Page 1
Boyce/DiPrima/Meade 11th ed, Ch 3.1: 2nd Order Linear
Homogeneous Equations-Constant Coefficients
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• A second order ordinary differential equation has the
general form
where f is some given function.
• This equation is said to be linear if f is linear in y and y':
Otherwise the equation is said to be nonlinear.
• A second order linear equation often appears as
• If g(t) or G(t) = 0 for all t, then the equation is called
homogeneous. Otherwise the equation is nonhomogeneous.
d2y
dt 2= f t,y,
dy
dt
æ
èçö
ø÷
y ''+ p(t)y '+ q(t)y = g(t)
P(t)y ''+Q(t)y '+ R(t)y =G(t)
Page 2
Homogeneous Equations, Initial Values
• In Sections 3.5 and 3.6, we will see that once a solution to a
homogeneous equation is found, then it is possible to solve
the corresponding nonhomogeneous equation, or at least
express the solution in terms of an integral.
• The focus of this chapter is thus on homogeneous equations;
and in particular, those with constant coefficients:
We will examine the variable coefficient case in Chapter 5.
• Initial conditions typically take the form
• Thus solution passes through (t0, y0), and the slope of solution
at (t0, y0) is equal to y0'.
ay ''+by '+ cy = 0
y t0( ) = y0, y ' t0( ) = y '0
Page 3
Example 1: Infinitely Many Solutions (1 of 3)
• Consider the second order linear differential equation
• Two solutions of this equation are
• Other solutions include
• Based on these observations, we see that there are infinitely
many solutions of the form
• It will be shown in Section 3.2 that all solutions of the
differential equation above can be expressed in this form.
y ''- y = 0
tt etyety )(,)( 21
tttt eetyetyety 53)(,5)(,3)( 543
tt ececty 21)(
Page 4
Example 1: Initial Conditions (2 of 3)
• Now consider the following initial value problem for our equation:
• We have found a general solution of the form
• Using the initial equations,
• Thus
y ''- y = 0, y(0) = 2, y '(0) = -1
tt ececty 21)(
y(0) = c1 + c2 = 2
¢y (0) = c1 - c2 = -1
üýþÞ c1 =
1
2, c2 =
3
2
y(t) =1
2et +
3
2e-t
Page 5
Example 1: Solution Graphs (3 of 3)
• Our initial value problem and solution are
• Graphs of both y(t) and y’(t) are given below. Observe that
both initial conditions are satisfied.
y ''- y = 0, y(0) = 2, y '(0) = -1 Þ y(t) =1
2et +
3
2e-t
0.5 1.0 1.5 2.0t
1
2
3
y t
0.5 1.0 1.5 2.0t
1
1
2
3
y' t
y(t) =1
2et +
3
2e-t y '(t) =
1
2et -
3
2e-t
Page 6
Characteristic Equation
• To solve the 2nd order equation with constant coefficients,
we begin by assuming a solution of the form y = ert.
• Substituting this into the differential equation, we obtain
• Simplifying,
and hence
• This last equation is called the characteristic equation of the differential equation.
• We then solve for r by factoring or using quadratic formula.
ay ''+by '+ cy = 0,
02 rtrtrt cebreear
0)( 2 cbrarert
02 cbrar
Page 7
General Solution
• Using the quadratic formula on the characteristic equation
we obtain two solutions, r1 and r2.
• There are three possible results:
– The roots r1, r2 are real and r1 ≠ r2.
– The roots r1, r2 are real and r1 = r2.
– The roots r1, r2 are complex.
• In this section, we will assume r1, r2 are real and r1 ≠ r2.
