Chapter 3
Second Order Differential Equations
3.1 Homogeneous Equations with constant
coefficients
A second order linear DE is written in the form
P (t ) y   Q (t ) y   R (t ) y  G (t )
y   p (t ) y   q (t ) y  g (t )
(1)
(2)
G (t )
Q (t )
R (t )
g
(
t
)

where p (t )  P (t ) , q (t )  P (t ) and
P (t ) .
Note: If the DE can't be written in the form (1)
and (2) then it is a nonlinear DE.
An initial value problem (IVP) consists of a DE
such as in (1) and (2) with a pair of initial
conditions
1
y (t 0 )  y 0, y (t 0 )  y 0
(3)
Where y 0 and y 0 are given numbers.
Definition: A second order linear equation is said
to be homogeneous if the term G (t ) in DE (1) or
the term g (t ) in DE (2) is zero for all t .
Otherwise the DE is called a nonhomogeneous.
Note: In this chapter, we will concentrate our
attention on DEs in which the functions P ,Q and
R in DE (1) are constants, that is in the form
ay   by   cy  0
(4)
Where a , b and c are constants.
rt
y

e
To solve DE (4), suppose
is a solution
of the DE (4).
y   re r t , y   r 2e r t
2
Now substitute in DE (4), we get
ar 2e r t  bre r t  ce r t  0
e r t (ar 2  br  c )  0
Since e r t  0 then
ar 2  br  c  0
(5)
Equation (5) is called the Characteristic equation
for the DE (4).
Assume the two roots of equation (5) are two
different real roots, r1 and r2 , r1  r2 .
r1 t
y

e
Then 1
and
y 2  e r2 t are two solutions of
the DE (4). The general solution of the DE (4) is
given by
y  c1 y 1  c 2 y 2  c1e r1 t  c 2e r2 t
3
(6)
Example 1: Solve the DE
y   y  0.
Example 2: Find the general solution of the DE
y   5 y   6 y  0.
4
Example 3: Find the solution of the IVP
y   5 y   6 y  0, y (0)  2, y (0)  3.
Example 4: Find the solution of the IVP
4 y   8 y   3 y  0, y (0)  2, y (0)  3.
5
H.W. Problems 1-18 Pages 142.
Example 5(Q.13, page 142). Solve the IVP
y   5 y   3 y  0, y (0)  1, y (0)  0.
6
Example 6(Q.18, page 142).
Find a DE whose general solution is
y  c1e

t
2
 c 2e  2t
3.2 Fundamental Solutions of Linear
Homogeneous Equations
Theorem 3.2.1:
Consider the initial value problem
y   p (t ) y   q (t ) y  g (t )
y (t 0 )  y 0, y (t 0 )  y 0
7
where p (t ), q (t ) and g (t ) are continuous on an
interval I that contains the point t 0 . Then there
is exactly one solution of this IVP, and the
solution exists throughout the interval I .
Note:
1.(Existence and Uniqueness)
Theorem 3.2.1 states that the given (IVP) has
a solution and it is unique.
It also states that the solution exists through any
interval I containing the initial point t 0 in which
p , q and g are continuous.
Example 1: Find the longest interval in which the
(IVP)
(t 2  3t ) y   ty   (t  3) y  0,
has a unique solution.
8
y (1)  2, y (1)  2
Theorem 3.2.2:
If y 1 and y 2 are two solutions of the DE
y   p (t ) y   q (t ) y  0
Then the linear combination c1 y 1  c 2 y 2 is also a
solution for any values of the constants c 1 and
c 2 on an interval I that contains the point t 0 .
Definition: Let y 1 and y 2 be two solutions of the
y   p (t ) y   q (t ) y  0 .
The wronskian determinate (or simply the
Wronskian) of the solutions y 1 and y 2 is given
by
y1
W ( y 1, y 2 ) 
y 1
Theorem 3.2.3:
9
y2
 y 1 y 2  y 2 y 1.
y 2
If y 1 and y 2 are two solutions of the DE
y   p (t ) y   q (t ) y  0
and the Wronskian W ( y 1 , y 2 )  y 1 y 2  y 2 y 1
is not zero at the point t 0 where the initial
conditions y (t 0 )  y 0, y (t 0 )  y 0 .
Then there are constants c 1 and c 2 for which
y  c1 y 1  c 2 y 2 the DE and the initial conditions.
Theorem 3.2.4:
If y 1 and y 2 are two solutions of the DE
y   p (t ) y   q (t ) y  0
and if there is a point t 0 where the Wronskian of
y 1 and
y 2 is nonzero, then the family of
solutions
y  c1 y 1  c 2 y 2
10
With arbitrary coefficients c 1 and c 2 includes
every solution of the DE.
Definition: The solution
y  c1 y 1  c 2 y 2 is
called the general solution of the DE
y   p (t ) y   q (t ) y  0 .
The solutions
y 1 and
y2
with a nonzero
Wronskian are said to form a fundamental set of
solutions of the DE.
r1 t
r t
y

