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1.2 Second Order Ordinary Linear
Differential Equations
Focus of Attention
 What is a second order linear
homogenous differential equation?
 What is the characteristic or auxiliary
equation?
 What are the general solutions for
different types of roots of the auxiliary
equation?
 What is an initial value problem?
1.2.1 Definition and terminologies
A general second order linear differential
equation is of the form
d2y
dy
P 2  Q  Ry  S
dx
dx
Or
Py  Qy  Ry  S
Where P, Q, R and S can be constants or
functions of x
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A second order linear differential equations
with constant coefficients is of the form
ay  by  cy  f (x)
(1.1)
where a, b, c are constants and S = f(x)
ay  by  cy  f (x) is
o called homogeneous if
ay  by  cy  0
f ( x)  0 ;
i.e.
(1.2)
o called non-homogeneous if f ( x)  0
ay  by  cy  f (x)
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Question 3: second order differential
equations with constant coefficients
For each of the following, compare the given
equations to ay  by  cy  F ( x) . Thus,
identify the constants, F(x) and state if it is
homogenous or non-homogenous.
ODE
Constants F ( x ) / Homogenous/
F (t )
(1) y   3 cos x
(2)
y   3 y   2 y  0
(3)
y  5 y  6 y  3x
(4)
s  2s  4  t 2  1
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nonhomogenous
1.2.2 Solution of the second order
differential linear homogeneous
different equations [equations with
constant coefficients and right side zero]
A second order homogenous linear
differential equation with constant
coefficients is given as
d2y
dy
a
 b  cy  0
2
or
dx
dx
ay  by  cy  0
(1.2)
1.2.2.1 Method of solution
mx
y

e
Let
be a solution of
ay  by  cy  0
then
Then,
y  emx satisfies the equation.
y  e mx
dy
 me mx
dx
d2y
2 mx

m
e
2
dx
12
Therefore, substitute
2
mx dy
mx d y
y  e ;  me ; 2  m2emx in (1.2):
dx
dx
am 2e mx  bmemx  ce mx  0
e mx (am 2  bm  c)  0
mx
e
 0 for all real values of x, then
Since
y  emx is a solution of equation (1.2) if and
only if
am 2  bm  c  0
(1.3)
(1.3) is called the characteristic equation or
the auxiliary equation for (1.2)
13
Note
(1) The auxiliary equation is derived from
the differential equation by substituting
y
dy
dx
d2y
dx 2
(2)
 m0  1
 m1
 m2
Notice that
am2 + bm + c = 0
(1.3)
is a quadratic equation and will have two
roots which we will call m1 and m2
Thus, we will get two solutions for
ay  by  cy  0 which are
y  em1x
and
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y  em2 x
Theorem 1: The Superposition/Linearity
Principle
If y1 ( x) and y2 ( x) are two solutions to the
linear homogeneous equation,
ay  by  cy  0 , then for any constants A
and B , the linear combination
y  Ay1 ( x)  B y2 ( x)
is also a solution to the Equation with A and B
are constants.
1.2.2.2 General solutions
m1 x
m2 x
y

e
and
y

e
Since
are solutions of the
equation
then the general solution of second order
linear homogenous differential equation is
y  Aem1x  Be m2 x
Can we find m1 and m2 ?
15
1.2.2.3 To find
m1
and
m2
 am2 + bm + c = 0 is a quadratic
equation
 Solving quadratic equations:
o Factorisation
o Using the formula for finding roots of
quadratic equations:
 b  b 2  4ac
m
2a
There are 3 types of roots:
2
b
 4ac ,
(i) Roots are real and different, if
2
(ii) Roots are real and equal, if b  4ac ,
(iii) Roots are not real or complex number
2
b
roots, if  4ac .
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2
(i) Roots are real and different if b  4ac
Example 1.4: Solution of differential
equation with real and different roots in
the auxiliary equation
Solve the equation y   3 y   2 y  0 .
Solution:
Find auxiliary equation:
Find roots:
General solution:
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2
b
 4ac  0
(ii) Roots real and equal, if
 b  b 2  4ac
From the formula: m 
;
2a
b
2
m

If b  4ac  0 , then
2a .
mx
mx
y

Ae

Be
 Solution of DE is:
mx
y

Ce
 one solution,
.

