Module 1 Introductory Topics

Introduction

In this module, we will discuss several introductory topics that are essential to the study
of differential equation. Several definitions and other necessary skills in solving differential
equations are discussed in this module. Student will learn different forms and structure of
a differential equation. Also, they will learn to create a differential equation based on a
problem as well as finding solutions of initial-value problem differential equation.

Topic Outcomes

1. Determine the type, order, degree and linearity of a differential equation by

identification of each features in a given differential equation correctly.

2. Verify that a function is a solution of a differential equation by means of substitution

to make a valid conclusion.

3. Express a parametric equation into its equivalent differential equation by eliminating

the arbitrary constants contained in that equation through the correct use of
differentiation and algebraic simplifications.

4. Express equations of families of curves into its equivalent differential equation by

eliminating the arbitrary constants contained in that equation through the correct
use of differentiation and algebraic simplifications.

5. Show the solution of an initial-value problem by imposing the initial condition to get
the value of an arbitrary constant.

6. Show all the intervals in which the solution of a given DE is unique through partial
derivative correctly.
Definitions and Terms

Many of the principles, or laws, underlying the behavior of the natural world are
statements or relations involving rates at which things happen. When expressed in
mathematical terms the relations are equations and the rates are derivatives.

The following are examples of differential equations:

dy
= cos x (1)
dx
d2 y
+ k2y = 0 (2)
dx2
x2 + y 2 dx − 2xy dy = 0


(3)
 2
∂ u ∂ 2u

∂u 2
=h + (4)
∂t ∂x2 ∂y 2
d2 i di 1
L 2
+ R + = Eω sin ωt (5)
dt dt C
2
∂ V ∂ 2V ∂ 2V
+ + =0 (6)
∂x2 ∂y 2 ∂z 2
 2 3
dw dw
2
− xy +w =0 (7)
dx dx
d3 x dx
3
+ x − 4xy = 0 (8)
dy dy
3
d2 y

dy
+7 − 8y = 0 (9)
dy 2 dx
d2 y d2 x
+ 2 =x (10)
dt2 dt
∂f ∂f
x +y = nf (11)
∂x ∂y

When an equation involves one or more derivatives with respect to a particular variable,
that variable is called an independent variable. A variable is called dependent if the derivative
of that variable occurs. In equation (5); i is the dependent variable, t the independent
variable and L, R, C, E, and ω are called parameters.
Differential Equation Defined

An equation containing the derivatives of one or more dependent variables, with respect
to one or more independent variables, is said to be a differential equation (DE).

Classifications of DE in terms of
• Type
• Order
• Linearity

Classifications by Type

If an equation contains only ordinary derivatives of one or more dependent variables

with respect to a single independent variable it is said to be an Ordinary Differential
Equation (ODE). Equations (1), (2), (3), (5), (7), (8), (9) and (10) are examples of
ordinary differential equations.

An equation containing partial derivatives of one or more dependent variables of two or

more independent variables is called Partial Differential Equation (PDE). Equations
(4), (6) and (11) are examples of partial differential equations.

Classifications by Order

The order of a differential equation (either ODE or PDE) is the order of the highest
derivative in the equation. For example, equation (1) is a first-order equation while equation
(2) is a second-order equation. First-order differential equations are written in the form
M (x, y)dx + N (x, y)dy = 0 for convenience.

In symbols, one can express an nth-order differential equation in one variable by the
general form
F x, y, y 0 , . . . , y (n) = 0


(12)
where F is areal-valued function of n + 2 variable: x, y, y 0 , . . . , y (n) .For both practical and
theoretical reasons it is assumed that it is possible to solve an ordinary differential equation
in the form (12) uniquely for the highest derivative of y (n) in the term of the remaining n + 1
variables. The differential equation:
dn y 0 (n)


= f x, y, y , . . . , y (13)
dxn
where f is a real-valued continuous function, is referred to as the normal form of equation
(12).
Classifications by Linearity

