Peter V. O’Neil
Advanced Engineering Mathematics, 7th Edition
CHAPTER 2
Linear Second-Order
Equations
© 2012 Cengage Learning Engineering. All Rights Reserved.
• A second-order differential equation is one containing
a second derivative but no higher derivative. The
theory of second-order differential equations is vast,
and we will focus on linear second-order equations,
which have many important uses.
CHAPTER 2 : Page 49
© 2012 Cengage Learning Engineering. All Rights Reserved.
2
2.1 The Linear Second-Order Equation
• This section lays the foundations for writing solutions
of the second-order linear differential equation.
Generally, this equation is
P( x) y  Q( x) y( x)  R( x) y ( x)  F ( x).
CHAPTER 2 : Page 49
© 2012 Cengage Learning Engineering. All Rights Reserved.
3
2.1 The Linear Second-Order Equation
• Notice that this equation “loses” its second derivative
at any point where P(x) is zero, presenting technical
difficulties in writing solutions. We will therefore
begin by restricting the equation to intervals (perhaps
the entire real line) on which P(x)  0. On such an
interval, we can divide the differential equation by
P(x) and confine our attention to the important case
(2.1)
y  p ( x) y  q ( x) y  f ( x).
We will refer to this as the second-order linear
differential equation.
CHAPTER 2 : Page 49
© 2012 Cengage Learning Engineering. All Rights Reserved.
4
2.1 The Linear Second-Order Equation
• Often, we assume that p and q (coefficient functions)
are continuous (at least on the interval where we seek
solutions). The function f is called a forcing function
for the differential equation, and in some applications,
it can have finitely many jump discontinuities.
CHAPTER 2 : Page 49
© 2012 Cengage Learning Engineering. All Rights Reserved.
5
2.1 The Linear Second-Order Equation
• To get some feeling for what we are dealing with,
consider a simple example
y − 12x = 0.
• Since y = 12x, we can integrate once to obtain
y( x)   12 x dx  6 x 2  c
and then once again to get
y( x)  2x  cx  k
3
with c and k as arbitrary constants.
CHAPTER 2 : Page 49-50
© 2012 Cengage Learning Engineering. All Rights Reserved.
6
2.1 The Linear Second-Order Equation
• It seems natural that the solution of a second-order
differential equation, which involves two
integrations, should contain two arbitrary constants.
For any choices of c and k, we can graph the
corresponding solution, obtaining integral curves.
Figure 2.1 shows integral curves for several choices
of c and k.
Usually we said those two arbitrary constants are
independent. Here “independent” means a different set of
constants will results in different solution functions.
CHAPTER 2 : Page 50
© 2012 Cengage Learning Engineering. All Rights Reserved.
7
2.1 The Linear Second-Order Equation
CHAPTER 2 : Page 50
© 2012 Cengage Learning Engineering. All Rights Reserved.
8
2.1 The Linear Second-Order Equation
• Unlike the first-order case, there may be many
integral curves through a given point in the plane. In
this example, if we specify that y(0) = 3, then we
must choose k = 3, leaving c still arbitrary. These
solutions through (0, 3) are
y(x) = 2x3 +cx + 3.
Some of these curves are shown in Figure 2.2.
CHAPTER 2 : Page 50
© 2012 Cengage Learning Engineering. All Rights Reserved.
9
2.1 The Linear Second-Order Equation
CHAPTER 2 : Page 50
© 2012 Cengage Learning Engineering. All Rights Reserved.
10
2.1 The Linear Second-Order Equation
• We single out exactly one of these curves if we
specify its slope at (0, 3). For example, if we specify
that y(0) = −1, then
y(0) = c = −1,
so y(x) = 2x3 −x + 3. This is the only solution passing
through (0, 3) with slope −1.
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
11
2.1 The Linear Second-Order Equation
• To sum up, in this example, we obtain a unique
solution by specifying a point that the graph must
pass through, together with the slope this solution
must have at this point.
• Thus, by specifying two conditions for a point
y( x0 )  A and y( x0 )  B,
• it is possible to get a unique solution. Such kind of
conditions are referred to initial conditions.
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
12
2.1 The Linear Second-Order Equation
• This leads us to define the initial value problem for
the linear second-order differential equation as the
problem
y  p( x) y  q( x) y  f ( x); y ( x0 )  A, y ( x0 )  B
in which x0, A, and B are given. We will state,
without proof, an existence theorem for this initial
value problem.
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
13
THEOREM 2.1
Existence of Solutions
• Let p, q, and f be continuous on an open interval I .
Then the initial value problem
y  p( x) y  q( x) y  f ( x); y( x0 )  A, y( x0 )  B,
has a unique solution on I .
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
14
2.1 The Linear Second-Order Equation
• We now have an idea of the kind of problem we will
be solving and of some conditions under which we
are guaranteed a solution. Now we want to develop a
strategy to follow to solve linear equations and initial
value problems. This strategy will be in two steps,
beginning with the case that f (x) is identically zero.
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
15
The Structure of Solutions
The second-order linear homogeneous equation has
the form
y + p(x)y + q(x)y = 0.
(2.2)
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
16
The Structure of Solutions
• If y1 and y2 are solutions and c1 and c2 are numbers,
we call c1y1 + c2y2 a linear combination of y1 and y2.
It is an important property of the homogeneous linear
equation (2.2) that a linear combination of solutions
is again a solution.
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
17
THEOREM 2.2
• Every linear combination of solutions of the
homogeneous linear equation (2.2) is also a solution.
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
18
THEOREM 2.2－Proof
• Let y1 and y2 be solutions, and let c1 and c2 be
numbers. Substitute c1y1 + c2y2 into the differential
equation:
(c1 y1  c2 y2 )  p ( x)(c1 y1  c2 y2 )  q ( x )(c1 y1  c2 y2 )
 c1 y1  c2 y2  c1 p ( x) y1  c2 p ( x ) y2  c1q ( x ) y1  c2q ( x ) y2
 c1[ y1  p ( x) y1  q ( x) y1 ]  c2 [ y2  p ( x) y2  q ( x ) y2 ]
 c1 (0)  c2 (0)  0
because of the assumption that y1 and y2 are both
solutions.
CHAPTER 2 : Page 51
© 2012 Cengage Learning Engineering. All Rights Reserved.
19
2.1 The Linear Second-Order Equation
• The point to taking linear combinations c1 y1 +c2 y2 is
to generate new solutions from y1 and y2. However, if
y2 is itself a constant multiple of y1, say y2 = ky1, then
the linear combination
c1 y1  c2 y2  c1 y1  c2 ky1  (c1  c2 k ) y1
is just a constant multiple of y1, so y2 does not
contribute any new information. This leads to the
following definition.
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
20
2.1 The Linear Second-Order Equation
Two functions are linearly independent on an open
interval I (which can be the entire real line) if neither
function is a constant multiple of the other for all x in
the interval. If one function is a constant multiple of
the other on the entire interval, then these functions
are called linearly dependent.
• The definition of “linear independency” is the same
as that you will learn in Linear Algebra.
 Anyone cannot be represented by the others.
• Now, it is simple to define for only two functions.
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
21
EXAMPLE 2.1
• cos(x) and sin(x) are solutions of y + y = 0 over the
entire real line. These solutions are linearly
independent, because there is no number k such that
cos(x) = k sin(x) or sin(x) = k cos(x) for all x. Because
these solutions are linearly independent, linear
combinations c1cos(x) + c2sin(x) give us new
solutions, not just constant multiples of one of the
known solutions.
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
22
2.1 The Linear Second-Order Equation
• There is a simple test to determine whether two
solutions of equation (2.2) are linearly independent or
dependent on an open interval I . Define the
Wronskian W(y1, y2) of two solutions y1 and y2 to be
the 2 × 2 determinant
y1 y2
W ( y1 , y2 ) 
 y1 y2  y2 y1.
y1 y2
Often we denote this Wronskian as just W(x).
Why need Wronskian? It is simple to test linear independence for two
functions. But for more functions or power series functions, it may not
be easy to check.
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
23
THEOREM 2.3
Properties of the Wronskian
•
Suppose y1 and y2 are solutions of equation (2.2) on
an open interval I .
1. Either W(x)=0 for all x in I, or W(x)  0 for all x
in I .
2. y1 and y2 are linearly independent on I if and only
if W(x)  0 on I .
• The proof will be given later.
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
24
2.1 The Linear Second-Order Equation
• Conclusion (2) is called the Wronskian test for linear
independence. Two solutions are linearly independent
on I exactly when their Wronskian is nonzero on I . In
