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Part I
Ordinary Differential Equations
Chapter 1Ordinary Differential Equations of theFirst Order Part 1 of this material deals with equations that contain one or more derivatives of a function ofa single variable and such equations are called ordinary differential equations, which can be usedto model a phenomena of interest in the sciences, engineering, economics, ecological studies, andother areas.In the first section we will see the basic concepts and ideas and in the remaining sections we willconsider equations which involve the first derivative of a given independent variable with respectto an independent variable, which are called Ordinary Differential Equations of the First Order. 1.
Basic Concepts and Ideas
The derivative
of a function
is itself another function
′^ (
)^
found by an appro-
priate rule of differentiation. For example, the function
(^2) 𝑥
is differentiable on the interval
,^
)^
and by the Chain Rule its derivative is
𝑥
2
. If we replace the right-hand
expression of the last equation by the symbol y, the equation becomes
In differentiation, the problem was ”Given a function
, find its derivative.”
1.1 Basic Concepts and Ideas
Now, the problem we face here is ”If we are given an equation such as (1.1), is there some wayor method by which we can find the unknown function
that satisfy the given equation,
without prior knowledge how it was constructed?” These kind of problems are the ones we aregoing to focus on in this part of the course. Definition 1.1.1.
An equation involving derivatives of one or more dependent variables with
respect to one or more independent variables is called a
Differential Equation
DE
Example 1.1.1.
3
2
= sin
.^
are all
Differential Equations
Differential equations can be classified by their
type, order,
and in term of
linearity.
We will
see these classifications before going to the solution concept. Classification by Type
If an equation contains only ordinary derivatives of one or more dependent variables withrespect to a single independent variable, then it is said to be an
ordinary differential
equation (ODE)
For example,
2
2
and
3
3
2
= sin
are all ordinary differential equations. ∙
If a function is defined in terms of two or more independent variables, the correspondingderivative will be a partial derivative with respect to each independent variable. An equationinvolving partial derivatives of one or more dependent variables of two or more independentvariables is called a
partial differential equation (PDE)
For example,
(^2)
2
2
2
and
are both partial differential equations.
In this part we will only consider the case of ordinary differential equations.
1.1 Basic Concepts and Ideas
Classification by Order The
order
of a differential equation (either ODE or PDE) is the order of the highest derivative
that appear in the equation. For example,
2
2
𝑥
are first and second-order ordinary differential equations respectively.The general
th
order ordinary differential equation in one dependent variable is given by the
general form
′^ , 𝑦
′′^
(𝑛
)^ ) = 0
,^
where F is a real-valued function of
variables
′^ , 𝑦
′′^
(𝑛
)^.
Remark 1.1.2.
For both practical and theoretical reasons we shall also make the assumption
hereafter that it is possible to explicitly solve the differential equation of the form (1.3) uniquelyfor the highest derivative
(𝑛
)^
in terms of the remaining
variables
′^ , 𝑦
′′
(𝑛
−
Then the differential equation (1.3) becomes
𝑛
𝑛
′^ , ..., 𝑦
(𝑛
−
,^
where
is a real-valued continuous function and this is referred to as the
normal form
of (1.3).
Example 1.1.2.
The normal form of the first-order equation
′^
is
′^
and the normal form of the second-order equation
′′
is
′′
′^
The first order ordinary differential equation is generally expressed as:
′^ ) = 0 or
′^
For example, the differential equation
′^
is equivalent to
′^
. If
′^ ) =
′^
then the given differential equation becomes of the form
′^ ) = 0
1.2 Separable Differential Equations
Many first-order
ODEs
can be reduced or transformed to the form
′^
where
and
are continuous functions. Then, from elementary calculus we have:
Such type of equations are called
separable equations
. Integrating both sides we get:
is the general solution of the given equation. Example 1.2.2.
Solve the DE
′^
Solution: The equation
′^
is equivalent to
and then
Integrating both sides,
gives
2
2
which is an implicit solution of the given first order differential equation. Example 1.2.3.