An nth-order differential equation (12) is said to be linear if F is linear in y, y 0 , . . . ,

(n)
y . This means that an nth-order differential equation is linear when (12) is in the form

an (x)y (n) + an−1 (x)y (n−1) + · · · + ax (x)y 0 + a0 (x)y − g(x) = 0

or
dn y d(n−1) y dy
an (x)
n
+ a n−1 (x) (n−1)
+ · · · + a1 (x) + a0 (x)y = g(x) (14)
dx dx dx
Two important cases of (14) are first-order differential equation (n=1) and second-order
differential equation (n=2):
dy
a1 (x) + a0 (x)y = g(x)
dx
and
d2 y dy
a1 (x) 2 + a1 (x) + a0 (x)y = g(x) (15)
dx dx
In the additive combination on the left-side of (14) it can be seen that the characteristic two
properties of a linear ordinary differential equations are as follows:

• the dependent variable y and all its derivatives y 0 , y 00 , . . . , y (n−1) are of the first degree

• the coefficients a0 , a1 , . . . , of y 0 , y 00 , . . . , y (n−1) depend at most on the independent

variable x.

The equations
(y − x) dx + 4x dy = 0
y 00 + 2y 0 + y = 0
and
d3 y dy
3
+ x − 5y = ex
dx dx
are, in turn, first-order linear differential equations. Note that the first equation can be
rewritten as 4xy 0 + y = x, thus it is linear in the variable y.

A non-linear ordinary differential equation is simply one that is not linear. Non-linear
functions of the dependent variable, or its derivatives, such as sin y or ey , cannot appear in
a linear equation.

The equation (1 − y)y 0 + 2y = ex is non-linear since it has a coefficient, (1 − y), that

d2 y
is dependent on y. The equation 2 + sin y = 0 is also non-linear since it has a term, sin y,
dx
d4 y
that is non-linear in y. Lastly, the equation 4 + y 2 = 0 is also a non-linear equation since
dx
it has a term, y 2 , that is not of power 1.
Solution of an Ordinary Differential Equation

Any function φ, defined on an interval I, which when substituted into an nth-order

differential equation reduces the equation to an identity, is said to be a solution of the
equation on the interval.
In other words, a solution of an nth-order ordinary differential equation in (12) is a
function φ that possesses at least n derivatives and for which

F x, φ(x), φ0 (x), . . . , φ(n) (x)

for all x in I.

Interval of Definition

One cannot think solution of an ordinary differential equation without simultaneously

thinking interval. The interval I in the definition of the solution of an ODE is variously called
the interval of definition, the interval of existence, the interval of validity or the domain of
the solution and can be an open interval (a, b), a closed interval [a, b], an infinite interval
(a, ∞) and so on.

Example 1. Verifying a solution of an Equation

1 4 dy 1
Verify if y = x is a solution of = xy 2 on the interval (−∞, ∞).
16 dx
Solution:
Substitute the function to the equation, then compare the left-hand and the right-hand side.
Left-hand side:
dy 1
1
= 4 · x3 = x3
dx 16 4
Right-hand side:
  12  
1 1 4 1 2 1
xy = x ·
2 x =x· x = x3
16 4 4

1 4 dy 1
Therefore, y = x is a solution of = xy 2 on the interval (−∞, ∞).
16 dx
Example 2. Verifying a solution of an Equation

Verify if y = xex is a solution of y 00 − 2y 0 + y = 0 on the interval (−∞, ∞).

Solution:
Substitute the function to the equation, then compare the left-hand and the right-hand side.
Left-hand side:

y 00 − 2y 0 + y = (xex + 2ex ) − 2 (xex + ex ) + xex

Right-hand side:
0
Therefore, y = xex is a solution of y 00 − 2y 0 + y = 0 on the interval (−∞, ∞).