view of conclusion (1), we need only check the
Wronskian at a single point of I , since the Wronskian
must be either identically zero on the entire interval
or nonzero on all of I . It cannot vanish for some x
and be nonzero for others in I .
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
25
THEOREM 2.3
Properties of the Wronskian
•
Note that the above conclusions are for y1 and y2 are
solutions of equation (2.2) on an open interval I .
•
Suppose y1 and y2 are two functions.
1. There is no entirely zero or nonzero property.
2. y1 and y2 are linearly independent on I if and only
if W(x)  0 for some x on I .
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
26
EXAMPLE 2.2
• Check by substitution that y1(x) = e2x and y2(x) = xe2x
are solutions of y − 4y + 4y = 0 for all x. The
Wronskian is
W ( x) 
e2 x
2e
2x
xe 2 x
e  2 xe
2x
 e  2xe  2xe  e ,
4x
2x
4x
4x
4x
and this is never zero, so y1 and y2 are linearly
independent solutions.
CHAPTER 2 : Page 52
© 2012 Cengage Learning Engineering. All Rights Reserved.
27
2.1 The Linear Second-Order Equation
• In many cases, it is obvious whether two functions
are linearly independent or dependent. However, the
Wronskian test is important, as we will see shortly
(for example, in Section 2.4.1 and in the proof of
Theorem 2.4).
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
28
2.1 The Linear Second-Order Equation
• We are now ready to determine what is needed to find
all solutions of the homogeneous linear equation y +
p(x)y + q(x)y = 0. We claim that, if we can find two
linearly independent solutions, then every solution
must be a linear combination of these two solutions.
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
29
THEOREM 2.4
• Let y1 and y2 be linearly independent solutions of y +
py + qy = 0 on an open interval I . Then every
solution on I is a linear combination of y1 and y2.
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
30
THEOREM 2.4
This provides a strategy for finding all solutions of the
homogeneous linear equation on an open interval.
1. Find two linearly independent solutions y1 and y2.
2. The linear combination
y = c1 y1 + c2 y2
contains all possible solutions.
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
31
THEOREM 2.4
For this reason, we call two linearly independent
solutions y1 and y2 a fundamental set of solutions on I ,
and we call c1 y1 + c2 y2 the general solution of the
differential equation on I .
Once we have the general solution c1 y1 + c2 y2 of the
differential equation, we can find the unique solution
of an initial value problem by using the initial
conditions to determine c1 and c2.
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
32
THEOREM 2.4－Proof
• Let φ be any solution of y  py  qy  0on I. We
want to show that there are numbers c1 and c2 such
that   c1 y1  c2 y2 . (not to show c1 y1  c2 y2 is a
solution, which is Theorem 2.2.)
• Choose any x0 in I. Let  ( x0 )  A and  ( x0 )  B .
Then φ is the unique solution of the initial value
problem
y  py  qy  0; y( x0 )  A, y( x0 )  B
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
33
THEOREM 2.4－Proof
• Now consider the two algebraic equations in the two
unknowns c1 and c2:
y1 ( x0 )c1  y2 ( x0 )c2  A
y'1 ( x0 )c1  y '( x0 )c2  B.
Because y1 and y2 are linearly independent, their
Wronskian W(x) is nonzero. These two algebraic
equations therefore yield
Ay2 ( x0 )  By2 ( x0 )
By1 ( x0 )  Ay1( x0 )
c1 
and c2 =
W ( x0 )
W ( x0 )
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
34
THEOREM 2.4－Proof
• With these choices of c1 and c2, it is easy to prove
c1 y1  c2 y2 is a solution of
y  py  qy  0; y( x0 )  A, y( x0 )  B
• φ is also a solution of the initial value problem. Since
this problem has the unique solution, then we can
claim   c1 y1  c2 y2, as we wanted to show.
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
35
EXAMPLE 2.3
• ex and e2x are solutions of y − 3y + 2y = 0.
Therefore, every solution has the form
y ( x)  c1e x  c2e 2x .
This is the general solution of y − 3y + 2y = 0.
CHAPTER 2 : Page 54
© 2012 Cengage Learning Engineering. All Rights Reserved.
36
EXAMPLE 2.3
• If we want to satisfy the initial conditions y(0) = −2,
y(0) = 3, choose the constants c1 and c2 so that
y(0) = c1 + c2 = −2
y'(0) = c1 + 2c2 = 3.
Then c1 = −7 and c2 = 5, so the unique solution of the
initial value problem is
y(x) = −7ex + 5e2x .
CHAPTER 2 : Page 54
© 2012 Cengage Learning Engineering. All Rights Reserved.
37
The Nonhomogeneous Case
• We now want to know what the general solution of
equation (2.1) looks like when f (x) is nonzero at least
for some x. In this case, the differential equation is
nonhomogeneous.
• The main difference between the homogeneous and
nonhomogeneous cases is that, for the
nonhomogeneous equation, sums and constant
multiples of solutions need not be solutions.
CHAPTER 2 : Page 54
© 2012 Cengage Learning Engineering. All Rights Reserved.
38
EXAMPLE 2.4
• We can check by substitution that sin(2x) +2x and
cos(2x) + 2x are solutions of the nonhomogeneous
equation y + 4y = 8x. However, if we substitute the
sum of these solutions, sin(2x) + cos(2x) + 4x, into
the differential equation, we find that this sum is not a
solution. Furthermore, if we multiply one of these
solutions by 2, taking, say, 2 sin(2x) + 4x, we find
that this is not a solution either.
CHAPTER 2 : Page 54
© 2012 Cengage Learning Engineering. All Rights Reserved.
39
2.1 The Linear Second-Order Equation
• However, given any two solutions Y1 and Y2 of
equation (2.1), we find that their difference Y1 − Y2 is
a solution, not of the nonhomogeneous equation, but
of the associated homogeneous equation (2.2). To see
this, substitute Y1 − Y2 into equation (2.2):
(Y1  Y2 )  p ( x)(Y1  Y2 )  q ( x)(Y1  Y2 )
 [Y1 p ( x)Y1  q ( x)Y1  Y2 p ( x)Y2  q ( x)Y2 ]
 f ( x)  f ( x)  0.
CHAPTER 2 : Page 54
© 2012 Cengage Learning Engineering. All Rights Reserved.
40
2.1 The Linear Second-Order Equation
• But the general solution of the associated
homogeneous equation (2.2) has the form c1 y1 + c2
y2, where y1 and y2 are linearly independent solutions
of the homogeneous equation (2.2). Since Y1 − Y2 is a
solution of this homogeneous equation, then for some
numbers c1 and c2:
Y1  Y2  c1 y1  c2 y2 ,
which means that
Y1  c1 y1  c2 y2  Y2 .
CHAPTER 2 : Page 54
© 2012 Cengage Learning Engineering. All Rights Reserved.
41
2.1 The Linear Second-Order Equation
• But Y1 and Y2 are any solutions of equation (2.1). This
means that, given any one solution Y2 of the
nonhomogeneous equation, any other solution has the
form c1 y1 + c2 y2 + Y2 for some constants c1 and c2.
• We will summarize this discussion as a general
conclusion, in which we will use Yp (for particular
solution) instead of Y2 of the discussion.
CHAPTER 2 : Page 54
© 2012 Cengage Learning Engineering. All Rights Reserved.
42
THEOREM 2.5
• Let Yp be any solution of the nonhomogeneous
equation (2.1). Let y1 and y2 be linearly independent
solutions of equation (2.2). Then the expression
c1 y1 + c2 y2 + Yp
contains every solution of equation (2.1).
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
43
THEOREM 2.5
•
For this reason, we call c1 y1 + c2 y2 + Yp the general
solution of equation (2.1).
• Theorem 2.5 suggests a strategy for finding all
solutions of the nonhomogeneous equation (2.1).
1. Find two linearly independent solutions y1 and y2
of the associated homogeneous equation y +
p(x)y + q(x)y = 0.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
44
THEOREM 2.5
2. Find any particular solution Yp of the
nonhomogeneous equation y + p(x)y + q(x)y = f (x).
3. The general solution of y + p(x)y + q(x)y = f (x) is
y(x) = c1 y1(x) + c2 y2(x) + Yp(x)
in which c1 and c2 can be any real numbers.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
45
THEOREM 2.5
• If there are initial conditions, use these to find the
constants c1 and c2 to solve the initial value problem.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
46
EXAMPLE 2.5
• We will find the general solution of
y + 4y = 8x.
• It is routine to verify that sin(2x) and cos(2x) are
linearly independent solutions of y + 4y = 0.
Observe also that Yp(x) = 2x is a particular solution of
the nonhomogeneous equation.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
47
EXAMPLE 2.5
• Therefore, the general solution of y + 4y =8x is
y = c1 sin(2x) + c2 cos(2x) + 2x.
This expression contains every solution of the given
nonhomogeneous equation by choosing different
values of the constants c1 and c2.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
48
EXAMPLE 2.5
• Suppose we want to solve the initial value problem
y  4y  8x; y    1, y    6.
• First we need
y    c2cos(2 )  2  c2  2  1,
so c2 = 1 − 2π.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
49
EXAMPLE 2.5
• Next, we need
y    2c1cos(2 )  2c2sin(2 )  2  2c1  2  6,
so c1 = −4. The unique solution of the initial value
problem is
y ( x)  4 sin(2x)  (1  2 )cos(2x)  2x.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