• In this case, the general solution has the form
,02 cbrar
y(t) = c1er1 t + c2e
r2 t
a
acbbr
2
42
Page 8
Initial Conditions
• For the initial value problem
we use the general solution
together with the initial conditions to find c1 and c2. That is,
• Since we are assuming r1 ≠ r2, it follows that a solution of the
form y = ert to the above initial value problem will always
exist, for any set of initial conditions.
ay ''+by '+ cy = 0, y(t0 ) = y0, y '(t0 ) = y '0,
c1er1 t0 + c2e
r2 t0 = y0
c1r1er1 t0 + c2r2e
r2 t0 = y '0
üýï
þïÞ c1 =
y '0 - y0r2
r1 - r2e-r1 t0 , c2 =
y0r1 - y '0
r1 - r2e-r2 t0
trtrececty 21
21)(
Page 9
Example 2 (General Solution)
• Consider the linear differential equation
• Assuming an exponential solution leads to the characteristic
equation:
• Factoring the characteristic equation yields two solutions:
r1 = –2 and r2 = –3
• Therefore, the general solution to this differential equation
has the form
y ''+ 5y '+ 6y = 0
032065)( 2 rrrrety rt
tt ececty 3
2
2
1)(
Page 10
Example 3 (Particular Solution)
• Consider the initial value problem
• From the preceding example, we know the general solution has the form:
• With derivative:
• Using the initial conditions:
• Thus
y ''+ 5y '+ 6y = 0, y 0( ) = 2, y ' 0( ) = 3
7,9332
121
21
21
cc
cc
cc
tt eety 32 79)(
tt ececty 3
2
2
1)(
tt ececty 3
2
2
1 32)('
0.0 0.5 1.0 1.5 2.0 2.5t
0.5
1.0
1.5
2.0
2.5
y t
tt eety 32 79)(
Page 11
Example 4: Initial Value Problem
• Consider the initial value problem
• Then
• Factoring yields two solutions,
• The general solution has the form
• Using initial conditions:
• Thus
4y ''- 8y '+ 3y = 0, y 0( ) = 2, y ' 0( ) =1
2
012320384)( 2 rrrrety rt
2/
2
2/3
1)( tt ececty
c1 + c2 = 2
3
2c1 +
1
2c2 =
1
2
ü
ýï
þïÞ c1 = -
1
2, c2 =
5
2
y(t) = -1
2e3t/2 +
5
2et/2
0.5 1.0 1.5 2.0 2.5t
1
1
2
3
y t
y(t) = -1
2e3t/2 +
5
2et/2
r1 =3
2 and r2 =
1
2
Page 12
Example 5: Find Maximum Value
• For the initial value problem in Example 3,
to find the maximum value attained by the
solution, we set y’(t) = 0 and solve for t:
y(t) = 9e-2 t - 7e-3t
y '(t) = -18e-2 t + 21e-3t =set
0
6e-2 t = 7e-3t
e t = 7 / 6
t = ln(7 / 6)
t » 0.1542
y » 2.204 0.0 0.5 1.0 1.5 2.0 2.5t
0.5
1.0
1.5
2.0
2.5
y t
tt eety 32 79)(
Page 13
Boyce/DiPrima/Meade 11th ed, Ch 3.2: Fundamental Solutions of
Linear Homogeneous Equations
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• Let p, q be continuous functions on an interval
which could be infinite. For any function y that is twice
differentiable on I, define the differential operator L by
• Note that L[y] is a function on I, with output value
• For example,
L y[ ] = y ''+ py '+ qy
L y[ ](t) = y ''(t)+ p(t)y '(t)+ q(t)y(t)
)sin(2)cos()sin()(
2,0),sin()(,)(,)(
22
22
tettttyL
Ittyetqttp
t
t
I = (a,b)
Page 14
Differential Operator Notation
• In this section we will discuss the second order linear
homogeneous equation L[y](t) = 0, along with initial
conditions as indicated below:
• We would like to know if there are solutions to this initial
value problem, and if so, are they unique.
• Also, we would like to know what can be said about the form
and structure of solutions that might be helpful in finding
solutions to particular problems.
• These questions are addressed in the theorems of this section.
L y[ ] = y ''+ p(t)y '+ q(t)y = 0
y(t0 ) = y0 , y '(t0 ) = y1
Page 15
Theorem 3.2.1 (Existence and Uniqueness)
• Consider the initial value problem
• where p, q, and g are continuous on an open interval I that
contains t0. Then there exists a unique solution on I.