Note that in this example, the differential equation has a constant solution y = 0 on
the interval −∞ < x < ∞. A solution of a differential equation that is identically zero on
an interval I is said to be a trivial solution.
Elimination of Arbitrary Constants

There are several ways a differential equation may come up, some will be discussed in
modeling. There is one way of obtaining a differential equation by stating some families of
relations defined by a parameter/s called arbitrary constants. These represent any random
value; however it is not a variable although it could take in any value but once it is given a
value it would no longer be given any more values like in the case of a variable.

Methods of elimination may vary with the way in which a constant is placed in a
relation. Methods that is effective with one problem may be poor for other problems.

A simple 3-step method can be applicable in eliminating arbitrary constants in a given

relation. These are, 1. count the number of arbitrary constants 2. perform differentiation
depending on the number of arbitrary constants 3. algebraically manipulate the equations.
Steps 2 and 3 can be interchanged accordingly.

Example 1.

Eliminate c from the relation

x3 − 3x2 y = c
Solution:
Since one constant is to be eliminated, obtain the first derivative,

x3 − 3x2 y =c
dy
3x2 − 3x2 − 6xy =0
dx
dy
−3x2 = 6xy − 3x2
dx
dy
x = (2y − x)
dx
xdy = (2y − x)dx
−(2y − x)dx + xdy =0
(2y − x)dx − xdy =0
Example 2.

Eliminate c from the relation

y sin x − xy 2 = c
Solution:
Obtain the first derivative to eliminate c since there is only one arbitrary constant to be
eliminated,

y sin x − xy 2 =c
dy dy
y cos x + sin x − 2xy − y2 =0
dx dx
dy dy
sin x − 2xy = y 2 − y cos x
dx dx
dy
(sin x − 2xy) = y(y − cos x)
dx
(sin x − 2xy)dy = y(y − cos x)dx
−y(y − cos x)dx + (sin x − 2xy)dy =0
y(y − cos x)dx − (sin x − 2xy)dy =0
Example 3.

Eliminate c from the relation

x2 y = 1 + cx
Solution:
Getting the first derivative,

dy
x2 + 2xy = c (1)
dx
Since c 6= 0, solving c from the main relation we have,

x2 y = 1 + cx
x2 y − 1 = cx
x2 y − 1
c=
x
then substitute this c to (1),

dy x2 y − 1
x2 + 2xy =
dx x
Simplifying,
dy
x3 + 2x2 y = x2 y − 1
dx
dy
x3 = x2 y − 2x2 y − 1
dx
dy
x3 = −x2 y − 1
dx
3
x dy = −(x2 y + 1)dx
(x y + 1)dx + x3 dy
2
=0
Example 4.

Eliminate c1 and c2 from the relation

x = c1 cos ωt + c2 sin ω

Solution:
Since there are two arbitrary constants to be eliminated, obtain the first and the second
derivatives,
dx
= −c1 ω sin ωt + c2 ω cos ωt
dt
d2 x
2
= −c1 ω 2 cos ωt − c2 ω 2 sin ωt
dt

Factor out −ω 2 from the second derivative gives,

d2 x
= −ω 2 (c1 cos ωt + c2 sin ωt)
dt2

Notice that, c1 cos ωt+c2 sin ωt is simply equal to x from the given relation. Then the second
derivative can be expressed as

d2 x
= −ω 2 x
dt2

or
d2 x
+ ω2x = 0
dt2
Example 5.

Eliminate a from the relation

y 2 = 4ax
Solution:
Obtain the first derivative to eliminate the arbitrary constant a,
dy
2y = 4a (1)
dx
Since a 6= 0 and notice that 4a from the given relation is

y2
4a =
x

then substitute this to (1), gives

dy y2
2y =
dx x

Clearing the fraction gives us the final result

2xydy = y 2 dx
2xydy − y 2 dx = 0
Example 6.

Eliminate c1 and c2 from the relation

y = c1 + c2 e3x

Solution:
Obtain the first and the second derivatives of the given relation,

y = c1 + c2 e3x
dy
= 3c2 e3x
dx
d2 y
= 3(3c2 e3x )
dx2

Notice that 3c2 e3x is the first derivative, then

d2 y dy
2
=3
dx dx
2
dy dy
2
−3 =0
dx dx

or

y 00 − 3y 0 = 0
Example 7.