50
2.1 The Linear Second-Order Equation
• We now have strategies for solving equations (2.1)
and (2.2) and the initial value problem. We must be
able to find two linearly independent solutions of the
homogeneous equation and any one particular
solution of the nonhomogeneous equation. We now
will develop important cases in which we can carry
out these steps.
CHAPTER 2 : Page 55
© 2012 Cengage Learning Engineering. All Rights Reserved.
51
Homework of Section 2.1
•1, 2, 8, 9, 11, 12, 13, 14, 15
© 2012 Cengage Learning Engineering. All Rights Reserved.
2.2 Reduction of Order
• Given y'' + p(x)y' + q(x)y = 0, we want two
independent solutions. Reduction of order is a
technique for finding a second solution, if we can
somehow produce a first solution.
CHAPTER 2 : Page 56
© 2012 Cengage Learning Engineering. All Rights Reserved.
53
2.2 Reduction of Order
• Suppose we know a solution y1, which is not
identically zero. We will look for a second solution of
the form y2(x) = u(x)y1(x). Compute
y'2 = u'y1 + uy'1, y''2= u''y1+2u'y'1 + uy''1.
• In order for y2 to be a solution we need
u''y1+2u'y'1 + uy''1 + p[u'y1 + uy'1] + quy1 = 0.
CHAPTER 2 : Page 57
© 2012 Cengage Learning Engineering. All Rights Reserved.
54
2.2 Reduction of Order
• Rearrange terms to write this equation as
u''y1 + u'[2y'1 + py1] + u[y''1 + py'1 + qy1] = 0.
• The coefficient of u is zero because y1 is a solution.
Thus we need to choose u so that
u''y1 + u' [2y'1 + py1] = 0.
• On any interval in which y1(x)  0, we can write
2 y '1  py1
u'' 
u'  0.
y1
CHAPTER 2 : Page 57
© 2012 Cengage Learning Engineering. All Rights Reserved.
55
2.2 Reduction of Order
• To help focus on the problem of determining u,
denote
2 y'1 ( x)  p( x) y1 ( x)
g ( x) 
y1 ( x)
• a known function because y1(x) and p(x) are known.
Then
u'' + g(x)u' = 0.
CHAPTER 2 : Page 57
© 2012 Cengage Learning Engineering. All Rights Reserved.
56
2.2 Reduction of Order
• Let v = u' to get
v' + g(x)v = 0.
• This is a linear first-order differential equation for v,
with general solution
v(x)= Ce g(x)dx.
• Since we need only one second solution y2, we will
take C = 1, so
v(x)= e g(x)dx.
CHAPTER 2 : Page 57
© 2012 Cengage Learning Engineering. All Rights Reserved.
57
2.2 Reduction of Order
• Finally, since v = u',
u ( x)   e 
 g ( x ) dx
.
• If we can perform these integrations and obtain u(x),
then y2 = uy1 is a second solution of y'' + py' + qy = 0.
Further,
W ( x)  y1 y '2  y '1 y2  y1 (uy '1  u ' y1 )  y '1 uy1  u ' y12  vy12 .
CHAPTER 2 : Page 57
© 2012 Cengage Learning Engineering. All Rights Reserved.
58
2.2 Reduction of Order
• Since v(x) is an exponential function, v(x)  0. And
the preceding derivation was carried out on an
interval in which y1(x)  0. Thus W(x)  0 and y1 and
y2 form a fundamental set of solutions on this
interval. The general solution of y'' + py' + qy = 0 is
c1y1 + c2y2.
• We do not recommend memorizing formulas for g, v
and then u. Given one solution y1, substitute y2 = uy1
into the differential equation and, after the
cancellations that occur because y1 is one solution,
solve the resulting equation for u(x).
CHAPTER 2 : Page 57
© 2012 Cengage Learning Engineering. All Rights Reserved.
59
EXAMPLE 2.6
• Suppose we are given that y1(x) = e2x is one solution
of y'' + 4y' + 4y = 0. To find a second solution, let
y2(x) = u(x)e2x. Then
y '2  u'e
2 x
2 x
 2e u
and y ''2  u '' e2 x  4e 2 xu  4u ' e 2 x .
• Substitute these into the differential equation to get
u '' e2 x  4e2 xu  4u ' e2 x  4(u ' e2 x  2e2 xu)  4ue2 x  0.
CHAPTER 2 : Page 58
© 2012 Cengage Learning Engineering. All Rights Reserved.
60
EXAMPLE 2.6
• Some cancellations occur because e2x is one solution,
leaving
u''e2x = 0,
• or
u'' = 0.
• Two integrations yield u(x) = cx + d. Since we only
need one second solution y2, we only need one u, so
we will choose c = 1 and d = 0.
CHAPTER 2 : Page 58
© 2012 Cengage Learning Engineering. All Rights Reserved.
61
EXAMPLE 2.6
• This gives u(x) = x and
y2(x) = xe2x .
• Now
W ( x) 
e
2 x
2e
2 x
xe
e
2 x
2 x
 2 xe
2 x
e
4 x
0
• for all x. Therefore, y1 and y2 form a fundamental set
of solutions for all x, and the general solution of y'' +
4y' + 4y = 0 is
y(x) = c1e2x + c2xe2x.
CHAPTER 2 : Page 58
© 2012 Cengage Learning Engineering. All Rights Reserved.
62
EXAMPLE 2.7
• Suppose we want the general solution of y''  (3 / x)y'
+ (4 / x2)y = 0 for x >0, and somehow we find one
solution y1(x) = x2. Put y2(x) = x2u(x) and compute
y’2 = 2xu + x2u’ and y''2 = 2u + 4xu' + x2u''.
• Substitute into the differential equation to get
3
4 2
2
2u  4 xu ' x u '' (2 xu  x u ')  2 ( x u )  0.
x
x
2
CHAPTER 2 : Page 58
© 2012 Cengage Learning Engineering. All Rights Reserved.
63
EXAMPLE 2.7
• Then
x2u'' + xu' = 0.
• Since the interval of interest is x >0, we can write this
as
xu'' + u' = 0.
• With v = u', this is
xv' + v = (xv)' = 0,
1
• so xv = c. We will choose c = 1. Then v  u ' 
x
CHAPTER 2 : Page 58
© 2012 Cengage Learning Engineering. All Rights Reserved.
64
EXAMPLE 2.7
• so
u = ln(x) + d,
• and we choose d = 0 because we need only one
suitable u. Then y2(x) = x2ln(x) is a second solution.
Further, for x >0,
 x2
x 2ln( x)  3
W ( x)  
  x  0.
 2 x 2 x ln( x)  x 
CHAPTER 2 : Page 59
© 2012 Cengage Learning Engineering. All Rights Reserved.
65
EXAMPLE 2.7
• Then x2 and x2ln(x) form a fundamental set of
solutions for x >0. The general solution is for x >0 is
y(x) = c1x2 + c2x2 ln(x) .
CHAPTER 2 : Page 59
© 2012 Cengage Learning Engineering. All Rights Reserved.
66
Homework of Section 2.2
•1, 4, 6, 9, 12(a)(d), 13(a)(d)(e)
© 2012 Cengage Learning Engineering. All Rights Reserved.
2.3 The Constant Coefficient Case
• We have outlined strategies for solving second-order
linear homogeneous and nonhomogeneous
differential equations. In both cases, we must begin
with two linearly independent solutions of a
homogeneous equation. This can be a difficult
problem. However, when the coefficients are
constants, we can write solutions fairly easily.
CHAPTER 2 : Page 60
© 2012 Cengage Learning Engineering. All Rights Reserved.
68
2.3 The Constant Coefficient Case
• Consider the constant-coefficient linear homogeneous
equation
y +ay +by = 0
(2.3)
in which a and b are constants (numbers). A method
suggests itself if we read the differential equation like
a sentence.
CHAPTER 2 : Page 60
© 2012 Cengage Learning Engineering. All Rights Reserved.
69
2.3 The Constant Coefficient Case
• We want a function y such that the second derivative,
plus a constant multiple of the first derivative, plus a
constant multiple of the function itself is equal to zero
for all x. This behavior suggests an exponential
function eλx , because derivatives of eλx are constant
multiples of eλx . We therefore try to find λ so that eλx
is a solution.
CHAPTER 2 : Page 60
© 2012 Cengage Learning Engineering. All Rights Reserved.
70
2.3 The Constant Coefficient Case
• Substitute eλx into equation (2.3) to get
 2 e x  a e x  be x  0.
Since eλx is never zero, the exponential factor cancels,
and we are left with a quadratic equation for λ:
λ2 + aλ + b = 0.
(2.4)
CHAPTER 2 : Page 60
© 2012 Cengage Learning Engineering. All Rights Reserved.
71
2.3 The Constant Coefficient Case
• The quadratic equation (2.4) is the characteristic
equation of the differential equation (2.3). Notice that
the characteristic equation can be read directly from
the coefficients of the differential equation, and we
need not substitute eλx each time. The characteristic
equation has roots
1
(a  a 2  4b ),
2
leading to the following three cases.
CHAPTER 2 : Page 60
© 2012 Cengage Learning Engineering. All Rights Reserved.
72
Case 1: Real, Distinct Roots
• This occurs when a2 − 4b > 0. The distinct roots are
1
1
2
1  (a  a  4b ) and 2  (a  a 2  4b )
2
2
e1x and e 2 x are linearly independent solutions, and
in this case, the general solution of equation (2.3) is
y  c1e1x  c2e 2 x .
CHAPTER 2 : Page 60
© 2012 Cengage Learning Engineering. All Rights Reserved.
73
EXAMPLE 2.8
• From the differential equation
y  y  6y  0,
we immediately read the characteristic equation
2    6  0
as having real, distinct roots 3 and −2. The general
solution is
y  c1e3x  c2 e 2x .
CHAPTER 2 : Page 60
© 2012 Cengage Learning Engineering. All Rights Reserved.
74
Case 2: Repeated Roots
• This occurs when a2 − 4b = 0 and the root of the
characteristic equation is λ= −a/2. One solution of the
differential equation is e−ax/2.
• We need a second, linearly independent solution. We
will invoke a method called reduction of order, which
will produce a second solution if we already have one
solution. Attempt a second solution y(x) = u(x)e−ax/2.
CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
75
Case 2: Repeated Roots
• Compute
y  u e
 ax /2
a  ax /2
 ue
2
and
2
a
y  ue ax /2  aue ax /2  ue ax /2
4
CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
76
Case 2: Repeated Roots
• Substitute these into equation (2.3) to get
2
a
 ax /2
 ax /2
 ax /2