• Note: While this theorem says that a solution to the initial
value problem above exists, it is often not possible to write
down a useful expression for the solution. This is a major
difference between first and second order linear equations.
y ''+ p(t)y '+ q(t)y = g(t)
y(t0 ) = y0 , y '(t0 ) = y '0
y = f(t)
Page 16
Example 1
• Consider the second order linear initial value problem
• Writing the differential equation in the form :
• The only points of discontinuity for these coefficients are
t = 0 and t = 3. So the longest open interval containing the
initial point t =1 in which all the coefficients are continuous
is 0 < t < 3
• Therefore, the longest interval in which Theorem 3.2.1
guarantees the existence of the solution is 0 < t < 3
y ''+ p(t)y '+ q(t)y = g(t)
(t 2 - 3t)y ''+ ty '- (t + 3)y = 0, y 1( ) = 2, y ' 1( ) =1
1000 )(,)(
)()()(
ytyyty
tgytqytpy
p(t) =1
t - 3, q(t) = -
t + 3
t(t - 3) and g(t) = 0
Page 17
Example 2
• Consider the second order linear initial value problem
where p, q are continuous on an open interval I containing t0.
• In light of the initial conditions, note that y = 0 is a solution
to this homogeneous initial value problem.
• Since the hypotheses of Theorem 3.2.1 are satisfied, it
follows that y = 0 is the only solution of this problem.
y ''+ p(t)y '+ q(t)y = 0, y 0( ) = 0, y ' 0( ) = 0
Page 18
Theorem 3.2.2 (Principle of Superposition)
• If y1and y2 are solutions to the equation
then the linear combination c1y1 + y2c2 is also a solution, for
all constants c1 and c2.
• To prove this theorem, substitute c1y1 + y2c2 in for y in the
equation above, and use the fact that y1 and y2 are solutions.
• Thus for any two solutions y1 and y2, we can construct an
infinite family of solutions, each of the form y = c1y1 + c2 y2.
• Can all solutions can be written this way, or do some
solutions have a different form altogether? To answer this
question, we use the Wronskian determinant.
L[y] = y ''+ p(t)y '+ q(t)y = 0
Page 19
The Wronskian Determinant (1 of 3)
• Suppose y1 and y2 are solutions to the equation
• From Theorem 3.2.2, we know that y = c1y1 + c2 y2 is a
solution to this equation.
• Next, find coefficients such that y = c1y1 + c2 y2 satisfies the
initial conditions
• To do so, we need to solve the following equations:
L[y] = y ''+ p(t)y '+ q(t)y = 0
y(t0 ) = y0, y '(t0 ) = y '0
c1y1(t0 )+ c2y2(t0 ) = y0
c1y '1(t0 )+ c2y '2 (t0 ) = y '0
Page 20
The Wronskian Determinant (2 of 3)
• Solving the equations, we obtain
• In terms of determinants:
0022011
0022011
)()(
)()(
ytyctyc
ytyctyc
)()()()(
)()(
)()()()(
)()(
02010201
0100102
02010201
0200201
tytytyty
tyytyyc
tytytyty
tyytyyc
)()(
)()(
)(
)(
,
)()(
)()(
)(
)(
0201
0201
001
001
2
0201
0201
020
020
1
tyty
tyty
yty
yty
c
tyty
tyty
tyy
tyy
c
Page 21
The Wronskian Determinant (3 of 3)
• In order for these formulas to be valid, the determinant W in
the denominator cannot be zero:
• W is called the Wronskian determinant, or more simply,
the Wronskian of the solutions y1and y2. We will sometimes
use the notation
)()()()()()(
)()(02010201
0201
0201tytytyty
tyty
tytyW
W
yty
yty
cW
tyy
tyy
c001
001
2
020
020
1
)(
)(
,)(
)(
021, tyyW
Page 22
Theorem 3.2.3
• Suppose y1 and y2 are solutions to the equation
with the initial conditions
Then it is always possible to choose constants c1, c2 so that
satisfies the differential equation and initial conditions if and
ony if the Wronskian
is not zero at the point t0
L[y] = y ''+ p(t)y '+ q(t)y = 0
2121 yyyyW
0000 )(,)( ytyyty
)()( 2211 tyctycy
Page 23
Example 3
• In Example 2 of Section 3.1, we found that
were solutions to the differential equation
• The Wronskian of these two functions is
• Since W is nonzero for all values of t, the functions
can be used to construct solutions of the differential
equation with initial conditions at any value of t
065 yyy
tt etyety 32
2
1 )(and)(
t
tt
tt
eee
eeW 5
32
32
32
21 and yy
Page 24
Theorem 3.2.4 (Fundamental Solutions)
• Suppose y1 and y2 are solutions to the equation
Then the family of solutions
y = c1y1 + c2 y2
with arbitrary coefficients c1, c2 includes every solution to
the differential equation if an only if there is a point t0 such
that W(y1,y2)(t0) ≠ 0, .