Eliminate c1 and c2 from the relation

y = x + c1 ex + c2 e−x

Solution:
Since there are two arbitrary constants to be eliminated, obtain the first and the second
derivatives,

y 0 = 1 + c1 ex − c2 e−x

y 00 = c1 ex + c2 e−x (1)
From the given relation,

y = x + c1 ex + c2 e−x
y − x = c1 ex + c2 e−x

Substituting to (1),

y 00 = c1 ex + c2 e−x
y 00 = y − x

or

y 00 − y + x = 0
Example 8.

Eliminate c1 and c2 from the relation

y = x2 + c1 e2x + c2 e3x

Solution:

y = x2 + c1 e2x + c2 e3x (1)

y 0 = 2x + 2c1 e2x + 3c2 e3x (2)

y 00 = 2 + 4c1 e2x + 9c2 e3x (3)

Multiply 2 in (1) and subtract (2) to eliminate c1 gives,

2y − y 0 = 2x2 − 2x − c2 e3x (4)

Multiply 2 in (2) and subtract (3) to eliminate c1 ,

2y 0 − y 00 = 4x − 2 − 3c2 e3x (5)

To eliminate c2 , multiply 3 to (4) and subtract (5) to obtain the result

y 00 − 5y 0 + 6y = 6x2 − 10x + 2
Example 9.

Eliminate A and B from the relation

y = Ae2x + Bxe2x

Solution:

y = Ae2x + Bxe2x (1)

y 0 = 2Ae2x + 2Bxe2x + Be2x (2)
y 00 = 4Ae2x + 4Bxe2x + 4Be2x (3)

To eliminate A, multiply 2 in (1) and subtract (2) which yields,

2y − y 0 = −Be2x (4)

Multiply 2 in (10) and subtract (11) gives,

2y 0 − y 00 = −2Be2x (5)

To eliminate B, multiply 2 in (4) and subtract (5) to obtain the result

y 00 − 4y 0 + 4y = 0
Families of Curves

As mentioned in the previous lecture, a differential equation is represented by a family of

solutions. These solutions involves parameters (arbitrary constants) along with them. Some
of the family of solution can be interpreted geometrically; now we shall apply elimination of
arbitrary constants to generate differential equations of some typical curves. These curves
may represent solutions of differential equations that will be studied in the future.

Some Formulas of Curves

Equations of Straight Lines

General Form: Ax + By + C = 0

Point-Slope Form: y − y1 = m (x − x1 )

Slope-Intercept Form: y = mx + b

y2 − y1
Two-Point Form: y − y1 = (x − x1 )
x2 − x1
x y
Intercept Form: + =1
a b
Normal Form: x cos α + y sin α = ρ

Equation of a Circle with center at (h, k) and radius r

(x − h)2 + (y − k)2 = r2
Equation of an Ellipse with center at (h, k), semi-minor axis a, and semi-minor axis b
(x − h)2 (y − k)2 (y − k)2 (x − h)2
+ = 1 or + =1
a2 b2 a2 b2
Equation of a Hyperbola with center at (h, k), semi-transverse axis a, and semi-conjugate
axis b
(x − h)2 (y − k)2 (y − k)2 (x − h)2
− = 1 or − =1
a2 b2 a2 b2
Equation of a Parabola with vertex at (h, k)and focal distant a
(y − k)2 = 4a (x − h) or (x − h)2 = 4a (y − k)
Equilateral Hyperbola with center at the origin
xy = c
Example 1.
Find the differential equation representing all lines passing through the origin.

Solution:
Some members of the family are drawn first.