Solve the DE
′^
2
−
𝑥^
First rewrite the equation as
2
2 𝑥
If
,^
this has the differential form
12 𝑦^
d
2 𝑥
d
where the variables have been separated. Integrating both sides we have
12 𝑦^
2
𝑥^
1.2 Separable Differential Equations
which implies
where
is a constant of integration. Then solve for
to get
2 𝑥
which is an explicit solution of the given first order differential equation. Remark 1.2.1.
It is recommended to write an explicit solution to the differential equation when
ever possible.
However, sometimes solving for the dependent variable (in our case
) may not
be possible.
In those cases one can represent the final solution by an implicit solution of the
differential equation. 1.2.
Equation Reducible to Separable Form
There are some differential equations which are not separable, but they can be transformed toa separable form by simple change of variables. We will see some of the possible substitutionshereunder. A. Linear Substitution Suppose we have a differential equation that can be written in the form:
′^
Such an equation is not in general separable. However, if we set
we get
Or
Thus (1.7) will be transformed into
where
and
can be separated.
Example 1.2.4.
Solve the differential equation
′^
1.2 Separable Differential Equations
Solution: Let
. Then
′^
′^
which implies
′^
′^
.^
With this substitution the equation
′^
2
is equivalent to
′^
2
2
Then
2
and integrate both sides,
2
to get
arctan
for an arbitrary constant
.^
Substituting back
in the last
equation gives us the general solution of the given DE to be
arctan(
Example 1.2.5.
Solve the differential equation
′^
Solution Let
.^
Then,
′^
′^
which implies
′^
12
′^ )
.^
Therefore, the equation
′^
becomes
1 2
′^ ) +
.^
Simplifying this we
get
′^
which implies
′^
Then
Now we integrate both sides
and we get
ln
∣^
1
But
. Then substituting this in the above equation gives us
ln
∣^
1
for an arbitrary constant
1
, or equivalently
∣^
where
𝑐^1
1.2 Separable Differential Equations
B. Quotient Substitution Suppose we have an equation that can be written in the form
′^
Let us substitute
Then
′^
2
′^
𝑦^2 𝑥^
This implies,
′^
′^
′^
Thus, the differential equation
′^
(^
is reduced to the equation
′^
)^
which is equivalent to the differential equation 𝑑𝑥𝑥
)^
Then by integrating we obtain a general solution. Example 1.2.6.
Solve
′^
2
2
Solution For
, the differential equation
′^
2
2
is equivalent to
′^
2
Let
𝑦𝑥^
. Then
2
and we get,
′^
which implies
2
We then integrate
2
and get
arctan
= ln
∣^
Now substituting
𝑦𝑥^
gives us
arctan(
) = ln
∣^
= ln
∣^
= ln
for some
1.3 Exact Differential Equations
Theorem 1.2.
(Existence and uniqueness of a solution)
.^
If
)^
is continuous function on
some rectangular region
in the
plane containing the point
)^
in its interior , then the
problem
′^
with
has at least one solution defined on some open interval of
containing
If, in addition, the function
is continuous on
R
, then the solution to the above equation (1.8) is unique on some open interval
containing
Remark 1.2.6.
The above condition for uniqueness can be eased by using a condition
piecewise
continuous
instead of the condition that “
∂𝑓 ∂𝑦
is continuous”.
Exact Differential Equations
Consider the differential equation:
′^
sin
cos
or equivalently
sin
cos
Notice that the left hand side is the
(total) differential
of the function
sin
2
Recall that the total differential of a function
)^
of two variables is
for all (x,y) in the domain of
Thus for
sin
(^2)
) = sin
cos
.^
This implies that
that is,
sin
(^2)
is the solution of the above DE.
Definition 1.3.1.
The expression
1.3 Exact Differential Equations
is called an
exact differential equation
in some domain
(an open connected set of points)
if there is a function
)^
such that^ ∂𝐹
)^
and
for all
If we can find a function
such that ∂𝐹∂𝑥
) and
then the differential equation
is just
.^
But recall that, if
,^
then
) = constant
.^
The equation
where
is an arbitrary constant, implicitly defines the general solution of the deferential equation
Now let us ask the following two fundamental questions. Given a Differential Equation
How can we determine the existence of such a function
)^
If it exists, how can we find it?