• The expression y = c1y1 + c2 y2 is called the general solution
of the differential equation above, and in this case y1 and y2
are said to form a fundamental set of solutions to the
differential equation.
L[y] = y ''+ p(t)y '+ q(t)y = 0.
Page 25
Example 4
• Consider the general second order linear equation below,
with the two solutions indicated:
• Suppose the functions below are solutions to this equation:
• The Wronskian of y1and y2 is
• Thus y1and y2 form a fundamental set of solutions to the
equation, and can be used to construct all of its solutions.
• The general solution is
2121 ,, 21 rreyeytrtr
. allfor 021
21
21
12
2121
21terr
erer
ee
yy
yyW
trr
trtr
trtr
0)()( ytqytpy
trtrececy 21
21
Page 26
Example 5: Solutions (1 of 2)
• Consider the following differential equation:
• Show that the functions below are fundamental solutions:
• To show this, first substitute y1 into the equation:
• Thus y1 is a indeed a solution of the differential equation.
• Similarly, y2 is also a solution:
1
2
2/1
1 , tyty
2t 2y ''+ 3t y '- y = 0, t > 0
012
3
2
1
23
42 2/12/1
2/12/32
ttt
tt
t
0134322 11232 tttttt
Page 27
Example 5: Fundamental Solutions (2 of 2)
• Recall that
• To show that y1 and y2 form a fundamental set of solutions,
we evaluate the Wronskian of y1 and y2:
• Since W ≠ 0 for t > 0, y1 and y2 form a fundamental set of
solutions for the differential equation
W =y1 y2
¢y1 ¢y2
=
t1/2 t-1
1
2t -1/2 -t -2
= -t -3/2 -1
2t -3/2 = -
3
2t-3/2
1
2
2/1
1 , tyty
2t 2y ''+ 3t y '- y = 0, t > 0
Page 28
Theorem 3.2.5: Existence of Fundamental Set
of Solutions
• Consider the differential equation below, whose coefficients
p and q are continuous on some open interval I:
• Let t0 be a point in I, and y1 and y2 solutions of the equation
with y1 satisfying initial conditions
and y2 satisfying initial conditions
• Then y1 and y2 form a fundamental set of solutions to the
given differential equation.
L[y] = y ''+ p(t)y '+ q(t)y = 0
y1(t0 ) =1, y '1(t0 ) = 0
y2(t0 ) = 0, y '2(t0 ) =1
Page 29
Example 6: Apply Theorem 3.2.5 (1 of 3)
• Find the fundamental set specified by Theorem 3.2.5 for the
differential equation and initial point
• In Section 3.1, we found two solutions of this equation:
The Wronskian of these solutions is W(y1, y2)(t0) = –2 ≠ 0 so
they form a fundamental set of solutions.