Need two different solutions so we
mx
mx
y

Ae
y

Bxe
choose 1
and 2

The general solution is
y  Aemx  Bxemx OR
y   A  Bx  emx .
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Example 1.5: Solution of differential
equation with real and equal roots in the
auxiliary equation
Solve the equation y   6 y  9 y  0 .
Solution:
Find auxiliary equation:
Find roots: factoring
General solution:
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Quick Revision: Complex numbers
2
Imaginary number: i  j where i  1 or
1  i .
Cartesian form: z  x  iy or z  x  jy
Polar form:
z  cos  i sin 
z  cos  i sin 
ei  cos   i sin 
Euler’s form: e i  cos  i sin 
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(iii)Roots are not real or roots are complex
2
numbers, if b  4ac
2
b
If  4ac ,
 b  b 2  4ac
m
has no real roots but we
2a
will have two complex number roots,
m    i
m    i
and
The general solution to the differential
equation becomes
y  Ae
 ix
 Be
Using exponential laws:


  i  x

y  A e x e  xi  B e x e   xi

y  e x Ae  xi  Be   xi


Using trigonometric representation: (see
revision notes)
y  e x  A  cos  x  i sin  x   B  cos  x  i sin  x 
y  ex (C cos x  D sin x)
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If m     i , the general solution to the
differential equation is
y  ex (C cos x  D sin x)
Example 1.6: Solution of differential
equation with complex number roots in the
auxiliary equation
Solve the equation y  4 y  13 y  0 .
Solution:
Find auxiliary equation:
Find roots:
General solution:
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Summary of the types of roots and general
solutions
Table 1.1
Roots of the auxiliary equation and the
general solution of the Differential Equation
ay  by  cy  0
2
am
 bm  c  0
auxiliary equation :
Case Roots are
General solution
1
real and unequal: y  Aem1x  Be m2 x
m1 and: m2
2
real and equal,
y  ( A  Bx )e mx
m  m1  m2
3
complex
numbers
m1    i ;
y  ex ( A cos x  B sin x)
m2    i
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Steps to solve a homogenous differential
equation:
1. Write the DE in the form of
a
d2y
dy

b
 cy  0
2
dx
dx
OR
ay   by   cy  0
2. Find the characteristic equation,
am2 + bm + c = 0.
3. Find the roots of the characteristic equation
4. Find the general solution.
5. If initial conditions are given, substitute
the values in the general solution to find
the values of the constants.
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Structured Examples: General solution of
second order linear homogenous
differential equations
 Solving second order linear homogenous
differential equations
 Finding roots of the characterisitc or
auxiliary equation
 Determining the general solution of the
differential equations
Question 4
(a) Solve the equation
Prompts/ Questions
 How do you solve a
second orer linear
homogenous
differential equation?
o What is an auxiliary
equation?
o How do you find its
roots?
o What types of roots
did you get?
2 y   3 y   y  0 .
(b)
Solve the equation
y  8 y  16 y  0 .
 What is the general
solution?
25
Question 5
Solve the equation
Prompts/ Questions
 How do you solve a
second orer linear
homogenous
differential
equation?
o What is an
auxiliary
equation?
o How do you find
its roots?
o What types of
roots did you get?
y  2 y  5 y  0 .
 What is the general
solution?
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Question 6
Prompts/ Questions
Find the general solutions of  How do you
the following differential
solve a second
equations:
orer linear
2
d y dy
homogenous


6
y

0
2
1. dx dx
differential
equation?
2. y   4 y  0
d s
ds
o What is an
3. dt  8 dt  16s  0
auxiliary

y

4
y

0
4.
equation?
d2y
dy
o How do you

2
5. dx 2 dx  4 y  0
find its roots?
6. 9 y  6 y  y  0
o What types of
roots did you
get?
 What is the
general solution?
2
2
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1.2.3 Initial Conditions and Initial Value
Problem
Definition
Initial conditions are the conditions set
upon the same independent variable.
Initial Value Problem is solving a
differential equation with given initial
conditions.
Particular solution
A particular solution is the solution of
differential equations that has satisfied the
initial conditions, i.e., the arbitrary constants
can be determined.
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Example 1.7:
Solve the equation y   4 y   3 y  0 if
y(1)  5;
y (1)  2
Solution:
Find auxiliary equation:
Find roots:
General solution:
Particular solution:
Given boundary conditions are
y  5 when x  1 and y  2 when x  1 .
Find: A and B.
3
y  2, x  1 2  Ae  Be
y   5 x  1 y  Ae x  3Be3 x
5  Ae  3Be3
Solve simultaneously:
 A = 0.1839; B = 0.0747
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Structured Examples: General solution of
differential equations with initial
conditions
 Solving initial value problems
 Finding roots of the characterisitc or
auxiliary equation
 Determining the general solution of the
differential equations
 Finding the particular solution of
differential equations
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Prompts/ Questions
 How do you solve a second orer linear
homogenous differential equation?
 What is the general solution?
 What is the particular solution?
o What are the initial conditions?
Question 7
Find the solution to each of the following
Differential Equations with the given initial
conditions.
a)
y  y  2 y  0,
b) y  6 y  9 y  0,
c) y  2 y  5 y  0,
y(0)  0
y (0)  3,
y (0)  1,
 
y   0,
2
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y(0)  2
 
y   1
2