ue
 au e
 ue
4
a  ax /2
 ax /2
 au e
 a ue
 bue  ax /2
2
2



a
 ax /2
e
u    b     0.
4 


CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
77
Case 2: Repeated Roots
• Since b − a2/4 = 0 in this case and e−ax/2 never
vanishes, this equation reduces to
u = 0.
• This has solutions u(x) = cx + d with c and d as
arbitrary constants. Therefore, any function y = (cx +
d)e−ax/2 is also a solution of equation (2.3) in this case.
CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
78
Case 2: Repeated Roots
• Since we need only one solution that is linearly
independent from e−ax/2, choose c = 1 and d = 0 to get
the second solution xe−ax/2. The general solution in
this repeated roots case is
y  c1e
 ax /2
 c2 xe
 ax /2
.
This is often written as y = e−ax/2(c1 + c2x).
CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
79
Case 2: Repeated Roots
• It is not necessary to repeat this derivation every time
we encounter the repeated root case. Simply write one
solution e−ax/2, and a second, linearly independent
solution is xe−ax/2.
CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
80
EXAMPLE 2.9
• We will solve y + 8y + 16y = 0. The characteristic
equation is
 2  8  16  0
with repeated root λ= −4. The general solution is
y  c1e 4x  c2 xe 4x .
CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
81
Case 3: Complex Roots
• The characteristic equation has complex roots when
a2 − 4b <0. Because the characteristic equation has
real coefficients, the roots appear as complex
conjugates α +iβ and α −iβ in which α can be zero but
β is nonzero.
CHAPTER 2 : Page 61
© 2012 Cengage Learning Engineering. All Rights Reserved.
82
Case 3: Complex Roots
• Now the general solution is
y  c1e(  i ) x  c2e( i ) x
or
y  e x  c1ei  x  c2e i  x  .
(2.5)
• This is correct, but it is sometimes convenient to have
a solution that does not involve complex numbers.
CHAPTER 2 : Page 61-62
© 2012 Cengage Learning Engineering. All Rights Reserved.
83
Case 3: Complex Roots
• We can find such a solution using an observation
made by the eighteenth century Swiss mathematician
Leonhard Euler, who showed that, for any real
number β,
ei  x  cos(  x)  i sin(  x).
• Problem 22 suggests a derivation of Euler’s formula.
By replacing x with −x, we also have
ei x  cos(  x)  i sin(  x).
CHAPTER 2 : Page 62
© 2012 Cengage Learning Engineering. All Rights Reserved.
84
Case 3: Complex Roots
• Then
y ( x)  e x  c1ei  x  c2e  i x 
 c1e x (cos(  x)  i sin(  x))  c2e x (cos(  x)  i sin(  x))
 (c1  c2 )e x cos(  x)  i (c1  c2 )e xsin(  x).
Here c1 and c2 are arbitrary real or complex numbers.
CHAPTER 2 : Page 62
© 2012 Cengage Learning Engineering. All Rights Reserved.
85
Case 3: Complex Roots
• If we choose c1 = c2 = 1/2, we obtain the particular
solution eαx cos(βx). And if we choose c1 = 1/2i = −c2,
we obtain the particular solution eαx sin(βx). Since
these solutions are linearly independent, we can write
the general solution in this complex root case as
(2.6)
y ( x)  c1e x cos(  x)  c2e xsin(  x)
in which c1 and c2 are arbitrary constants. We may
also write this general solution as
y ( x)  e x (c1cos(  x)  c2sin(  x)).
(2.7)
CHAPTER 2 : Page 62
© 2012 Cengage Learning Engineering. All Rights Reserved.
86
Case 3: Complex Roots
• Either of equations (2.6) or (2.7) is the preferred way
of writing the general solution in Case 3, although
equation (2.5) also is correct.
• We do not repeat this derivation each time we
encounter Case 3. Simply write the general solution
(2.6) or (2.7), with α ±iβ the roots of the characteristic
equation.
CHAPTER 2 : Page 62
© 2012 Cengage Learning Engineering. All Rights Reserved.
87
EXAMPLE 2.10
• Solve y + 2y + 3y =0. The characteristic equation is
 2  2  3  0
with complex conjugate roots 1  i 2 . With α = −1
and
  , the
2 general solution is
y  c1e x cos( 2 x)  c2e xsin( 2 x).
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
88
EXAMPLE 2.11
• Solve y + 36y = 0. The characteristic equation is
 2  36  0
with complex roots λ= ±6i. Now α = 0 and β = 6, so
the general solution is
y( x)  c1cos(6x)  c2sin(6x).
CHAPTER 2 : Page 53
© 2012 Cengage Learning Engineering. All Rights Reserved.
89
2.3 The Constant Coefficient Case
• We are now able to solve the constant coefficient
homogeneous equation
y  ay  by  0
in all cases. Here is a summary.
• Let λ1 and λ2 be the roots of the characteristic
equation
 2  a  b  0.
CHAPTER 2 : Page 62-63
© 2012 Cengage Learning Engineering. All Rights Reserved.
90
2.3 The Constant Coefficient Case
•
Then:
1. 1. If λ1 and λ2 are real and distinct,
y ( x)  c1e 1 x  c2e 2 x .
2. If λ1 =λ2,
y ( x)  c1e1 x  c2 xe1 x .
3. If the roots are complex α ± iβ,
y ( x)  c1e x cos(  x)  c2e xsin(  x).
CHAPTER 2 : Page 63
© 2012 Cengage Learning Engineering. All Rights Reserved.
91
Homework of Section 2.3
•1, 4, 9, 12, 14, 20, 21
© 2012 Cengage Learning Engineering. All Rights Reserved.
2.4 The Nonhomogeneous Equation
• From Theorem 2.4, the keys to solving the
nonhomogeneous linear equation (2.1) are to find two
linearly independent solutions of the associated
homogeneous equation and a particular solution Yp
for the nonhomogeneous equation. We can perform
the first task at least when the coefficients are
constant. We will now focus on finding Yp,
considering two methods for doing this.
CHAPTER 2 : Page 64
© 2012 Cengage Learning Engineering. All Rights Reserved.
93
2.4.1 Variation of Parameters
• Suppose we know two linearly independent solutions
y1 and y2 of the associated homogeneous equation.
One strategy for finding Yp is called the method of
variation of parameters. Look for functions u1 and u2
so that
Yp ( x)  u1 ( x) y1 ( x)  u2 ( x ) y2 ( x ).
CHAPTER 2 : Page 64
© 2012 Cengage Learning Engineering. All Rights Reserved.
94
2.4.1 Variation of Parameters
• To see how to choose u1 and u2, substitute Yp into the
differential equation. We must compute two
derivatives. First,
Yp  u1 y1  u2 y2  u1 y1  u2 y2 .
• Simplify this derivative by imposing the condition
that
u1 y1  u2 y2  0.
(2.8)
Why we can have this extra condition?
CHAPTER 2 : Page 64
© 2012 Cengage Learning Engineering. All Rights Reserved.
95
2.4.1 Variation of Parameters
• We have two known u1 and u2.
• But, when substitute them into the differential
equation y + p(x)y + q(x)y = f (x), we only have one
equation. It is impossible to find exact solutions.
• There are lots of solution for u1 and u2 and we cannot
find them.
• If we can introduce one extra condition, we may be
able to find solutions easily.
• Any conditions are fine. But u1 y1  u2 y2  0 can
work well.
CHAPTER 2 : Page 64
© 2012 Cengage Learning Engineering. All Rights Reserved.
96
2.4.1 Variation of Parameters
• Now
so
Yp  u1 y1  u2 y2 ,
Yp  u1 y1  u2 y2  u1 y1  u2 y2.
• Substitute Yp into the differential equation to get
u1 y1  u2 y2  u1 y1  u2 y2
 p( x)(u1 y1  u2 y2 )  q( x)(u1 y1  u2 y2 )  f ( x).
CHAPTER 2 : Page 64
© 2012 Cengage Learning Engineering. All Rights Reserved.
97
2.4.1 Variation of Parameters
• Rearrange terms to write
u1[ y1  p ( x) y1  q ( x) y1 ]
u2 [ y2  p ( x) y2  q ( x) y2 ]
u1 y1  u2 y2  f ( x).
• The two terms in square brackets are zero, because y1
and y2 are solutions of y + p(x)y + q(x)y = 0. The
last equation therefore reduces to
(2.9)
u1 y1  u2 y2  f ( x).
CHAPTER 2 : Page 64
© 2012 Cengage Learning Engineering. All Rights Reserved.
98
2.4.1 Variation of Parameters
• Equations (2.8) and (2.9) can be solved for u'1 and u'2
to get
y2 ( x) f ( x)
y1 ( x) f ( x)
u1( x)  
and u2 ( x) 
(2.10)
W ( x)
W ( x)
where W(x) is the Wronskian of y1(x) and y2(x).
CHAPTER 2 : Page 64-65
© 2012 Cengage Learning Engineering. All Rights Reserved.
99
2.4.1 Variation of Parameters
• We know that W(x)  0, because y1 and y2 are
assumed to be linearly independent solutions of the
associated homogeneous equation. Integrate
equations (2.10) to obtain
y2 ( x) f ( x)
y1 ( x) f ( x)
u1 ( x)   
dx and u2 ( x)  
dx
W ( x)
W ( x)
(2.11)
Be sure that this formula is obtained from the use of
y + p(x)y + q(x)y = f (x)
Note that the leading coefficient is 1 (to check what f (x) is).
CHAPTER 2 : Page 65
© 2012 Cengage Learning Engineering. All Rights Reserved.
100
2.4.1 Variation of Parameters
• Once we have u1 and u2, we have a particular solution
Yp = u1 y1 + u2 y2, and the general solution of y +
p(x)y + q(x)y = f (x) is
y  c1 y1  c2 y2  Yp .
CHAPTER 2 : Page 65
© 2012 Cengage Learning Engineering. All Rights Reserved.
101
EXAMPLE 2.12
• Find the general solution of
y  4y  sec( x)
for −π/4 < x <π/4.
• The characteristic equation of y+4y=0 is λ2+4=0
with complex roots λ= ±2i. Two linearly independent
solutions of the associated homogeneous equation
y+4y = 0 are
y1 ( x)  cos(2x)
CHAPTER 2 : Page 65
and y2 ( x)  sin(2x).
© 2012 Cengage Learning Engineering. All Rights Reserved.
102
EXAMPLE 2.12
• Now look for a particular solution of the
nonhomogeneous equation. First compute the
Wronskian
cos(2 x)
sin(2 x)
W ( x) 
 2(cos 2 ( x)  sin 2 ( x))  2.
2sin(2 x) 2cos(2 x)
CHAPTER 2 : Page 65
© 2012 Cengage Learning Engineering. All Rights Reserved.
103
EXAMPLE 2.12
• Use equations (2.11) with f (x)=sec(x) to obtain
sin(2 x) sec( x)
u1 ( x)   
dx
2
2sin( x) cos( x) sec( x)
 
dx
2
sin( x) cos( x)
 
dx
cos( x)
   sin( x)dx  cos( x)
CHAPTER 2 : Page 65
© 2012 Cengage Learning Engineering. All Rights Reserved.
104
EXAMPLE 2.12
and
cos(2 x) sec( x)
u2 ( x )  
dx
2
2
(2 cos ( x)  1)

dx
2 cos( x)
1


   cos( x)  sec( x) dx
2


1
 sin(x)  ln|sec(x)  tan( x) | .
2
CHAPTER 2 : Page 65
© 2012 Cengage Learning Engineering. All Rights Reserved.
105
EXAMPLE 2.12
• This gives us the particular solution
Yp ( x)  u1 ( x) y1 ( x)  u2 ( x) y2 ( x)
1