The following theorem will answer the first question. Theorem 1.3.
(Test for Exactness)
.^
Let
,^
∂𝑀∂𝑦
and
∂𝑁∂𝑥
be all continuous func-
tions within a rectangle
(or some domain) in the
-plane. Then
is an exact differential in
if and only if
every where in
Example 1.3.1.
Consider the equation
3
2
(^2)
4
𝑦^
1.3 Exact Differential Equations
Then write
3
2
4 𝑦
Let
3
and
3
2
4 𝑦
.^
Then
Therefore, the given differential equation is exact. Example 1.3.2.
Consider the equation
ln
−
𝑥𝑦
ln
Let
ln
−
𝑥𝑦
and
1 𝑦^
ln
Then
∂𝑀∂𝑦
= ln
−
𝑥𝑦
and
∂𝑁∂𝑥
= ln
which
implies
∂𝑀∂𝑦
∂𝑁∂𝑥
Therefore, the given differential equation is not exact.
After knowing the exactness of a differential equation, the next question is ”How can we solvethe given equation?” The method for this is described here below.Suppose a differential equation
is exact. Then, there exists a function
)^
such that
and
From
∂𝐹 ∂𝑥
, we have (by integrating with respect to
,^
where
)^
is only a function of
but constant with respect to
Now to determine
(the constant of integration), differentiate equation (1.9) with respect
to
to get
′^ (
which implies
′^ (
) and hence
′^ (
by exactness. Therefore,
)^
Example 1.3.3.
Solve the differential equation
sin
cos
1.3 Exact Differential Equations
Solution Let
) = sin
and
cos
. Then
𝑦
= cos
𝑥^
.^
Since
𝑦
𝑥
are all continuous in
R
2 , the given equation is exact. Thus, there exists a function
such
that
= sin
and
cos
which implies
sin
sin
)^
and
∂𝐹∂𝑦
cos
′^ (
.^
That is,
cos
cos
′^ (
which implies
′^ (
and hence
2
Therefore,
sin
2
constant, which implies 𝑥
sin
2
determines
)^
implicitly.
Example 1.3.4.
Solve the differential equation
3
(^2)
2
(^3)
SolutionStep 1: Checking Exactness Let
3
(^2) and
2
(^3)
.^
Then
Therefore the given equation is exact. Step 2: Finding Implicit Solution Then to find
)^
we use
3
2
4
2
(^2)
where
is a function of
only. To find A(y); ∂𝐹∂𝑦
2
′^ (
3
1.3 Exact Differential Equations
This idea works correctly if the ratio
𝑦
𝑥
is a function of x only, that is,
𝑦
𝑥
a function of
In this case
𝑦^
𝑥
which implies
∫^
𝑝
(𝑥
)𝑑𝑥
If the quotient
𝑦^
𝑥
is not a function of
alone, then the integrating factor
can not be
obtained using the above procedure, but we can try to find
as a function of
alone,
Then when
)^
is only a function of
, equation (1.11) will be reduced to 𝑑𝜇𝑑𝑦
𝑦^
𝑥
which implies
𝑦^
𝑥
,^
which is a separable differential equation
If the fraction
𝑦
𝑥
is a function of
alone, then
−
∫ 𝑞
(𝑦
)𝑑𝑦
Example 1.3.6.
Consider the equation
−
2 𝑦
.^
Let
and
−
2 𝑦
.^
Then
∂𝑀∂𝑦
and
∂𝑁∂𝑥
and hence
∂𝑀∂𝑦
∂𝑁∂𝑥
which implies
that the given differential equation is not exact.Assume that the given equation has an integrating factor. But
𝑦^
𝑥
−
2 𝑦
2
𝑦
2
𝑦
which is not a function of
alone. Hence obtaining
)^
is not possible.