• But these two solutions do not satisfy the initial conditions
stated in Theorem 3.2.5, and thus they do not form the
fundamental set of solutions mentioned in that theorem.
• Let y3 and y4 be the fundamental solutions of Thm 3.2.5.
tt eyey 21 ,
y ''- y = 0, t0 = 0
1)0(,0)0(;0)0(,1)0( 4433 yyyy
Page 30
Example 6: General Solution (2 of 3)
• Since y1 and y2 form a fundamental set of solutions,
• Solving each equation, we obtain
• The Wronskian of y3 and y4 is
• Thus y3, y4 form the fundamental set of solutions indicated in
Theorem 3.2.5, with general solution in this case
1)0(,0)0(,
0)0(,1)0(,
44214
33213
yyededy
yyececy
tt
tt
)sinh(2
1
2
1)(),cosh(
2
1
2
1)( 43 teetyteety tttt
01sinhcoshcoshsinh
sinhcosh22
21
21
tt
tt
tt
yy
yyW
)sinh()cosh()( 21 tktkty
Page 31
Example 6:
Many Fundamental Solution Sets (3 of 3)
• Thus
both form fundamental solution sets to the differential
equation and initial point
• In general, a differential equation will have infinitely many
different fundamental solution sets. Typically, we pick the
one that is most convenient or useful.
ttSeeS tt sinh,cosh,, 21
y ''- y = 0, t0 = 0
Page 32
Theorem 3.2.6
Consider again the equation (2):
where p and q are continuous real-valued functions.
If y = u(t) + iv(t) is a complex-valued solution of Eq. (2), then its real part u and its imaginary part v are also solutions of this equation.
L[y] = y ''+ p(t)y '+ q(t)y = 0
Page 33
Theorem 3.2.7 (Abel’s Theorem)
• Suppose y1 and y2 are solutions to the equation
where p and q are continuous on some open interval I. Then
the W[y1,y2](t) is given by
where c is a constant that depends on y1 and y2 but not on t.
• Note that W[y1,y2](t) is either zero for all t in I (if c = 0) or else
is never zero in I (if c ≠ 0).
L[y] = y ''+ p(t)y '+ q(t)y = 0
W[y1,y2 ](t) =ce- p(t )dtò
Page 34
Example 7 Apply Abel’s Theorem
• Recall the following differential equation and its solutions: with solutions
• We computed the Wronskian for these solutions to be
• Writing the differential equation in the standard form
• So and the Wronskian given by Thm.3.2.6 is
• This is the Wronskian for any pair of fundamental solutions. For the solutions given above, we must let c = –3/2
p(t) =3
2t
2t 2y ''+ 3t y '- y = 0, t > 01
2
2/1
1 , tyty
W =y1 y2
¢y1 ¢y2
= -3
2t -3/2 = -
3
2 t 3
y ''+3
2ty '-
1
2t 2y = 0, t > 0
W[y1,y2 ](t) =ce-
3
2tdtò
= ce-
3
2lnt
= ct - 3 / 2
Page 35
Summary
• To find a general solution of the differential equation
we first find two solutions y1 and y2.
• Then make sure there is a point t0 in the interval such that
W[y1, y2](t0) ≠ 0.
• It follows that y1 and y2 form a fundamental set of solutions
to the equation, with general solution y = c1y1 + c2 y2.
• If initial conditions are prescribed at a point t0 in the interval
where W ≠ 0, then c1 and c2 can be chosen to satisfy those
conditions.
y ''+ p(t)y '+ q(t)y = 0, a < t < b
Page 36
Boyce/DiPrima/Meade 11th ed, Ch 3.3:
Complex Roots of Characteristic Equation
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley
& Sons, Inc.
• Recall our discussion of the equation
where a, b and c are constants.