 cos( x) cos(2 x)   sin( x)  ln|sec(x)  tan( x) |  sin(2 x).
2


• The general solution of y + 4y = sec(x) is
y ( x)  c1cos(2x)  c2sin(2x)
1


 cos( x) cos(2 x)   sin( x)  ln|sec(x)  tan( x) |  sin(2 x).
2


CHAPTER 2 : Page 66
© 2012 Cengage Learning Engineering. All Rights Reserved.
106
2.4.2 Undetermined Coefficients
• We will discuss a second method for finding a
particular solution of the nonhomogeneous equation,
which, however, applies only to the constant
coefficient case y + ay + by = f (x).
• The idea behind the method of undetermined
coefficients is that sometimes we can guess a general
form for Yp(x) from the appearance of f (x).
CHAPTER 2 : Page 66
© 2012 Cengage Learning Engineering. All Rights Reserved.
107
EXAMPLE 2.13
• We will find the general solution of y − 4y = 8x2 −
2x.
• It is routine to find the general solution c1e2x + c2e−2x
of the associated homogeneous equation. We need a
particular solution Yp(x) of the nonhomogeneous
equation.
CHAPTER 2 : Page 66
© 2012 Cengage Learning Engineering. All Rights Reserved.
108
EXAMPLE 2.13
• Because f (x) = 8x2 − 2x is a polynomial and
derivatives of polynomials are polynomials, it is
reasonable to think that there might be a polynomial
solution. Furthermore, no such solution can include a
power of x higher than 2. If Yp(x) had an x3 term, this
term would be retained by the −4y term of y−4y, and
8x2 − 2x has no such term.
CHAPTER 2 : Page 66
© 2012 Cengage Learning Engineering. All Rights Reserved.
109
EXAMPLE 2.13
• This reasoning suggests that we try a particular
solution Yp(x) = Ax2 + Bx + C. Compute y(x) = 2Ax +
B and y(x) = 2A. Substitute these into the differential
equation to get
2A  4( Ax 2  Bx  C )  8x 2  2x.
• Write this as
(4A  8) x 2  (4B  2) x  (2A  4C )  0.
CHAPTER 2 : Page 66
© 2012 Cengage Learning Engineering. All Rights Reserved.
110
EXAMPLE 2.13
• This second-degree polynomial must be zero for all x
if Yp is to be a solution. But a second-degree
polynomial has only two roots, unless all of its
coefficients are zero. Therefore,
−4A − 8 = 0,
−4B + 2 = 0,
and
2A − 4C = 0.
CHAPTER 2 : Page 66
© 2012 Cengage Learning Engineering. All Rights Reserved.
111
EXAMPLE 2.13
• Solve these to get A = −2, B = 1/2, and C = −1. This
gives us the particular solution
1
2
Yp ( x)  2x  x  1.
2
• The general solution is
y ( x)  c1e  c2e
2x
CHAPTER 2 : Page 66
2x
1
 2x  x  1.
2
2
© 2012 Cengage Learning Engineering. All Rights Reserved.
112
EXAMPLE 2.14
• Find the general solution of y + 2y − 3y = 4e2x .
• The general solution of y + 2y − 3y = 0 is c1e−3x
+c2ex .
CHAPTER 2 : Page 66
© 2012 Cengage Learning Engineering. All Rights Reserved.
113
EXAMPLE 2.14
• Now look for a particular solution. Because
derivatives of e2x are constant multiples of e2x , we
suspect that a constant multiple of e2x might serve.
Try Yp(x)= Ae2x . Substitute this into the differential
equation to get
4Ae2x  4Ae2x  3Ae2x  5Ae 2x  4e 2x .
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
114
EXAMPLE 2.14
• This works if 5A = 4, so A = 4/5. A particular solution
is Yp(x) = 4e2x/5.
• The general solution is
4 2x
3 x
x
y ( x)  c1e  c2 e  e
5
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
115
EXAMPLE 2.15
• Find the general solution of y − 5y + 6y = −3sin(2x).
• The general solution of y − 5y + 6y = 0 is c1e3x
+c2e2x .
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
116
EXAMPLE 2.15
• We need a particular solution Yp of the
nonhomogeneous equation. Derivatives of sin(2x) are
constant multiples of sin(2x) or cos(2x). Derivatives
of cos(2x) are also constant multiples of sin(2x) or
cos(2x). This suggests that we try a particular solution
Yp(x) = A cos(2x) + B sin(2x). Notice that we include
both sin(2x) and cos(2x) in this first attempt, even
though f (x) just has a sin(2x) term.
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
117
EXAMPLE 2.15
• Compute
Yp ( x)  2A sin(2x)  2B cos(2x)
and
Yp( x)  4A cos(2x)  4B sin(2x).
• Substitute these into the differential equation to get
4A cos(2x)  4B sin(2x)  5[ 2A sin(2x)  2Bcos(2x)]
6[ Acos(2x)  B sin(2x)]  3 sin(2x).
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
118
EXAMPLE 2.15
• Rearrange terms to write
[2B + 10A + 3]sin(2x) = [−2A + 10B]cos(2x).
• But sin(2x) and cos(2x) are not constant multiples of
each other unless these constants are zero. Therefore,
2B +10A + 3 = 0 = −2A + 10B.
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
119
EXAMPLE 2.15
• Solve these to get A = −15/52 and B = −3/52. A
particular solution is
15
3
Yp ( x)   cos(2 x)  sin(2x)
52
52
• The general solution is
15
3
y ( x)  c1e  c2e  cos(2 x)  sin(2 x).
52
52
3x
CHAPTER 2 : Page 67
2x
© 2012 Cengage Learning Engineering. All Rights Reserved.
120
2.4.2 Undetermined Coefficients
• The method of undetermined coefficients has a trap
built into it. Consider the following.
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
121
EXAMPLE 2.16
• Find a particular solution of y + 2y − 3y = 8ex .
• Reasoning as before, try Yp(x)= Aex . Substitute this
into the differential equation to obtain
Aex + 2Aex − 3Aex = 8ex .
But then 8ex = 0, which is a contradiction.
CHAPTER 2 : Page 67
© 2012 Cengage Learning Engineering. All Rights Reserved.
122
2.4.2 Undetermined Coefficients
• The problem here is that ex is also a solution of the
associated homogeneous equation, so the left side
will vanish when Aex is substituted into y + 2y − 3y
= 8ex .
• Whenever a term of a proposed Yp(x) is a solution of
the associated homogeneous equation, multiply this
proposed solution by x. If this results in another
solution of the associated homogeneous equation,
multiply it by x again.
CHAPTER 2 : Page 68
© 2012 Cengage Learning Engineering. All Rights Reserved.
123
EXAMPLE 2.17
• Revisit Example 2.16. Since f (x)=8ex , our first
impulse was to try Yp(x)= Aex . But this is a solution
of the associated homogeneous equation, so multiply
by x and try Yp(x)= Axex. Now
Yp  Ae x  Axe x and Yp  2Ae x  Axe x .
CHAPTER 2 : Page 68
© 2012 Cengage Learning Engineering. All Rights Reserved.
124
EXAMPLE 2.17
• Substitute these into the differential equation to get
2Ae x  Axe x  2( Ae x  Axe x )  3Axe x  8e x .
• This reduces to 4Aex = 8ex, so A = 2, yielding the
particular solution Yp(x) = 2xex . The general solution
is
y ( x)  c1e 3x  c2e x  2xe x .
CHAPTER 2 : Page 68
© 2012 Cengage Learning Engineering. All Rights Reserved.
125
EXAMPLE 2.18
• Solve y − 6y + 9y = 5e3x .
• The associated homogeneous equation has the
characteristic equation (λ − 3)2 = 0 with repeated
roots λ = 3. The general solution of this associated
homogeneous equation is c1e3x +c2xe3x .
CHAPTER 2 : Page 68
© 2012 Cengage Learning Engineering. All Rights Reserved.
126
EXAMPLE 2.18
• For a particular solution, we might first try Yp(x) =
Ae3x , but this is a solution of the homogeneous
equation. Multiply by x and try Yp(x) = Axe3x . This is
also a solution of the homogeneous equation, so
multiply by x again and try Yp(x) = Ax2e3x . If this is
substituted into the differential equation, we obtain A
= 5/2, so a particular solution is Yp(x)=5x2e3x/2. The
general solution is
5 2 3x
y  c1e  c2 xe  x e .
2
3x
CHAPTER 2 : Page 68
3x
© 2012 Cengage Learning Engineering. All Rights Reserved.
127
2.4.2 Undetermined Coefficients
• The method of undetermined coefficients is limited
by our ability to “guess” a particular solution from
the form of f (x), and unlike variation of parameters,
requires that the coefficients of y and y be constant.
• Here is a summary of the method. Suppose we want
to find the general solution of
y  ay  by  f ( x).
CHAPTER 2 : Page 68
© 2012 Cengage Learning Engineering. All Rights Reserved.
128
2.4.2 Undetermined Coefficients
• Step 1.
– Write the general solution
yh ( x)  c1 y1 ( x)  c2 y2 ( x)
of the associated homogeneous equation
y  ay  by  0
with y1 and y2 linearly independent. We can always
do this in the constant coefficient case.
CHAPTER 2 : Page 68
© 2012 Cengage Learning Engineering. All Rights Reserved.
129
2.4.2 Undetermined Coefficients
• Step 2.
– We need a particular solution Yp of the
nonhomogeneous equation. This may require
several steps. Make an initial attempt of a general
form of a particular solution using f (x) and
perhaps Table 2.1 as a guide. If this is not possible,
this method cannot be used. If we can solve for the
constants so that this first guess works, then we
have Yp.
CHAPTER 2 : Page 68
© 2012 Cengage Learning Engineering. All Rights Reserved.
130
2.4.2 Undetermined Coefficients
• Step 3.
– If any term of the first attempt is a solution of the
associated homogeneous equation, multiply by x.
If any term of this revised attempt is a solution of
the homogeneous equation, multiply by x again.
Substitute this final general form of a particular
solution into the differential equation and solve for
the constants to obtain Yp.
CHAPTER 2 : Page 69
© 2012 Cengage Learning Engineering. All Rights Reserved.
131
2.4.2 Undetermined Coefficients
• Step 4.
– The general solution is
y  y1  y2  Yp .
CHAPTER 2 : Page 69
© 2012 Cengage Learning Engineering. All Rights Reserved.
132
2.4.2 Undetermined Coefficients
CHAPTER 2 : Page 69
© 2012 Cengage Learning Engineering. All Rights Reserved.
133
2.4.2 Undetermined Coefficients
• Table 2.1 provides a list of functions for a first try at
Yp(x) for various functions f (x) that might appear in
the differential equation. In this list, P(x) is a given
polynomial of degree n, Q(x) and R(x) are
polynomials of degree n with undetermined
coefficients for which we must solve, and c and β are
constants.
• If there is a homogeneous solution in f (x), we need to
multiply Q(x) (or R(x)) with x, where  is the
smallest nonnegative integer that makes xQ(x) does
not have any homogeneous solution.
CHAPTER 2 : Page 69
© 2012 Cengage Learning Engineering. All Rights Reserved.
134
2.4.3 The Principle of Superposition
• Suppose we want to find a particular solution of
y  p( x) y  q( x) y  f1 ( x)  f 2 ( x) 
 f N ( x).
• It is routine to check that, if Yj is a solution of
y  p( x) y  q ( x) y  f j ( x),
then Y1 + Y2 + ··· + YN is a particular solution of the
original differential equation.
CHAPTER 2 : Page 69
© 2012 Cengage Learning Engineering. All Rights Reserved.
135
EXAMPLE 2.19
• Find a particular solution of
y + 4y = x + 2e−2x .
• To find a particular solution, consider two problems:
– Problem 1: y + 4y = x
– Problem 2: y + 4y = 2e−2x
CHAPTER 2 : Page 69
© 2012 Cengage Learning Engineering. All Rights Reserved.
136
EXAMPLE 2.19
• Using undetermined coefficients, we find a particular
solution Yp ( x)  x / 4 of Problem 1 and a particular
solution Yp2 ( x)  e2x / 4 of Problem 2. A particular
solution of the given differential equation is
1
1 2 x
Yp ( x )  x  e .
4
4
• Using this, the general solution is
1
1
y ( x)  c1cos(2x)  c2sin(2x)  ( x  e 2 x ).
4
CHAPTER 2 : Page 69-70
© 2012 Cengage Learning Engineering. All Rights Reserved.
137
Homework of section 2.4
•1, 2, 5, 6, 8, 9, 14, 15, 17, 18
•Go to p. 201
© 2012 Cengage Learning Engineering. All Rights Reserved.
2.5 Spring Motion
• A spring suspended vertically and allowed to come to
rest has a natural length L. An object (bob) of mass m
is attached at the lower end, pulling the spring d units
past its natural length. The bob comes to rest in its
equilibrium position and is then displaced vertically a
distance y0 units (Figure 2.3) and released from rest
or with some initial velocity. We want to construct a
model allowing us to analyze the motion of the bob.
CHAPTER 2 : Page 70
© 2012 Cengage Learning Engineering. All Rights Reserved.
139
2.5 Spring Motion
• Let y(t) be the displacement of the bob from the
equilibrium position at time t, and take this
equilibrium position to be y = 0. Down is chosen as
the positive direction. Now consider the forces acting
on the bob. Gravity pulls it downward with a force of
magnitude mg. By Hooke’s law, the spring exerts a
force ky on the object. k is the spring constant, which
is a number quantifying the “stiffness" of the spring.
CHAPTER 2 : Page 70
© 2012 Cengage Learning Engineering. All Rights Reserved.
140
2.5 Spring Motion
• At the equilibrium position, the force of the spring is
−kd, which is negative because it acts upward. If the
object is pulled downward a distance y from this
position, an additional force −ky is exerted on it. The
total force due to the spring is therefore −kd − ky. The
total force due to gravity and the spring is mg −kd
−ky. Since at the equilibrium point this force is zero,
then mg = kd. The net force acting on the object due
to gravity and the spring is therefore just −ky.
CHAPTER 2 : Page 70
© 2012 Cengage Learning Engineering. All Rights Reserved.
141
2.5 Spring Motion
• There are forces tending to retard or damp out the
motion. These include air resistance or perhaps
viscosity of a medium in which the object is
suspended. A standard assumption (verified by
observation) is that the retarding forces have
magnitude proportional to the velocity y. Then for
some constant c called the damping constant, the
retarding forces equal cy. The total force acting on
the bob due to gravity, damping, and the spring itself
is −ky − cy.
CHAPTER 2 : Page 70-71
© 2012 Cengage Learning Engineering. All Rights Reserved.
142
2.5 Spring Motion
• Finally, there may be a driving force f (t) acting on
the bob. In this case, the total external force is
F  ky  cy  f (t ).
CHAPTER 2 : Page 71
© 2012 Cengage Learning Engineering. All Rights Reserved.
143
2.5 Spring Motion
CHAPTER 2 : Page 71
© 2012 Cengage Learning Engineering. All Rights Reserved.
144
2.5 Spring Motion
• Assuming that the mass is constant, Newton’s second
law of motion gives us
my  ky  cy  f (t )
or
c
k
1
y  y  y  f (t ).
(2.12)
m
m
m
CHAPTER 2 : Page 71
© 2012 Cengage Learning Engineering. All Rights Reserved.
145
2.5 Spring Motion
• This is the spring equation. Solutions give the
displacement of the bob as a function of time and
enable us to analyze the motion under various
conditions.
CHAPTER 2 : Page 71
© 2012 Cengage Learning Engineering. All Rights Reserved.
146
2.5.1 Unforced Motion
• The motion is unforced if f (t) = 0. Now the spring
equation is homogeneous, and the characteristic
equation has roots
c
1