1.4 Linear First Order Differential Equations
However,
𝑦^
𝑥
can be considered as a function of
alone. Therefore, it is possible to solve for
)^
and is given
by
−
∫ (
−
𝑑𝑦
3 𝑦
Now to solve the problem in (1.13), multiplying the given equation by
3 𝑦
we get the
equation
3 𝑦
−
2
𝑦^ )
3 𝑦
which is an exact differential equation. Thus, there exists
such that
3 𝑦
and
−
2 𝑦
3
𝑦
which implies that
3 𝑦
3 𝑦
To determine
we use
)^
which is obtained above and differentiate it with respect to
and equate the result with
−
2 𝑦
3 𝑦
. Hence we have
−
2 𝑦
3 𝑦
3 𝑦
′^ (
Then
3 𝑦
3
𝑦^
′^ (
)^
which implies that
′^ (
𝑦^
and hence
𝑦^
𝑦^
Therefore,
3 𝑦
𝑦^
That means
3 𝑦
𝑦^
where
is an
arbitrary constant, defines the implicit solution of the Differential Equation in (1.13). 1.
Linear First Order Differential Equations
Consider the general first-order linear differential equation
1
′^
0
1
By dividing both sides by
1
,^
we get
′^
where
)^
and
1.4 Linear First Order Differential Equations
Here we assume that
)^
and
)^
are continuous.
There is a general approach to solve linear equations. To solve for
from the given equation
we start with the simplest case, when
.^
That is, (1.14) becomes
′^
.^
This problem is called a homogeneous version of (1.14). Now to solve (1.15) first we get
′^
and we divide both sides by
and get
′ 𝑦^
Then by integrating
we get
ln
∣^
which implies
∣^
𝑐−
∫^
𝑝(
𝑥)
𝑑𝑥
−
∫ 𝑝
(𝑥
)𝑑𝑥
,^
for
Therefore,
−
∫ 𝑝
(𝑥
)𝑑𝑥
,^
where
is an arbitrary constant,
is a general solution of (1.15). Example 1.4.1.
Solve the following differential equations.
′^
′^
Solution:
- If
′^
,^
then
′𝑦 (^) 𝑦
We integrate
to get
ln
∣^
2
and hence
−
𝑥
2
is the general solution.
1.4 Linear First Order Differential Equations
- If
′^
,^
then
′ 𝑦
We integrate
to get
(𝑥
−
2 ln(
𝑥
+2)
), or
2
𝑥^
which is the general solution of the
given equation.
Now we want to solve the general first - order linear ordinary differential equation
′^
)^
This can be done in two steps. Step 1. Consider the homogeneous version of (1.16) and find the solution to be
ℎ
−
∫ 𝑝
(𝑥
)𝑑𝑥
where
indicate the general solution for the homogeneous part of the equation
Step 2. To get the solution for the non-homogeneous part of the equation we vary the constant
with
different value of
Hence we assume that
−
∫ 𝑝
(𝑥
)𝑑𝑥
is a solution for (1.16). Then (1.17) must satisfy (1.16). i.e.
−
∫ 𝑝
(𝑥
)𝑑𝑥
′^
−
∫ 𝑝
(𝑥
)𝑑𝑥
which implies
′^ (
−
∫ 𝑝
(𝑥
)𝑑𝑥
−
∫ 𝑝
(𝑥
)𝑑𝑥
−
∫ 𝑝
(𝑥
)𝑑𝑥
Simplifying this gives us,
′^ (
∫^
𝑝(
𝑥)
𝑑𝑥
1.5 *Nonlinear Differential Equations of the First Order
Solution Since
,^
let
(^1) −
2
−
Then
′^
−
′^
and substituting
2
for
′^
we get
′^
−
2
2
−
1
Then we get the differential equation
′^ +
which is equivalent to the equation
.^
This implies
and we integrate
and get
𝐵 𝐴
−
𝐴𝑥
Therefore, the general solution of the original differential equation is
𝐵𝐴
−
𝐴𝑥
The Riccati Equation.
A differential equation of the form
′^
(^2)
is called
Riccati
equation. If
,^
then the equation is linear.
If we can obtain one particular solution
)^
of the
Riccati
equation, then the change of variables
transforms the
Riccati
equation in to a linear equation in
and
Then we find the general
solution of this linear equation and we use it to write the general solution of the original
Riccati
equation. Example 1.5.2.
Solve the
Riccati
equation
′^
12 𝑥^
2
(Hint:
is one solution.)