e
y

e
1
Example 2: Suppose
and 2
are
2
two solutions of y   p (t ) y   q (t ) y  0 . Show
that they form a fundamental set of solutions if
r1  r2 .
11
1
2
y 1  t and y 2  t 1
Example 3: Show that
form a fundamental set of solutions of
2t 2 y   3ty   y  0, t  0.
12
H.W. Problems 1-13, 17-20, 23-26 (Pages
151-152).
Example 4(Q.12, page 151). Find the longest
interval in which the (IVP)
(x  2) y   y   (x  2)(tan x ) y  0,
has a unique solution.
13
y (3)  1, y (3)  2
Example 5(Q.18, page 151).
2 t
If the Wronskian of f and g is t e and if
f (t )  t find g (t ) .
Example 6(Q.25, page 151)
x
y

xe
y

x
2
Is the functions 1
and
form a
fundamental set of solutions of the DE
x 2 y   x (x  2) y   (x  2) y  0, x  0.
14
3.3 Linear independence and Wronskian
Definition:
Two functions f (t ) and g (t ) are said to be
linearly dependent on an interval I if there exist
two constants k 1 and k 2 no both zero, such that
k 1 f (t )  k 2 g (t )  0
For all t  I .
15
(1)
The functions f (t ) and g (t ) are said to be linearly
independent on an interval I if they are not
linearly dependent.
Theorem 3.3.1
If f (t ) and g (t ) are differentiable functions on
an open interval I and if W (f , g )(t 0 )  0 for some
point t 0  I then f (t ) and g (t ) are linearly
independent on an interval I .
Moreover, if f (t ) and g (t ) are linearly dependent
on an interval I then W (f , g )(t )  0 for every
t I .
Theorem 3.3.2 (Abel's Theorem)
If y 1 and y 2 are two solutions of the DE
16
y   p (t ) y   q (t ) y  0
(2)
Where p (t ) and q (t ) are continuous on an open
interval I , then the Wronskian W ( y 1 , y 1 )(t ) is
given by
W ( y 1 , y 1 )(t )  c exp    p (t ) dt 
(3)
Where c is a constant that depends on y 1 and
y 2 but not on t .
Note:
 0,
W ( y 1 , y 1 )(t ) 
 0,
c  0,
c  0.
1
2
1
Example 1: Assume that y 1  t and y 2  t are
solution of the DE
2t 2 y   3ty   y  0, t  0.
17
Use Abel's Theorem to find the constant c in (3).
Theorem 3.3.3:
Let y 1 and y 2 be solutions of the DE
y   p (t ) y   q (t ) y  0
where p (t ) and q (t ) are continuous on an open
interval
I,
then
y 1 and
y2
are linearly
dependent on I if and only if W ( y 1 , y 1 )(t ) is zero
for all t  I .
18
Alternatively, y 1 and y 2 are linearly independent
on I if and only if W ( y 1 , y 1 )(t ) is never zero for
all t  I .
Summary: We can summarize the facts about
fundamental set of solutions, Wronskian and
linear independence in the following
Let y 1 and y 2 be solutions of the DE
y   p (t ) y   q (t ) y  0
19
where p (t ) and q (t ) are continuous on an open
interval I . Then the following statements are
equivalent.
1- The
functions
y 1 and
y2
are
a