• Assuming an exponential soln leads to characteristic equation:
• Quadratic formula (or factoring) yields two solutions, r1 and r2:
• If b2 – 4ac < 0, then complex roots:
Thus
0 cyybya
0)( 2 cbrarety rt
a
acbbr
2
42
titi etyety )(,)( 21
r1 = l + im and r2 = l - im
Page 37
Euler’s Formula; Complex Valued Solutions
• Substituting it into Taylor series for et, we obtain Euler’s
formula:
• Generalizing Euler’s formula, we obtain
• Then
• Therefore
tit
n
ti
n
t
n
ite
n
nn
n
nn
n
nit sincos
!12
)1(
!2
)1(
!
)(
1
121
0
2
0
eimt = cosmt + isinmt
el+im( )t = elteimt = elt cosmt + isinmt[ ] = elt cosmt + ielt sinmt
tieteety
tieteety
ttti
ttti
sincos)(
sincos)(
2
1
Page 38
Real Valued Solutions
• Our two solutions thus far are complex-valued functions:
• We would prefer to have real-valued solutions, since our
differential equation has real coefficients.
• To achieve this, recall that linear combinations of solutions
are themselves solutions:
• Ignoring constants, we obtain the two solutions
tietety
tietety
tt
tt
sincos)(
sincos)(
2
1
tietyty
tetyty
t
t
sin2)()(
cos2)()(
21
21
tetytety tt sin)(,cos)( 43
Page 39
Real Valued Solutions: The Wronskian
• Thus we have the following real-valued functions:
• Checking the Wronskian, we obtain
• Thus y3 and y4 form a fundamental solution set for our ODE,
and the general solution can be expressed as
tetytety tt sin)(,cos)( 43
0
cossinsincos
sincos
2
t
tt
tt
e
ttette
teteW
tectecty tt sincos)( 21
Page 40
Example 1 (1 of 2)
• Consider the differential equation
• For an exponential solution, the characteristic equation is
• Therefore, separating the real and imaginary components,
and thus the general solution is
)( 3sin3cos3sin3cos)( 21
2/2/
2
2/
1 tctcetectecty ttt
025.9 yyy
ii
rrrety rt
2
3
2
1
2
31
2
41101)( 2
l = -1
2, m = 3
Page 41
Example 1 (2 of 2)
• Using the general solution just determined
• We can determine the particular solution that satisfies the
initial conditions
• So
• Thus the solution of this IVP is
• The solution is a decaying oscillation
)( 3sin3cos)( 21
2/ tctcety t
8)0('and2)0( yy
y(0) = c1 = 2
¢y (0) = -1
2c1 + 3c2 = 8
üýï
þïÞ c1 = 2, c2 = 3
2 4 6 8 10t
2
1
1
2
3
y t
)2( 3sin33cos)( 2/ ttety t
)2( 3sin33cos)( 2/ ttety t
Page 42
Example 2
• Consider the initial value problem
• Then
• Thus the general solution is
• And
• The solution of the IVP is
• The solution is displays a
growing oscillation
1)0(',2)0(,0145816 yyyyy
irrrety rt 34
10145816)( 2
tectecty tt 3sin3cos)( 4/
2
4/
1
y(0) = c1 = -2
¢y (0) = -1
4c1 + 3c2 = 1
üýï
þïÞ c1 = -2, c2 =
1
2
y(t) = -2et /4 cos 3t( ) +1
2et /4sin 3t( )
2 4 6 8t
10
5
5
10
y t
)2( 3sin2/13cos)( 4/ ttety t
Page 43
2 4 6 8t
2
2
4
y t
Example 3
• Consider the equation
• Then
• Therefore
• and thus the general solution is
• Because there is no exponential
factor in the solution, so the amplitude
of each oscillation remains constant.
The figure shows the graph of two
typical solutions
09 yy
irrety rt 309)( 2
3,0
tctcty 3sin3cos)( 21
,0 tty
tty
3sin2/13cos:dashed
3sin23cos2:solid
Page 44
Boyce/DiPrima/Meade 11th ed, Ch 3.4:
Repeated Roots; Reduction of Order
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley & Sons, Inc.
• Recall our 2nd order linear homogeneous ODE
• where a, b and c are constants.