1.5 *Nonlinear Differential Equations of the First Order
The Clairuat Equation
A nonlinear differential equation of the form
′^
′^ )
is called
Clairuat
equation.
To solve such equation, let us differentiate both sides of the equation with respect to
Then
we get,
′^
′^
′′
′^ (
′^ )
′′
which implies that
′′
′^ (
′^ )) = 0
and hence
′′
or
′^ (
′^ ) = 0
Solving
′′^
gives us the general solution
and solving
′^ (
′^ ) = 0
gives us a
singular solution (include definition of singular solution in the basic section ). Example 1.5.3.
Solve the
Clairuat
equation
′^
1 ′ 𝑦^
Solution Differentiating both sides with respect to
to get
′^
′^
′′
′′ (𝑦
′^ )
2
This implies
′′
′^ )
2
and then solving
′′
gives us a general solution
and solving
′^ )
2
gives us a singular solution. Then
′^ )
2
(^1) 𝑥^
which implies
′^
1 ±
√
𝑥^
.^
Hence
is a
singular solution. Remark 1.5.1.
The general solution of the
Clairuat
equation is
Therefore, our
main focus for such equation is the singular solution.
1.6 Exercises
Exercises
Exercise 1.6.1.
Solve each of the following differential equations.
′^
2
′^
with inital condition
.^
Chapter 2Ordinary Differential Equations of TheSecond and Higher Order A second-order differential equation is a differential equation containing a second derivative ofa dependent variable with respect to the independent variable but no higher derivative.
The
theory of second-order differential equations is vast, and here we will focus on linear second-orderequations, which have many important uses.
Most of the results are given for a higher order
ODEs and second order ODEs are special cases, but most of our examples are for second orderODEs. 2.
Basic Theory
In this section, we will focus on the general theory of linear ordinary differential equations beforewe start to discuss about solving such problems. Definition 2.1.1.
A linear ordinary differential equation of order
in the dependent variable
and independent variable
is an equation which can be expressed in the form:
𝑛^
(𝑛
)^
𝑛
−
(𝑛
−
1
′^
where
𝑛
)^
and the functions
𝑛^
are continuous real- valued functions of
[
].
The function
)^
is called the non-homogeneous term and all the points
𝜖^
[
]^
in which
𝑛
) = 0𝜖^
are called singular points of the DE (2.1).
2.2 General Solution of Homogeneous Linear ODEs
The following theorem guarantees that any
th
order Linear Homogenous Ordinary Differential
Equation has
linearly independent solutions.
Theorem 2.2.
(Existence of Linearly Independent Solutions for a LHODE)
.^
The Linear Ho-
mogenous Differential Equation (LHODE) (2.2) always has
Linearly Independent (LI) solutions.
Furthermore, if
1
2
𝑛
are
LI solutions of (2.2), then every solution of (2.2)
can be expressed as a linear combination of these solution functions. i.e. If y is a solution for(2.2), then
𝑛 �^ 𝑖
=
𝑓𝑖^
(𝑖^
for some
𝑛^
R
Example 2.2.2.
Consider the second order linear homogenous DE
′′
Then
) = sin
) = cos
are LI solutions of the given equation.
Then
sin
cos
is the fundamental set of solutions of the given DE and hence the general solution of the DE isgiven by
1
sin
2
cos
for constants
1
2
R
Definition 2.2.4.
If
2
𝑛
)^
are
linearly independent solutions of (2.2) on
[
],
then the set
𝑛
is called the
Fundamental Set of Solutions
of (2.2) and
the function
1
2
𝑛
𝑛^
[
],
where
1
2
𝑛
are arbitrary constants is called a
General Solution
of (2.2) on
[
].
and
each
2
𝑛
are called particular solutions.
Example 2.2.3.
Consider the third order linear homogenous DE
′′′
′′
′^
a) The functions
𝑥
−
𝑥
2
𝑥^
are (particular) solutions (check!)
b)
𝑥^
−
𝑥^
and
2 𝑥
are LI (check!)
c) Therefore, the general solution of the given equation is given by:
𝑥
2
−
𝑥^
3
2 𝑥
2.2 General Solution of Homogeneous Linear ODEs
There is a simple test to determine whether a given set of functions is linearly independent ordependent on an open interval
, for some real numbers
, by using the idea of
determinant of a matrix. Definition 2.2.5.