fundamental set of solutions on I .
2- The functions
y 1 and
y 2 are lineary
independent on I
3- W ( y 1 , y 1 )(t 0 )  0 for some t 0  I .
4- W ( y 1 , y 1 )(t )  0 for all t  I .
H.W. 1-13, 15-23 Page 158
Example 2(Q.4, page 158) Determine whether
the following functions are linearly independent
or linearly dependent
f (x )  e 3x and g (x )  e 3( x 1)
20
Example 3(Q.18, page 158) Find the Wronskian
of two solutions of the given DE
Example 4(Q.4, page 158)
21
3.4 Complex roots of the Characteristic
Equations
Consider the DE
ay   by   cy  0
where a , b and c are constants.
The Characteristic equation of the DE(1)
22
(1)
ar 2  br  c  0
Suppose
(2)
b 2  4ac  0 . Then the roots of Eq.(2)
are conjugate complex numbers, we denote them
by
r1    i , r2    i
(3)
Then the two solutions of DE (1), are
y 1  e (    i )t ,
y 2  e (    i )t
(4)
Euler,s formula
e ti  cos t  i sin t
(5)
e ti  cos t  i sin t
Now, using Euler,s formula, we have
y 1  e (   i )t  e t e ti  e t (cos t  i sin t )
(6)
y 2  e (   i )t  e t e  ti  e t (cos t  i sin t )
(7)
23
Note, the two solutions y 1 and y 2 are complexvalued functions. We want a real–valued
solutions. To get real solutions from y 1 , and y 2 ,
we use their sum and difference
y 1  y 2  e t (cos t  i sin t )  e t (cos t  i sin t )  2e t cos t
y 1  y 2  e t (cos t  i sin t )  e t (cos t  i sin t )  2ie t sin t
Let
W (u ,v )(t ) 
u (t )  e t cos t , v (t )  e t sin t
e  t cos  t
t
(8)
e  t sin  t
t
t
t
 e cos  t   e sin  t  e sin  t   e cos  t
If   0.
This implies that u (t ) and v (t ) form a
fundamental set of solutions of DE(1).
24
 e 2 t  0,
From the above, we conclude that, the general
solution of DE(1) is
y  c1e t cos t  c 2e t sin t
where
c 1 and c 2 are
arbitrary constants.
Example 1: Find the general solution of
y   y   y  0.
25
(9)
Example 2: Find the general solution of
y   9 y  0.
Example 3: Find the solution of the IVP
16 y   8 y   145 y  0, y (0)  2, y (0)  1.
H.W. 7-22, Page 164.
26
3.5 Repeated roots; Reduction of order
Consider the DE
ay   by   cy  0
(1)
where a , b and c are constants.
The Characteristic equation of the DE(1)
ar 2  br  c  0
Suppose
(2)
b 2  4ac  0 .
Then Eq.(2) has two equal roots
b
r1  r2 
2a
(3)
Then we have one solution of DE (1),
y1  e
27
b
t
2a
To find a second solution
y2
that is linearly
independent with the first solution
y 1,
assume the general solution of DE(1)
y  v (t ) y 1  v (t )e
y   v e
b
t
2a
y   v e
b
 ve
2a
b
t
2a
b
t
2a
b
t
2a
b
b
t
t
b
b2
2a
2a
 v e  2 v e
a
4a
Substitute in DE(1), we get
28
we
b
b
b
b
b
2
t
t
t
t
t



b
b
b
2a
2a
2a
2a
2a
a v e  v  e  2 v e   b v e  v e



a
4a
2a



b
t 

2a
c v (t )e   0.