c 2  4km .
2m 2m
• As we might expect, the solution for the
displacement, and hence the motion of the bob,
depends on the mass, the amount of damping, and the
stiffness of the spring.
CHAPTER 2 : Page 71
© 2012 Cengage Learning Engineering. All Rights Reserved.
147
Case 1: c2 − 4km > 0
• Now the roots of the characteristic equation are real
and distinct:
c
1
1  

c 2  4km .
2m 2m
and
c
1
2  

c 2  4km .
2m 2m
CHAPTER 2 : Page 72
© 2012 Cengage Learning Engineering. All Rights Reserved.
148
Case 1: c2 − 4km > 0
• The general solution of the spring equation in this
case is
y (t )  c1e 1 t  c2e 2 t .
Clearly, λ2 < 0.
CHAPTER 2 : Page 72
© 2012 Cengage Learning Engineering. All Rights Reserved.
149
Case 1: c2 − 4km > 0
• Since m and k are positive, c2 − 4km < c2,
so c2  4km < c and λ1 < 0. Therefore,
lim y (t )  0
t 
regardless of initial conditions. In the case that
c2 − 4km > 0, the motion simply decays to zero as
time increases. This case is called overdamping.
CHAPTER 2 : Page 72
© 2012 Cengage Learning Engineering. All Rights Reserved.
150
EXAMPLE 2.20
Overdamping
• Let c = 6, k = 5, and m = 1. Now the general solution
is y(t) = c1e−t + c2e−5t . Suppose the bob was initially
drawn upward 4 feet from equilibrium and released
downward with a speed of 2 feet per second. Then
y(0) = −4 and y(0) = 2, and we obtain
1 t
4 t
y (t )  e  9  e  .
2
CHAPTER 2 : Page 72
© 2012 Cengage Learning Engineering. All Rights Reserved.
151
EXAMPLE 2.20
Overdamping
• Figure 2.4 is a graph of this solution. Keep in mind
here that down is the positive direction. Since −9 +
e−4t < 0 for t > 0, then y(t) < 0, and the bob always
remains above the equilibrium point. Its velocity y(t)
= e−t (9 − 5e−4t)/2 decreases to zero as t increases, so
the bob moves downward toward equilibrium with
decreasing velocity, approaching arbitrarily close to
but never reaching this position and never coming
completely to rest.
CHAPTER 2 : Page 72
© 2012 Cengage Learning Engineering. All Rights Reserved.
152
EXAMPLE 2.20
Overdamping
CHAPTER 2 : Page 72
© 2012 Cengage Learning Engineering. All Rights Reserved.
153
Case 2: c2 − 4km = 0
• In this case, the general solution of the spring
equation is
y (t )  (c1  c2t )e  ct /2m .
This case is called critical damping. While y(t)→0 as
t→∞, as with overdamping, now the bob can pass
through the critical point, as the following example
shows.
CHAPTER 2 : Page 73
© 2012 Cengage Learning Engineering. All Rights Reserved.
154
EXAMPLE 2.21
Critical Damping
• Let c =2 and k =m =1. Then y(t) = (c1 +c2t)e−t .
Suppose the bob is initially pulled up four feet above
the equilibrium position and then pushed downward
with a speed of 5 feet per second. Then y(0) = −4, and
y(0) = 5. So
y(t )  (4  t )et .
CHAPTER 2 : Page 73
© 2012 Cengage Learning Engineering. All Rights Reserved.
155
EXAMPLE 2.21
Critical Damping
• Since y(4) = 0, the bob reaches the equilibrium four
seconds after it was released and then passes through
it. In fact, y(t) reaches its maximum when t = 5
seconds, and this maximum value is y(5) = e−5, which
is about 0.007 units below the equilibrium point. The
velocity y(t) = (5−t)e−t is negative for t >5, so the
bob’s velocity decreases after the five second point.
CHAPTER 2 : Page 73
© 2012 Cengage Learning Engineering. All Rights Reserved.
156
EXAMPLE 2.21
Critical Damping
• Since y(t)→0 as t→∞, the bob moves with decreasing
velocity back toward the equilibrium point as time
increases. Figure 2.5 is a graph of this displacement
function for 2 ≤ t ≤ 8.
• In general, when critical damping occurs, the bob
either passes through the equilibrium point exactly
once, as in Example 2.21, or never reaches it at all,
depending on the initial conditions.
CHAPTER 2 : Page 73
© 2012 Cengage Learning Engineering. All Rights Reserved.
157
EXAMPLE 2.21
Critical Damping
CHAPTER 2 : Page 73
© 2012 Cengage Learning Engineering. All Rights Reserved.
158
Case 3: c2 − 4km < 0
• Here the spring constant and mass of the bob are
sufficiently large that c2 < 4km and the damping is
less dominant. This is called underdamping. The
general underdamped solution has the form
 ct /2m
y (t )  e
[c1cos(  t )  c2sin(  t )]
in which
1

4km  c 2 .
2m
CHAPTER 2 : Page 73-74
© 2012 Cengage Learning Engineering. All Rights Reserved.
159
Case 3: c2 − 4km < 0
• Since c and m are positive, y(t)→0 as t→∞, as in the
other two cases. This is not surprising in the absence
of an external driving force. However, with
underdamping, the motion is oscillatory because of
the sine and cosine terms in the displacement
function. The motion is not periodic however because
of the exponential factor e−ct/2m, which causes the
amplitudes of the oscillations to decay as time
increases.
CHAPTER 2 : Page 74
© 2012 Cengage Learning Engineering. All Rights Reserved.
160
EXAMPLE 2.22
Underdamping
• Let c = k = 2 and m = 1. The general solution is
y (t )  e  t [c1cos(t )  c2sin(t )].
• Suppose the bob is driven downward from a point
three feet above equilibrium with an initial speed of
two feet per second. Then y(0) = −3, and y(0) = 2.
The solution is
y(t )  et (3 cos(t )  sin(t )).
CHAPTER 2 : Page 74
© 2012 Cengage Learning Engineering. All Rights Reserved.
161
EXAMPLE 2.22
Underdamping
• The behavior of this solution is visualized more easily
if we write it in phase angle form. Choose C and δ so
that
3 cos(t )  sin(t )  C cos(t   ).
CHAPTER 2 : Page 74
© 2012 Cengage Learning Engineering. All Rights Reserved.
162
EXAMPLE 2.22
Underdamping
• For this, we need
3 cos(t )  sin(t )  C cos(t )cos    C sin(t )sin   .
• Then
C cos    3 and C sin    1,
so
C sin( )
1
 tan ( )   .
C cos( )
3
CHAPTER 2 : Page 74
© 2012 Cengage Learning Engineering. All Rights Reserved.
163
EXAMPLE 2.22
Underdamping
• Now
 1
1
  arctan      arctan   .
 3
 3
• To solve for C, write
C 2cos2    C 2sin 2    C 2  32  1  10.
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
164
EXAMPLE 2.22
Underdamping
• Then C  10 , and the solution can be written in
phase angle form as
y (t )  10e  t cos(t  arctan(1/ 3)).
• The graph is a cosine curve with decaying amplitude
t
y