Let
1
𝑛
be
real valued functions each of which has an
𝑡ℎ
derivative on the interval
[
].
The determinant:
W
[f
1
,^
f^2
f
n
] =
)^
)^
𝑛
)^
)^
′ 𝑛
...^
...^
( 𝑛
−
1
( 𝑛
−
2
( 𝑛
−
𝑛
W
is called the
Wronskian
of these
functions.
Example 2.2.4.
The function
2 𝑥
and
2
4 𝑥
are solutions of the second order
linear homogenous differential equation
′′
′^ + 4
.^
Then the Wronskian,
W
(x
)^
of
1
and
2
is
W
x
2 𝑥
4
𝑥
2
𝑥^
4 𝑥
4 𝑥
�����^
4 𝑥
4 𝑥
4 𝑥
4 𝑥
Question: Are the two functions
2 𝑥
and
4
𝑥^
linearly independent?
The above question can be easily answered using the following theorem. Theorem 2.2.
(Wronskian Test for Linearly Independence)
.^
The
functions
𝑛
are
Linearly Independent on an interval
[
]^
if and only if the
Wronskian
of
1
2
𝑛
is different
from zero for some
[
].
That is,
1
2
𝑛
are LI if and only if there exists
[
]
such that
W
)^
Example 2.2.5.
- Show that
and
2
are Linearly Independent.
Solution Consider the
Wronskian
of
and
2
W
(x
,^ x
2
2
�����^
2
2
2
This implies
W
2
2
,^
and hence
and
2
are LI.
- Show that
𝑥
−
𝑥
2 𝑥
are Linearly Independent.
2.2 General Solution of Homogeneous Linear ODEs
Solution Consider the
Wronskian
of
𝑥^ , 𝑒
−
𝑥
and
2 𝑥
W
(x
𝑥^
−
𝑥^
2 𝑥
𝑥^
−
𝑥
2 𝑥
𝑥^
−
𝑥^
2 𝑥
which is equal to
𝑥
𝑥
𝑥^
)^
−
𝑥
3 𝑥
3 𝑥
2 𝑥
2 𝑥
2 𝑥
2
𝑥
2 𝑥
,^
R
Hence,
𝑥^
−
𝑥^
and
2 𝑥
are Linearly Independent.
- Show that the functions
1
𝑥^
2
−
𝑥
) = sinh
are linearly dependent on
R
, since
3
) = sinh
𝑥^
−
𝑥
Solution Consider the
Wronskian
of
𝑥^ , 𝑒
−
𝑥
and
sinh
W
x
𝑥
−
𝑥
𝑥𝑒 −
−𝑒
𝑥 2
𝑥
−
𝑥^
𝑥𝑒
−𝑒
𝑥 2
𝑥
−
𝑥
𝑥𝑒 −
−𝑒
𝑥 2
,^
R
Hence
𝑥
−
𝑥
and
sinh
are linearly dependent.
Remark 2.2.7.
The Wronskian of
solutions
1
2
𝑛
of the DE (2.2) is either identically
zero on
[
]^
or else is never zero on
[
].
That is, if
1
2
𝑛^
are solutions of the DE (2.2),
then
,^
[
]^
if
1
2
𝑛
are LD. or
,^
[
]^
if
1
𝑛
are
LI.
2.2 General Solution of Homogeneous Linear ODEs
Reduction of Order
In the preceding section we saw that the general solution of a homogeneous linear second-orderdifferential equation
′′
is a linear combination
1
1
where
1
and
2
are linearly independent solu-
tions on some interval I.In this method we can construct a second solution
2
of a homogeneous equation (2.5) (even
when the coefficients in (2.5) are variable) provided that we know a nontrivial solution
1
of the
DE. The basic idea described in this section is that equation (2.5) can be reduced to a linearfirst-order DE by means of a substitution involving the known solution
. A second solution 1
2
of (2.5) is apparent after this first-order differential equation is solved.The method is described bellow.Suppose that
1
denotes a nontrivial solution of (2.5) and that
1
is defined on an interval I. We
want to find a second solution
2
so that the set consisting of
1
and
2
is linearly independent on I.