b2 b2
av   (b  b )v   ( 
 c )v  0
4a 2a
v   0
v  c1t  c 2
y  v (t ) y 1  v (t )e
y1  e
b
t
2a
W ( y 1, y 2 ) 
and
e
b
t
2a
 (c1t  c 2 )e
y 2  te
b
t
2a
te
b
b 2a t
e
2a
This implies that
e
b
t
2a
b
t
2a
 c1te
b
t
2a
 c 2e
b
t
2a
b
t
2a
b
t
2a
b
bt 2a t

e
2a
e
b
t
a
 0.
y 1 and y 2 form a fundamental
set of solutions of DE(1).
29



Example 1: Find the general solution of the DE
y   4 y   4 y  0.
Example 2: Find the solution of the IVP
1
y   y   0.25 y  0, y (0)  2, y (0)  .
3
30
Summary
Now, we can summarize the results that
we have obtain for the second order linear
homogeneous
DEs
with
constant
coefficients as follows:
Consider the DE
ay   by   cy  0
Let
r1 and r2 be two roots of the
corresponding characteristic equation
ar 2  br  c  0
1- If r1 and r2 are different real roots
r1  r2 ,then the general solution is
given by
y  c1e r1t  c 2e r2t
31
2- If r1 and r2 are complex conjugate roots
r1     i ,
r2     i ,then the
general solution is given by
y  c1e t cos  t  c 2e t sin  t
.
3- If r  r1  r2 , then the general solution
is given by
y  c1e rt  c 2te rt .
32
Reduction of order
Suppose we know one solution y 1 (t ) ,
not every where zero of the DE
y   p (t ) y   q (t ) y  0
(4)
To find a second solution, let
y  v (t ) y 1 (t )
(5)
Then
y   v y 1 vy 1, y   v y 1  2v y 1 vy 1
Substitute y , y , y  in DE(4) and colleting,
we have
y 1v   (2 y 1  py 1 )v   ( y 1 py 1  qy )v  0
y 1v   (2 y 1  py 1 )v   0
33
(6)
Equation (6) is a first order equation for the
function v  and can be solved as a linear DE
or a separable DE. After finding v  , then v
can be obtained by an integration.
Note: This method is called the method of
reduction of order because the main step is
to solve DE (6) which is a first order DE for
v  rather than solving DE (4) which is
second order DE for y .
34
1
y

