚𝐧 (𝐭)𝐲 𝐧 (𝐭) + 𝐚𝐧−𝟏 (𝐭)𝐲 𝐧−𝟏 (𝐭) + ⋯ + 𝐚𝟏 (𝐭)𝐲 ′ (𝐭) + 𝐚𝟎 (𝐭)𝐲(𝐭) = 𝐠(𝐭) (1)
The important thing to note about linear differential equations is that there are no products
of the function, 𝐲(𝐭), and its derivatives and neither the function or its derivatives occur to
any power other than the first power. Also note that neither the function or its derivatives
are “inside” another function, for example, √𝐲′ or 𝐞𝐲
The coefficients 𝐚𝟎 (𝐭), 𝐚𝐧 (𝐭) and 𝐠(𝐭) can be zero or non-zero functions, constant or non-
constant functions, linear or non-linear functions. Only the function 𝐲(𝐭), and its
derivatives are used in determining if a differential equation is linear.
If a differential equation cannot be written in the form of equation 1 then its called a non
– linear differential equation.
In the ODE group from c to e, c and e are considered as linear while d is non – linear.
SOLUTION OF DIFFERENTIAL EQUATIONS
When we first performed integrations, we obtained a general solution (involving a
constant, C).
We obtained a particular solution by substituting known values for 𝐱 and 𝐲. These known
conditions are called boundary conditions (or initial conditions).
It is the same concept when solving differential equations - find general solution first, then
substitute given numbers to find particular solutions.
Example 2
Solve for the a.) general solution of 𝐝𝐲 + 𝟕𝐱𝐝𝐱 = 𝟎 and b.) its particular solution if 𝐲(𝟎) =
𝟑
For the general solution, primarily we can transfer either 𝐝𝐲 or 𝟕𝐱𝐝𝐱 at the left side, either
will have the same outcome.
𝐝𝐲 = −𝟕𝐱𝐝𝐱
Then at this point we will integrate the two terms individually,
∫ 𝐝𝐲 = ∫ −𝟕𝐱𝐝𝐱
Which will give us an answer of,
𝟕
𝐲 = − 𝐱𝟐 + 𝐂
𝟐
At this point we have solved the general solution of the given differential equation, now
we can easily solve the particular solution by substituting the values of 𝐱 and 𝐲 from the
given.
The values of 𝐲 = 𝟑 , while the value of 𝐱 = 𝟎 that will give us a value for our arbitrary
constant,
𝟕
𝟑 = − ( 𝟎) + 𝐂
𝟐
𝐂=𝟑
The complete the solution simply plug in the value of 𝐂 to the general solution which gives
us an answer of,
𝟕
𝐲𝐩 = − 𝐱 𝟐 + 𝟑
𝟐
We denoted the particular solution as 𝐲𝐩 to clarity purposes.
𝟕
In summary, our general solution is 𝐲 = − 𝟐 𝐱 𝟐 + 𝐂 while our particular solution is 𝐲𝐩 =
𝟕
− 𝟐 𝐱 𝟐 + 𝟑.
Example 3
Find the particular solution of 𝐲 ′ = 𝟓 given that when 𝐱 = 𝟎, 𝐲 = 𝟐
Here we expand 𝐲′ to 𝐝𝐲/𝐝𝐱 and multiplying both side by 𝐝𝐱 we will have,
𝐝𝐲 = 𝟓𝐝𝐱
We integrate both sides which gives us,
𝐲 = 𝟓𝐱 + 𝐂
Plugging in the values of 𝐱 and 𝐲 results
𝐲𝐩 = 𝟓𝐱 + 𝟐
In summary, general solution 𝐲 = 𝟓𝐱 + 𝐂 and 𝐲𝐩 = 𝟓𝐱 + 𝟐.
NOTE: You can refer to other resources below for further understanding
Cioranescu , Doina. (2012). Introduction to classical and variational partial differential
equations Quezon City: The University of the Philippines Press
Zill, Dennis G.(2009). A first course in differential equations with modeling applications.
9th Australia: Brooks/Cole Cengage Learning