• Assuming an exponential solution leads to characteristic
equation:
• Quadratic formula (or factoring) yields two solutions, r1 and r2:
• When b2 – 4ac = 0, r1 = r2 = –b/(2a), since method only gives
one solution:
0 cyybya
0)( 2 cbrarety rt
a
acbbr
2
42
y1(t) = ce-bt /(2a)
Page 45
Second Solution: Multiplying Factor v(t)
• We know that
• Since y1 and y2 are linearly dependent, we generalize this
approach and multiply by a function v, and determine
conditions for which y2 is a solution:
• Then
solution a )()(solution a )( 121 tcytyty
y1(t) = e-bt /(2a) a solution Þ try y2(t) = v(t)e-bt /(2a)
y2(t) = v(t)e-bt /(2a)
¢y2(t) = ¢v (t)e-bt /(2a) -b
2av(t)e-bt /(2a)
¢¢y2(t) = ¢¢v (t)e-bt /(2a) -b
2a¢v (t)e-bt /(2a) -
b
2a¢v (t)e-bt /(2a) +
b2
4a2v(t)e-bt /(2a)
Page 46
Finding Multiplying Factor v(t)
• Substituting derivatives into ODE, we seek a formula for v:
0 cyybya
e-bt /(2a) a ¢¢v (t) -b
a¢v (t) +
b2
4a2v(t)
é
ëê
ù
ûú + b ¢v (t) -
b
2av(t)
é
ëêù
ûú+ cv(t)
ìíî
üýþ
= 0
a ¢¢v (t) - b ¢v (t) +b2
4av(t) + b ¢v (t) -
b2
2av(t) + cv(t) = 0
a ¢¢v (t) +b2
4a-b2
2a+ c
æ
èçö
ø÷v(t) = 0
a ¢¢v (t) +b2
4a-
2b2
4a+
4ac
4a
æ
èçö
ø÷v(t) = 0 Û a ¢¢v (t) +
-b2
4a+
4ac
4a
æ
èçö
ø÷v(t) = 0
a ¢¢v (t) -b2 - 4ac
4a
æ
èçö
ø÷v(t) = 0
¢¢v (t) = 0 Þ v(t) = k3t + k4
Page 47
General Solution
• To find our general solution, we have:
• Thus the general solution for repeated roots is
y(t) = k1e-bt /(2a) + k2v(t)e
-bt /(2a)
= k1e-bt /(2a) + k3t + k4( )e-bt /(2a)
= c1e-bt /(2a) + c2te
-bt /(2a)
y(t) = c1e-bt /(2a) + c2te
-bt /(2a)
Page 48
Wronskian
• The general solution is
• Thus every solution is a linear combination of
• The Wronskian of the two solutions is
• Thus y1 and y2 form a fundamental solution set for equation.
y(t) = c1e-bt /(2a) + c2te
-bt /(2a)
y1(t) = e-bt /(2a), y2(t) = te-bt /(2a)
W (y1, y2 )(t) =
e-bt /(2a) te-bt /(2a)
-b
2ae-bt /(2a) 1-
bt
2a
æ
èçö
ø÷e-bt /(2a)
= e-bt /a 1-bt
2a
æ
èçö
ø÷+ e-bt /a bt
2a
æ
èçö
ø÷
= e-bt /a ¹ 0 for all t
Page 49
Example 1 (1 of 2)
• Consider the initial value problem
• Assuming exponential soln leads to characteristic equation:
• So one solution is and a second solution is found:
• Substituting these into the differential equation and
simplifying yields
where are arbitrary constants.
044 yyy
20)2(044)( 22 rrrrety rt
ttt
tt
t
etvetvetvty
etvetvty
etvty
222
2
22
2
2
2
)(4)(4)()(
)(2)()(
)()(
tety 21 )(
211 )(,)(',0)(" ktktvktvtv
21 and cc
Page 50
Example 1 (2 of 2)
• Letting
• So the general solution is
• Note that both tend to 0 as regardless of the values of • Here are three solutions of this equation with different sets of initial conditions.