10
e
squeezed between graphs of
and
y   10et . Figure 2.6 shows y(t) and these two
exponential functions as reference curves.
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
165
EXAMPLE 2.22
Underdamping
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
166
EXAMPLE 2.22
Underdamping
• The bob passes back and forth through the
equilibrium point as t increases. Specifically, it passes
through the equilibrium point exactly when y(t) = 0,
which occurs at times
 1  2n  1
t  arctan   

2
3
for n = 0, 1, 2, ···.
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
167
Case 3: c2 − 4km < 0
• Next we will pursue the effect of a driving force on
the motion of the bob.
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
168
2.5.2 Forced Motion
• Different driving forces will result in different
motion. We will analyze the case of a periodic
driving force f (t)= A cos(ωt). Now the spring
equation (2.12) is
c
k
A
y  y  y 
cos (t ).
m
m
m
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
(2.13)
169
2.5.2 Forced Motion
• We have solved the associated homogeneous
equation in all cases on c, k, and m. For the general
solution of equation (2.13), we need only a particular
solution. Application of the method of undetermined
coefficients yields the particular solution
mA(k  m 2 )
Yp (t ) 
cos(t )
2 2
2 2
(k  m )   c
A c

sin(t ).
2 2
2 2
(k  m )   c
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
170
2.5.2 Forced Motion
• It is customary to denote 0  k / m to write
mA(02   2 )
Yp (t )  2 2
cos(t )
2 2
2 2
m (0   )   c
Ac
 2 2
sin(t ).
2 2
2 2
m (0   )   c
• We will analyze some specific cases to get some
insight into the motion with this forcing function.
CHAPTER 2 : Page 75
© 2012 Cengage Learning Engineering. All Rights Reserved.
171
Case 1: Overdamped Forced Motion
• Overdamping occurs when c2 − 4km > 0. Suppose c =
6, k = 5, m = 1 A  6 5 and   5.
• If the bob is released from rest from the equilibrium
position, then y(t) satisfies the initial value problem
y  6y  5y  6 5 cos( 5t );
The solution is
y(0)  y(0)  0
5
y (t ) 
(e  t  e 5t )  sin( 5t ).
4
CHAPTER 2 : Page 75-76
© 2012 Cengage Learning Engineering. All Rights Reserved.
172
Case 1: Overdamped Forced Motion
• A graph of this solution is shown in Figure 2.7. As
time increases, the exponential terms decay to zero,
and the displacement behaves increasingly like sin( 5t ),
oscillating up and down through the equilibrium point
with approximate period 2 / 5 . Contrast this with
the overdamped motion without the forcing function
in which the bob began above the equilibrium point
and moved with decreasing speed down toward it but
never reached it.
CHAPTER 2 : Page 76
© 2012 Cengage Learning Engineering. All Rights Reserved.
173
Case 1: Overdamped Forced Motion
CHAPTER 2 : Page 76
© 2012 Cengage Learning Engineering. All Rights Reserved.
174
Case 2: Critically Damped Forced
Motion
• Let c = 2, m = k = 1, ω = 1, and A = 2. Assume that
the bob is released from rest from the equilibrium
point. Now the initial value problem is
y  2y  y  2 cos(t ); y (0)  y(0)  0
with the solution
t
y(t )  te  sin(t ).
CHAPTER 2 : Page 76
© 2012 Cengage Learning Engineering. All Rights Reserved.
175
Case 2: Critically Damped Forced
Motion
• Figure 2.8 is a graph of this solution, which is a case
of critically damped forced motion. As t increases,
the term with the exponential factor decays (although
not as fast as in the overdamping case where there is
no factor of t). Nevertheless, after sufficient time, the
motion settles into nearly (but not exactly because
−te−t is never zero for t > 0) a sinusoidal motion back
and forth through the equilibrium point.
CHAPTER 2 : Page 76
© 2012 Cengage Learning Engineering. All Rights Reserved.
176
Case 2: Critically Damped Forced
Motion
CHAPTER 2 : Page 76
© 2012 Cengage Learning Engineering. All Rights Reserved.
177
Case 3: Underdamped Forced Motion
• Let c = k = 2, m = 1,   2 , and A  2 2 , so c2 −
4km < 0. Suppose the bob is released from rest at the
equilibrium position. The initial value problem is
y  2y  2y  2 2cos( 2t ) ; y (0)  y(0)  0
with the solution
y (t )   2e t sin(t )  sin( 2t ).
CHAPTER 2 : Page 76
© 2012 Cengage Learning Engineering. All Rights Reserved.
178
Case 3: Underdamped Forced Motion
• This is underdamped forced motion. Unlike
overdamping and critical damping, the exponential
term e−t has a trigonometric factor sin(t). Figure 2.9 is
a graph of this solution. As time increases, the term
 2et sin(t ) becomes less influential and the motion
settles nearly into an oscillation back and forth
through the equilibrium point with a period of nearly
2 / 2.
CHAPTER 2 : Page 77
© 2012 Cengage Learning Engineering. All Rights Reserved.
179
Case 3: Underdamped Forced Motion
CHAPTER 2 : Page 77
© 2012 Cengage Learning Engineering. All Rights Reserved.
180
2.5.3 Resonance
• In the absence of damping, an important phenomenon
called resonance can occur. Suppose c = 0, but there
is still a driving force A cos(ωt). Now the spring
equation (2.12) is
k
A
y  y  cos t  .
m
m
CHAPTER 2 : Page 77
© 2012 Cengage Learning Engineering. All Rights Reserved.
181
2.5.3 Resonance
• From the particular solution Yp found in Section
2.5.2, with c = 0, we find that this spring equation has
general solution
A
y (t )  c1cos(t )  c2sin(t ) 
cos(t )
2
2
m(0   )
in which 0  k / m.
CHAPTER 2 : Page 77
© 2012 Cengage Learning Engineering. All Rights Reserved.
182
2.5.3 Resonance
• This number is called the natural frequency of the
spring system, and it is a function of the stiffness of
the spring and the mass of the bob. ω is the input
frequency and is contained in the driving force. This
general solution assumes that the natural and input
frequencies are different. Of course, the closer we
choose the natural and input frequencies, the larger
the amplitude of the cos(ωt) term in the solution.
CHAPTER 2 : Page 77
© 2012 Cengage Learning Engineering. All Rights Reserved.
183
2.5.3 Resonance
• Resonance occurs when the natural and input
frequencies are the same. Now the differential
equation is
k
A
y  y 
cos (0t ).
m
m
CHAPTER 2 : Page 77
© 2012 Cengage Learning Engineering. All Rights Reserved.
(2.14)
184
2.5.3 Resonance
• The solution derived for the case when ω  ω0 does
not apply to equation (2.14). To find the general
solution in the present case, first find the general
solution of the associated homogeneous equation
k
y   y  0
m
• This has the general solution
yh (t )  c1cos(0t )  c2sin(0t ).
CHAPTER 2 : Page 77
© 2012 Cengage Learning Engineering. All Rights Reserved.
185
2.5.3 Resonance
• Now we need a particular solution of equation (2.14).
To use the method of undetermined coefficients, we
might try a function of the form a cos(ω0t) + b
sin(ω0t). However, these are solutions of the
associated homogeneous equation, so instead we try
Yp (t )  at cos(0t )  bt sin(0t ).
CHAPTER 2 : Page 78
© 2012 Cengage Learning Engineering. All Rights Reserved.
186
2.5.3 Resonance
• Substitute Yp(t) into equation (2.14) to get
A
2a0sin(0t )  2b0cos(0t )  cos(0t ).
m
• Thus, choose
A
a  0 and 2b0  .
m
• This gives us the particular solution
A
Yp (t ) 
t sin (0t ).
2m0
CHAPTER 2 : Page 78
© 2012 Cengage Learning Engineering. All Rights Reserved.
187
2.5.3 Resonance
• The general solution is
A
y (t )  c1cos(0t )  c2sin(0t ) 
t sin (0t ).
2m0
• This solution differs from the case ω ω0 in the factor
of t in the particular solution. Because of this,
solutions increase in amplitude as t increases. This
phenomenon is called resonance.
CHAPTER 2 : Page 78
© 2012 Cengage Learning Engineering. All Rights Reserved.
188
2.5.3 Resonance
• As an example, suppose c1 = c2 =ω0 = 1 and A/2m =
1. Now the solution is
y (t )  cos(t )  sin(t )  tsin(t ).
Figure 2.10 displays the increasing amplitude of the
oscillations with time.
CHAPTER 2 : Page 78
© 2012 Cengage Learning Engineering. All Rights Reserved.
189
2.5.3 Resonance
CHAPTER 2 : Page 78
© 2012 Cengage Learning Engineering. All Rights Reserved.
190
2.5.3 Resonance
• While there is always some damping in the real
world, if the damping constant is close to zero when
compared to other factors and if the natural and input
frequencies are (nearly) equal, then oscillations can
build up to a sufficiently large amplitude to cause
resonance-like behavior. This caused the collapse of
the Broughton Bridge near Manchester, England, in
1831 when a column of soldiers marching across
maintained a cadence (input frequency) that happened
to closely match the natural frequency of the material
of the bridge.
CHAPTER 2 : Page 78
© 2012 Cengage Learning Engineering. All Rights Reserved.
191
2.5.3 Resonance
• More recently the Tacoma Narrows Bridge in the
state of Washington experienced increasing
oscillations driven by high winds, causing the
concrete roadbed to oscillate in sensational fashion
until it collapsed into Puget Sound. This occurred on
November 7, 1940. At one point, one side of the
roadbed was about twenty-eight feet above the other
as it thrashed about. Unlike the Broughton Bridge,
local news crews were on hand to film this, and
motion pictures of the collapse are available in many
engineering and science schools.
CHAPTER 2 : Page 78-79
© 2012 Cengage Learning Engineering. All Rights Reserved.
192
2.5.4 Beats
• In the absence of damping, an oscillatory driving
force can also cause a phenomenon called beats.
Suppose ω ω0, and consider
y  02 y 
CHAPTER 2 : Page 79
A
cos (t ).
m
© 2012 Cengage Learning Engineering. All Rights Reserved.
193
2.5.4 Beats
• Assume that the object is released from rest from the
equilibrium position, so y(0) = y(0) = 0. The solution
is
A
y (t ) 
[cos(t )  cos(0t )].
2
2
m(0   )
• The behavior of this solution reveals itself more
clearly if we write it as
2A
1