t
1
Example 3: Given that
is a solution of
2t 2 y   3ty   y  0, t  0,
Find a second linearly independent solution.
H.W. 1-14, 23-30 Page 172-174.
35
3.6 Nonhomogeneous Equations, Method of
undetermined coefficients
Consider the nonhomogeneous DE
y   p (t ) y   q (t ) y  g (t )
(1)
where p , q and g are given continuous functions
on an open interval I .
The DE
y   p (t ) y   q (t ) y  0
(2)
in which g (t )  0 and p , q as in DE(1) is called
the corresponding homogeneous DE to DE(1).
Theorem 3.6.1
If Y 1 and Y 2 are two solutions of
the nonhomogeneous DE(1), then their
difference Y 1 Y 2 is a solution of the
corresponding homogeneous DE(2).
36
In addition if y 1 and y 2 are a fundamental set of
solutions of DE(2), then
Y 1 (t ) Y 2 (t )  c1 y 1 (t )  c 2 y 2 (t )
Where c 1 and
c2
are constants.
Theorem 3.6.2 The general solution of the
nonhomogeneous DE(1) can be written in the
form
y (t )  c1 y 1 (t )  c 2 y 2 (t ) Y (t )
(3)
Where y 1 and y 2 are a fundamental set of the
corresponding homogeneous DE(2), c 1 and
c2
are arbitrary constants and Y is some specific
solution of the nonhomogeneous DE(1).
37
Note: The solution of the nonhomogeneous
DE(1) can be obtained using the following three
steps.
1.Find the complementary solution
y c  c1 y 1  c 2 y 2
which is the general solution of corresponding
homogeneous DE(2).
2.Find a particular solution y p of the
nonhomogeneous DE(1).
3.Find the general solution of the
nonhomogeneous DE(1) by adding y c and y p
From the first two steps
y  yc  y p
38
The method of undetermined coefficients
This method usually is used only for problems in
which
the homogeneous DE
has constant
coefficients and the nonhomogeneous terms that
consist of polynomials, exponential functions,
sines and cosines.
Example 1: Find a particular solution of the DE
y   3 y   4 y  3e 2t .
39
Example 2: Find a particular solution of the DE
y   3 y   4 y  2sin t .
Example 3: Find a particular solution of the DE
y   3 y   4 y  4t 2  1.
40
Note:
g (t )
e t
yp
Ae t
sin  t or cos  t
a0t n  a1t n 1   an
A sin  t  B cos  t
A 0t n  A1t n 1   A n
Remark 1:
When g (t ) is a product of any two or all three of
these types of functions, we use the product the
same functions.
Example 4: Find a particular solution of the DE
y   3 y   4 y  8e t cos 2t .
41
Remark 2:
Suppose that g (t )  g 1 (t )  g 2 (t ) and suppose
that Y p1 (t ) and Y p (t ) are particular solutions of
2
the DEs
ay   by   cy  g 1 (t ).
ay   by   cy  g 2 (t ).
respectively, then Y p1 (t ) +Y p (t ) is a particular
2
solution of the DE
ay   by   cy  g (t ).
Example 5: Find a particular solution of the DE
y   3 y   4 y  3e 2t  2sin t  8e t cos 2t .
42
Example 6: Find a particular solution of the DE
y   3 y   4 y  2e t .
Remark 3: If the form of the particular
solution that we choose is a solution of the
corresponding homogeneous DE, then we
s
multiply it by t , s  1, 2.
H.W. 1-26 Page 184.
43
Example 7(Q. 14 page 184): Solve the following
DE
y   4 y  t 2  3e t , y (0)  0, y (0)  1.
Example 8(Q. 26 page 184): Determine a sutible
form of the particular solution of the following
DE
y   2 y   5 y  3te t cos 2t  2te 2t cos t .
44
3.7 Variation of parameters
This method is more general the method of
undetermined coefficients, since it can enable us
to get a particular solution of an arbitrary second
order linear nonhomogeneous DE.
y   p (t ) y   q (t ) y  g (t )
(1)
where p , q and g are given continuous functions
on an open interval I .
If g (t ) is not one of the functions as in Section
3.6 ( exponential, polynomial, sines, cosines,
their product or sums) then we can't use the
method of undetermined coefficients, therefore
we
will
use
the
method
parameters.
45
of
variation
of
Suppose
y 1 , y 2 are
two linearly independent
solutions of the corresponding homogeneous
DE(2).
y   p (t ) y   q (t ) y  0
(2)
Then
y c  c1 y 1 (t )  c 2 y 2 (t ) ,
where c 1 and
c2
are constants is the solution of
DE(2).
In the method of Variation to find a particular
solution of DE(1), we replace the constants c 1
and
c2
by two functions u 1 (t ) and
u 2 (t )
particular solution
y p  u (t )1 y 1 (t )  u 2 (t ) y 2 (t )
46
to get a
where,
u1 (t )  
 y 2 (t ) g (t )
dt ,
W ( y 1 , y 2 )(t )
u 2 (t )  
y 1 (t ) g (t )
dt .
W ( y 1 , y 2 )(t )
The general solution of the nonhomogeneous
DE(1) can be written in the form
y (t )  c1 y 1 (t )  c 2 y 2 (t )  y p (t )
(3)
where y 1 and y 2 are a fundamental set of the
corresponding homogeneous DE(2), c 1 and
are arbitrary constants and
y p (t )
c2
is some specific
solution of the nonhomogeneous DE(1).
Example 1: Find a particular solution of the DE
y   4 y  3csc t .
47
H.W. 1-20, 29-32 Page 190, Page 192.
Example 2(Q. 5 page 190): Solve the following
DE
y   y  tan t , 0  t 
48

2
Example 3(Q. 17 page 190):
Verify that
y1  x 2
and
y 2  x 2 ln x
are solutions of
the corresponding homogeneous DE of the
following nonhomogeneous DE
x 2 y   3xy   4 y  x 2 ln x , x  0.
49
Example 4(Q. 29 page 192):
Use the method of reduction of order to solve
the DE
t 2 y   2ty   2 y  4t 2 , t  0,
50
y1 t.