The quotient
2
1
is nonconstant on I, that is,
or
1
.^
The function
)^
can be found by substituting
𝑦
2
1
into the given differential equation.Consider the derivatives
′ 2
′^ 𝑦
1
′ 1
and
′′ 2
′′^ 𝑦
1
′^ 𝑦
′ 1
′′ 1
and substituting these in
(2.5) we get
′′^
1
′^ 𝑦
′ 1
′′ 1
′^ 𝑦
1
′ 1
1
and simplifying this gives us
′′
1
′^ (
′ 1
1
′′ 1
′ 1
1
2.3 Homogeneous LODE with Constant Coefficients
Case 1. Distinct Real Roots Suppose that (2.7) has
distinct roots,
2
𝑛
where
𝑖^
,𝑗^
for
𝑖^
Then, the
solutions
𝜆^1
𝑥
𝜆^2
𝑥
𝜆
𝑛^
𝑥^
are linearly independent. (Use the Wronskian to prove this.)
If
𝑛
are the
distinct real roots of (2.7), then the general solution of (2.6) is:
𝑦
𝜆
𝑥 1
𝜆
𝑥 2
𝑛
𝜆^ 𝑛
𝑥
𝑛 �^ 𝑖
=
𝜆
𝑥𝑖
where
2
𝑛
are arbitrary constants.
Example 2.3.1.
- For the differential equation
′′
′^
,^
the characteristic equation
is:
2
,^
and
1
and
2
are the two distinct real roots of this
characteristic equation.
Hence, the general solution of the given differential equation is
2 𝑥
2
𝑥
- For the differential equation
′′′
′′
′^
,^
the corresponding characteristic
equation is:
3
2
with distinct real roots
1
2
and
3
Therefore, the general solution of the give equation is
1
2 𝑥
3
𝑥^
3
−
𝑥^
Case 2. Repeated Real Roots To understand the situation let us consider the following example. Example 2.3.2.
Consider the DE
′′^
′^
.^
Then, its characteristic equation is
2
,^
which implies
2
.^
Therefore,
1
2
,^
which is a repeated real root.
One of the solutions of the given linear differential equation is
3 𝑥
Let
3 𝑥
.^
The given equation will have two linearly independent solutions and the second
solution can be found by using the method of
reduction of order.
Let
2
be another solution
so that
1
and
2
are linearly independent. Then
2
where
−
∫ −
6 𝑑𝑥
3 𝑥
2
6
𝑥 𝑒
6
𝑥
Therefore
2
3
𝑥^
and
1
3 𝑥
2
3 𝑥
is a general solution for constants
1
and
2
Remark 2.3.3.
Given a differential equation:
- if the characteristic equation has double real root
then
𝜆𝑥
and
𝜆𝑥
are two linearly
independent solutions and;
2.3 Homogeneous LODE with Constant Coefficients
- if the characteristic equation has triple root
then the corresponding linearly independent
solutions are
𝜆𝑥
𝜆𝑥
and
2
𝜆𝑥
Let us proof the first part of the above remark for a second order linear homogenous differentialequation .If the given DE is
′′
′^ +
,^
then its characteristic equation is
2
and then
1
2
−
𝑏 2 𝑎
One of the solution of the given DE is
1
𝜆𝑥
.^
Then we can use the method
of reduction of order to find a second solution
2
so that
1
and
2
are linearly independent.
The given equation is equivalent to
′′
′^
and
2
where
−
∫ 𝑏𝑎
𝑑𝑥
𝜆𝑥
2
−
𝑏𝑥𝑎 𝑒
2 𝜆𝑥
since
−
𝑏𝑎
and hence
2
𝜆𝑥
The following theorem is a generalization for the above remark. Theorem 2.3.4.