As seen from the sketch, the lines have a common y-intercept b = 0 and have different slopes,
we can use the slope-intercept form.

y = mx + b

Since b = 0, we have
y = mx
which has one arbitrary constant m.
To eliminate m, we need to find the only the first derivative.

y = mx

Get the first derivative,

y0 = m
Substitute y 0 in y,
y = xy 0
Rearranging,
xy 0 − y = 0
Example 2.

Find the differential equation representing all lines with a slope 2/3.

Solution:
Some members of the family are drawn first.

As seen from the sketch, there is only one parameter that is changing and it is the y-intercept.
We can use the slope-intercept form,
2
y= +b
3
Differentiating,
2
y0 =
3
Example 3.

Find the differential equation representing all circles with center at the origin.

Solution:
The equation to use is
x2 + y 2 = r 2
wherein it only has one arbitrary constant r.

Get the first derivative to eliminate r2 .

2x dx + 2y dy = 0

Rearranging,
x dx + y dy = 0
Example 4.
Find the differential equation representing all parabolas with vertex at the origin and focus
located on the y-axis.

Solution:
Some members of the family are drawn first.

It can be seen that the parabolas have the y-axis to be the principal axis, thus the family’s
equation is
(x − h)2 = 4a (y − k)
wherein, the vertex is the origin then h = 0 and k = 0. The equation to use now is
x2 = 4ay
To eliminate a, we need to do one differentiation.
x2 = 4ay (6)
Get the first derivative
4ay 0 = 2x
Solve for 4a
2x
4a =
y0
Substitute 4a in (1),
2x
x2 = ·y
y0
Simplifying,
x0 − 2y = 0
Example 5.

Find the differential equation representing all circles with center at the x-axis.

Solution:
Some members of the family are drawn.

From the figure, it can be seen that the center may have the coordinates (h, k) wherein h
could be any value and k = 0. The radii of the circles are also varying.
Hence the equation to use is
(x − h)2 + y 2 = r2 (1)
which has two arbitrary constants.
First differentiation,
2 (x − h) + 2yy 0 = 0
Simplify,
yy 0 + x = h
Second differentiation,
2
yy 00 + (y 0 ) + 1 = 0
Example 6.

Find the differential equation representing ellipses and hyperbolas with axes on the
coordinate axes.

Solution:
We can write the equation as
ax2 + by 2 = 1. (1)

Isolate a in Equation (1)

x−2 1 − by 2 = a


(2)
Differentiating Equation (2)

−2x−3 1 − by 2 + x−2 (−2byy 0 )

(3)

Simplify Equation (3) then isolate b,

1
=b (4)
y2 − xyy 0

For an easier computation instead of differentiating Equation (4), we can rewrite it as

1
y 2 − xyy 0 = (5)
b
Differentiating Equation (5) and simplifying we have,

x(y 0 )2 − yy 0 + xyy 00 = 0
Example 7.
Find the differential equation representing all circles with center on the line y = x and
passing through the origin.

Solution:
Some members of the families are drawn.

The center form of the equation of the circle is (x − h)2 + (y − k)2 = r2 .

Now, let us focus on the relationship of h, k and r.
Since, C(h, k) is in y = x, therefore h = k. And by using the Pythagorean Theorem,
r2 = h2 + k 2 which in turn is r2 = h2 + h2 = 2h2 .

Now, our equation becomes

(x − h)2 + (y − h)2 = 2h2 (1)
Isolating h in Equation (1).
x2 + y 2
= 2h (2)
x+y
Differentiate Equation (2)

(x + y) (2x dx + 2y dy) − (x2 + y 2 ) (dx + dy)

=0 (3)
(x + y)2

Simplifying Equation (3)

x2 + 2xy − y 2 dx + y 2 + 2xy − x2 dy = 0

Example 8.

Find the differential equation of all straight lines.

Solution:
We can use the standard forms of the equation of a line with the least number of arbitrary
constant. So we will use the slope-intercept form.

y = mx + b

First derivative,
y0 = m
Second derivative,
y 00 = 0
No more arbitrary constant.
Example 9.

Find the differential equation of all sine waves with a period of 2π and a phase shift of 0◦ .