1

y(t ) 
sin  (0   )t  sin  (0   )t  .
2
2
m(0   )  2

2

CHAPTER 2 : Page 79
© 2012 Cengage Learning Engineering. All Rights Reserved.
194
2.5.4 Beats
• This formulation exhibits a periodic variation of
amplitude in the solution, depending on the relative
sizes of ω0 + ω and ω0 − ω. This periodic variation is
called a beat. As an example, suppose ω0 +ω =5 and
ω0 −ω =1, and the constants are chosen so that
2A / m(02   2 )  1.
CHAPTER 2 : Page 79
© 2012 Cengage Learning Engineering. All Rights Reserved.
195
2.5.4 Beats
• Now the displacement function is
 5t   t 
y (t )  sin   sin   .
 2  2
• Figure 2.11 is a graph of this solution.
CHAPTER 2 : Page 79
© 2012 Cengage Learning Engineering. All Rights Reserved.
196
2.5.4 Beats
CHAPTER 2 : Page 79
© 2012 Cengage Learning Engineering. All Rights Reserved.
197
2.5.5 Analogy with an Electrical Circuit
• In an RLC circuit with electromotive force E(t), the
differential equation for the current is
1
Li(t )  Ri (t )  q (t )  E (t ).
C
• Since i = q, this is a second-order differential
equation for the charge:
R
1
1
q  q 
q  E (t ).
L
LC
L
CHAPTER 2 : Page 79-80
© 2012 Cengage Learning Engineering. All Rights Reserved.
198
2.5.5 Analogy with an Electrical Circuit
• Assuming that the resistance, inductance, and
capacitance are constant, this equation is exactly
analogous to the spring equation with a driving force,
which has the form
c
k
1
y  y  y  f (t ).
m
m
m
CHAPTER 2 : Page 80
© 2012 Cengage Learning Engineering. All Rights Reserved.
199
2.5.5 Analogy with an Electrical Circuit
• This means that solutions of the spring equation
immediately translate into solutions of the circuit
equation with the following identifications:
– Displacement function y(t) ⇐⇒ charge q(t)
– Velocity y(t) ⇐⇒ current i(t)
– Driving force f (t) ⇐⇒ electromotive force E(t)
– Mass m ⇐⇒ inductance L
– Damping constant c ⇐⇒ resistance R
– Spring modulus k ⇐⇒ reciprocal 1/C of the
capacitance.
CHAPTER 2 : Page 80
© 2012 Cengage Learning Engineering. All Rights Reserved.
200
2.6 Euler’s Differential Equation
If A and B are constants, the second-order differential
equation
x2 y + Axy + By = 0
(2.15)
is called Euler’s equation. Euler’s equation is defined
on the half-lines x > 0 and x < 0.We will find
solutions on x > 0, and a simple adjustment will yield
solutions on x < 0.
CHAPTER 2 : Page 81
© 2012 Cengage Learning Engineering. All Rights Reserved.
201
2.6 Euler’s Differential Equation
• A change of variables will convert Euler’s equation to
a constant coefficient linear second-order
homogeneous equation, which we can always solve.
Let
x  et
or, equivalently, t = ln(x). If we substitute x = et into
y(x), we obtain a function of t as
Y (t )  y (e ).
t
CHAPTER 2 : Page 81
© 2012 Cengage Learning Engineering. All Rights Reserved.
202
2.6 Euler’s Differential Equation
• To convert Euler’s equation to an equation in t, we
need to convert derivatives of y(x) to derivatives of
Y(t). First, by the chain rule, we have
d
d
( y ( x))  ( y (t ))
dx
dx
dY dt 1

 Y (t ).
dt dx x
y( x) 
CHAPTER 2 : Page 81
© 2012 Cengage Learning Engineering. All Rights Reserved.
203
2.6 Euler’s Differential Equation
• Next,
CHAPTER 2 : Page 81
d
( y( x))
dx
d 1

  Y (t ) 
dx  x

1
1 d
  2 Y (t ) 
(Y (t ))
x
x dx
1
1 dY (t ) dt
  2 Y (t ) 
x
x dt dx
1
11
  2 Y (t ) 
Y (t )
x
xx
1
 2 (Y (t )  Y (t )).
x
y( x) 
© 2012 Cengage Learning Engineering. All Rights Reserved.
204
2.6 Euler’s Differential Equation
• Therefore,
x 2 y( x)  Y (t )  Y (t ).
• Substitute these into Euler’s equation to obtain the
transformed differential equation
Y (t )  Y (t )  AY (t )  BY (t )  0
or
Y (t )  ( A  1)Y (t )  BY (t )  0.
(2.16)
This is a constant coefficient equation which we
know how to solve.
CHAPTER 2 : Page 82
© 2012 Cengage Learning Engineering. All Rights Reserved.
205
2.6 Euler’s Differential Equation
• We need not go through this derivation whenever we
encounter an Euler equation. The coefficients of
equation (2.16) can be read directly from those of
the Euler equation. Solve this transformed equation
for Y (t), then replace t = ln(x) to obtain the solution
y(x) of the Euler equation.
CHAPTER 2 : Page 82
© 2012 Cengage Learning Engineering. All Rights Reserved.
206
2.6 Euler’s Differential Equation
• In doing this, it is useful to recall that, for any number
r and for x > 0,
x r  e r ln ( x ) .
• Furthermore,
e ln ( k )  k
for any positive quantity k. Thus, for example,
3 ln( x )
e
CHAPTER 2 : Page 82
e
ln( x3 )
x .
3
© 2012 Cengage Learning Engineering. All Rights Reserved.
207
EXAMPLE 2.23
• We will find the general solution of
x 2 y  2xy  6y  0.
• With A = 2 and B = −6, this Euler equation
transforms to
Y (t )  Y (t )  6Y (t )  0.
CHAPTER 2 : Page 82
© 2012 Cengage Learning Engineering. All Rights Reserved.
208
EXAMPLE 2.23
• This constant coefficient linear homogeneous
equation has general solution
Y (t )  c1e 3t  c2e 2t .
• Replace t = ln(x) to obtain the general solution of the
Euler equation:
y ( x)  c1e 3 ln ( x )  c2e 2 ln ( x )  c1 x 3  c2 x 2
for x > 0.
CHAPTER 2 : Page 82
© 2012 Cengage Learning Engineering. All Rights Reserved.
209
EXAMPLE 2.24
• Consider the Euler equation x2 y − 5xy + 9y = 0.
The transformed equation is
Y   6Y   9Y  0,
with the general solution Y(t) = c1e3t +c2te3t . The
Euler equation has the general solution
y ( x)  c1e3 ln ( x )  c2 ln( x)e3 ln ( x )  c1 x 3  c2 x 3ln( x)
for x > 0.
CHAPTER 2 : Page 82
© 2012 Cengage Learning Engineering. All Rights Reserved.
210
EXAMPLE 2.25
• Solve x2 y + 3xy + 10y = 0.
• The transformed equation is Y + 2Y + 10Y = 0 with
the general solution
Y (t )  c1e  t cos(3t )  c2e  t sin(3t ).
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
211
EXAMPLE 2.25
• Then
y ( x)  c1e  ln ( x ) cos(3 ln( x))  c2e  ln ( x )sin(3 ln( x))
1
 (c1cos(3 ln( x))  c2sin(3 ln( x))).
x
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
212
2.6 Euler’s Differential Equation
• Another way of finding solutions for Euler’s DE is to
check the solution of the form xr for x2 y + Axy +
By = 0 and to find r.  r2 + (A-1)r + Br = 0 .
• As usual, we solve an initial value problem by finding
the general solution of the differential equation and
then using the initial conditions to determine the
constants.
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
213
EXAMPLE 2.26
• Solve
x 2 y  5xy  10 y  0; y(1)  4, y(1)  6.
• The Euler equation transforms to Y − 6Y + 10Y = 0
with the general solution
Y (t )  c1e3t cos(t )  c2e3t sin(t )
for x > 0.
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
214
EXAMPLE 2.26
• Then
y ( x)  x3 (c1cos(ln( x))  c2sin(ln( x))) .
• Then
y(1)  4  c1.
• Thus far,
y ( x)  4x3cos(ln( x))  c2 x 3sin(ln( x)).
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
215
EXAMPLE 2.26
• Compute
y( x)  12x 2cos(ln( x))  4x 2sin(ln( x))
3c2 x 2sin(ln( x))  c2 x 2cos(ln( x)).
• Then
y(1)  12  c2  6,
so c2 = −18. The solution of the initial value problem
is
y( x)  4x3cos(ln( x))  18x3sin(ln( x)).
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
216
2.6 Euler’s Differential Equation
• Note when you apply variation parameter for Euler’
DE, the formula
y2 ( x) f ( x)
y1 ( x) f ( x)
u1 ( x)   
dx and u2 ( x)  
dx
W ( x)
W ( x)
be sure that the equation is
A
B
y  y  2 y  f (t ).
x
x
because in the transformation, the original form is
y + p(x)y + q(x)y = f (x).
CHAPTER 2 : Page 82
© 2012 Cengage Learning Engineering. All Rights Reserved.
217
2.6 Euler’s Differential Equation
• But when you apply underdetermined coefficient for
Euler’ DE, to guess the possible particular forms, the
equation must be considered in a constant coefficient
form, then the equation must be written as
x 2 y  Axy  By  f (t ).
and then we can check the form of f (x) to find
possible particular solution forms.
CHAPTER 2 : Page 82
© 2012 Cengage Learning Engineering. All Rights Reserved.
218
Homework of Section 2.6
•2, 3, 5, 8, 12, 15, 17
© 2012 Cengage Learning Engineering. All Rights Reserved.
Higher Order Differential Equations
• Basically, those methodologies used for second order
DSs can also be applied to higher order DEs.
• Concept definitions for higher order DEs:
--Linear dependency/independency: y1, y2, …, yn are
said to be linearly independent if it is impossible to
write any one of those functions as a linear
combination of the others.
Or c1y1+c2y2+ …+cnyn =0 only if c1=c2= …=cn =0
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
220
Higher Order Differential Equations
--General solution: y1, y2, …, yn are linearly independent
solutions of an n-th order homogeneous linear DE,
then the general solution is a linear combination of
those functions.
-- The constant coefficient linear DE can also be solved
in a similar way (finding roots for the associated
characteristic equation) and so does Euler’s
equations.
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
221
Higher Order Differential Equations
--For nonhomogeneous DEs, the variation parameter
and undetermined coefficient approaches can also be
employed to find a particular solution.
--For variation parameter, for an n-th order DE, there
are n unknown functions, but only have one equation.
Thus, we need to introduce n-1 extra equations. The
similar idea can be used.
CHAPTER 2 : Page 83
© 2012 Cengage Learning Engineering. All Rights Reserved.
222