- If the characteristic equation (2.7) has the real root
occurring k times
1
2
𝑘^
)^
where
then the part of the general solution for (2.6) corresponding to
this k fold repeated root is
1
3
2
𝑘
𝑘
−
𝜆𝑥
- If further, the remaining roots are the distinct real roots
𝑘
𝑘+
𝑛
,^
the general
solution of (2.6) will be:
1
𝜆𝑥
2
𝜆𝑥
3
2
𝜆𝑥
𝑘
𝑘
−
𝜆𝑥
𝑘+
𝜆
𝑘+
𝑥
𝑛
𝜆
𝑛
𝑥^
Example 2.3.3.
- Consider the Differential Equation
(4)
′′′
′′^
′^
The corresponding characteristic equation is
4
3
2
and the roots are
1
2
3
4
Therefore, the general solution is given by
1
2 𝑥
2
2 𝑥
2
2 𝑥
4
−
𝑥^
,^
where
𝑐^1
3
and
4
are constants.
2.3 Homogeneous LODE with Constant Coefficients
- Consider the Differential Equation
′′′
′′
′^
The roots of the characteristic
equation are,
1
2
and
3
and hence the general solution of the equation is:
3 𝑥
2
3 𝑥
−
2
𝑥^
,^
where
1
2
and
3
are constants.
Case 3. Conjugate Complex Roots Suppose the equation (2.7) has a complex root
R
.^
Then (we know from the
theory of algebraic equations that) the conjugate
is also a root of (2.7) and the
corresponding part of the general solution of (2.6) will be:
(𝑎
𝑖𝑏
)𝑥
(𝑎
−
𝑖𝑏
)𝑥
But
𝑎
𝑖𝑏
𝑎^
𝑖𝑏
𝑎^
(cos
sin
(by applying Euler’s formula) and then
(𝑎
𝑖𝑏
)𝑥
2
(𝑎
−
𝑖𝑏
)𝑥
1
𝑎𝑥
(cos
sin
2
𝑎𝑥
(cos
𝑥^
sin
𝑎𝑥
[(
1
) cos 2
1
2
) sin
]
𝑎𝑥
𝑐^1
cos
2
sin
where
1
1
2
and
2
1
are arbitrary constants from the set of complex numbers
C
On the other hand if
and
are each k fold roots of (2.7), then the part of
the general solution that corresponds to this part is
𝑎𝑥
( [
1
𝑘
𝑘
−
cos
𝑘
𝑘+
2 𝑘
𝑘−
1
sin
].
Example 2.3.4.
Solve
′′
′^
Solution The characteristic equation of the given equation is
2
with roots
1
and
2
Then
1
1+
𝑖^
and
2
1 −
3 𝑖
are two independent solutions of the given
equation. Therefore,
1
1
where
1
and
2
are constants, is a general solution of the
given equation. That means
𝑥
1
cos 3
2
sin 3
2.4 Nonhomogeneous Equations with Constant Coefficients
Exercises
Exercise 2.3.5.
Solve each of the following Differential Equation.
′′
′′
′^
(4)
′′′
′′
′^
,^
where
1
2
𝑖^
and
3
4
Nonhomogeneous Equations with Constant Coefficients
Consider the equation
2
.^
which governs the displacement
of a mechanical oscillator. Here
)^
is the forcing function
and the equation is a non-homogeneous linear ODE with constant coefficients. There are severalpractical problems which can be modeled in this form.Recall that differential equations of the form
𝑛
(𝑛
)^
′^
,^
where
)^
are called nonhomogeneous differential equations.
In the previous sections we have seen how
to solve homogeneous differential equations.
In this section we are going to see how to solve
differential equations of the form
𝑛
(𝑛
)^
𝑛−
1
(𝑛
−
′^
where
𝑛^
0
are constants is called a nonhomogeneous differential equation with constant
coefficients. The following theorem is very important in such cases. Theorem 2.4.
(Homogeneous-Nonhomogeneous Solution Relation)
.^
Consider the nonhomoge-
neous differential equation
𝑛
(𝑛
)^
′^
0
,^
where
If
,^
then the equation becomes a homogeneous equation.
- If
1
and
2
are solutions of the nonhomogeneous equation on an interval I, then
1
2
is also a solution of the homogeneous equation in the interval I.