Solution:
Some members of the families are drawn.

The equation of a sine wave is

2πx
y = A sin( + θ)
T
The period is T = 2π and the phase shift θ = 0◦ . Now, we have the working equation,

y = A sin x

Getting the first derivative.

y 0 = A cos x
Divide y by y 0 ,
y A sin x
=
y0 A cos x
y
= tan x
y0
y = y 0 tan x or
y0 = y cot x
Initial-Value Problems

We are often interested in problems which we seek a solution y(x) of a differential

equation so that y(x) satisfies some prescribed conditions, and these conditions are imposed
on y(x) and its derivative. On some interval I containing x0 , the problem

Solve:
dn y
= f x, y, y 0 , y 00 , . . . , y (n−1)


n
(1)
dx

Subject to:

y(x0 ) = y0 y 0 (x0 ) = y1 · · · y (n−1) (x0 ) = yn−1

where y0 , y1 , · · · , y(n−1) are arbitrary specified values, is called an initial-value problem.
The values of y(x) and its first (n − 1) derivatives are called initial conditions.

Example 1. Two First-order Initial-Value Problems

The family y = c ex is a one-parameter family of solutions of a simple first-order
equation y 0 = y. All the solutions of this family are defined on the interval (−∞, ∞). If we
impose an initial condition, e.g. y(0) = 3.
Solving for c,
y = c ex
3 = c e0
c = 3.
Thus, y = 3 ex is a solution of the initial-value problem
y 0 = y; y(0) = 3.
Now, if we demand a solution that passes through the point (1, −2) the constant of the
family of solution is
y = c ex
−2 = c e1
c = −2e−1 .
In this case y = −2 ex−1 is a solution of the initial value problem
y 0 = y; y(1) = −2.
The solution curves are shown on Figure (1).
 Figure 1: Some families of y 0 = y and the solutions of the two IVP

Example 2. Interval of Definition of Solutions

−1
The first-order differential equation y 0 + 2xy 2 = 0 has a solution y = x2 + c . If we
impose the initial condition y(0) = −1, then the family of solution will give c = −1. Thus,

−1
y = x2 + c .

Now, let us focus on some of the following distinctions:

−1
• y = x2 + c is considered a function, the domain of which is the set of all real values
x 6= 0. See Figure (2).

−1
• y = x2 + c is considered a solution of the differential equation y 0 +2xy 2 = 0, wherein
the interval I could be taken over which
−1y is defined and differentiable. From Figure
2
(2), the intervals on which y = x + c is a solution of the differential equation are
−∞ < x < −1, −1 < x < 1, ans 1 < x < ∞.

−1
• y = x2 + c is considered a solution of the initial-value problem y 0 +2xy 2 = 0 subject
to y(0) = −1 for the interval I containing the point (0, −1), i.e. (−1 < x < 1). See
Figure (3).

−1
Figure 2: Graph of the function y = x2 − 1 .

Figure 3: Solution of y 0 + 2xy 2 = 0 containing the point (0, −1).

Example 3. Second-order Initial Value Problem
The differential equation ẍ + 16x = 0, has the family of solution x = c1 cos 4t + c2 sin 4t.
Find a solution to the initial value problem
π  π 
ẍ + 16x = 0; x = −2, ẋ = 1.
2 2
1
Applying the initial conditions, we will find that c1 = −2 and c2 = . Hence,
4
1
x = −2 cos 4t + sin 4t
4
is the solution of the initial value problem.
Example 4. An Initial-Value Problem can have several solutions
1 4 1
Each of the functions y = 0 and y = x satisfies the differential equation y 0 = xy 2
16
subject to the initial condition y(0) = 0, and so the initial-value problem
1
y 0 = xy 2 ; y(0) = 0

has at least two solutions. As illustrated in Figure (4), the graphs of both passes throught
the point (0, 0).

1
Figure 4: Solutions of y 0 = xy 2 subject to the initial condition y(0) = 0.
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