• y(0) = 2, y’(0) = 1 (top)
• y(0) = 1, y’(0) = 1 (middle)
• y(0) = ½, y’(0) = 1 (bottom)
tt tececty 2
2
2
1)(
ttetyttvkk 2221 )(and)(,0 and1
21 and yy t
21 and cc
Page 51
Example 2 (1 of 2)
• Consider the initial value problem
• Assuming exponential solution leads to characteristic equation:
• Thus the general solution is
• Using the initial conditions:
• Thus
¢¢y - ¢y +1
4y = 0, y 0( ) = 2, ¢y 0( ) =
1
3
y(t) = ert Þ r2 - r +1
4= 0 Û (r -
1
2)2 = 0 Û r =
1
2
2/
2
2/
1)( tt tececty
3
2,2
3
1
2
1
2
2121
1
cccc
c
2/2/
3
22)( tt teety
1 2 3 4t
2
1
1
2
3
4
y t
)3/22()( 2/ tety t
Page 52
Example 2 (2 of 2)
• Suppose that the initial slope in the previous problem was
increased
• The solution of this modified problem is
• Notice that the coefficient of the second
term is now positive. This makes a big
difference in the graph, since the
exponential function is raised to a
positive power:
1 2 3 4t
2
2
4
6
y t
20,20 yy
l =1
2> 0
2/2/2)( tt teety
)3/22()(:blue
)2()(:red
2/
2/
tety
tety
t
t
Page 53
Reduction of Order
• The method used so far in this section also works for
equations with nonconstant coefficients:
• That is, given that y1 is solution, try y2 = v(t)y1:
• Substituting these into ODE and collecting terms,
• Since y1 is a solution to the differential equation, this last
equation reduces to a first order equation in v :
0)()( ytqytpy
)()()()(2)()()(
)()()()()(
)()()(
1112
112
12
tytvtytvtytvty
tytvtytvty
tytvty
02 111111 vqyypyvpyyvy
02 111 vpyyvy
Page 54
Example 3: Reduction of Order (1 of 3)
• Given the variable coefficient equation and solution y1,
use reduction of order method to find a second solution:
• Substituting these into the ODE and collecting terms,
,)(;0,032 1
1
2 ttytyytyt
321
2
21
2
1
2
)(2 )(2 )()(
)( )()(
)()(
ttvttvttvty
ttvttvty
ttvty
)()( where,02
02
033442
03222
111
1213212
tvtuuut
vvt
vtvtvvtvtv
vtvttvtvttvtvt
Page 55
Example 3: Finding v(t) (2 of 3)
• To solve
for u, we can use the separation of variables method:
• Thus
and hence
.0 since,
ln2/1ln2
102
2/12/1
tctuetu
Ctudttu
duu
dt
dut
C
)()(,02 tvtuuut
2/1ctv
v(t) =2
3ct 3/ 2 + k
Page 56
Example 3: General Solution (3 of 3)
• Since
• Recall
• So we can neglect the second term of y2 to obtain
• The Wronskian of can be computed
• Hence the general solution to the differential equation is
y2(t) =2
3ct 3/ 2 + k
æ
èçö
ø÷t -1 =
2
3ct1/ 2 + k t -1
2/12 )( tty
1
1 )( tty
2/12
1
1)( tctcty
v(t) =2
3ct 3/ 2 + k
)(and)( 21 tyty
W [y1,y2](t) =3
2t -3/2 ¹ 0, t > 0
Page 57
Boyce/DiPrima/Meade 11th ed, Ch 3.5: Nonhomogeneous
Equations; Method of Undetermined Coefficients
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, , and Doug Meade ©2017 by John Wiley & Sons, Inc.
• Recall the nonhomogeneous equation
where p, q, g are continuous functions on an open interval I.
• The associated homogeneous equation is
• In this section we will learn the method of undetermined
coefficients to solve the nonhomogeneous equation, which
relies on knowing solutions to the homogeneous equation.
0)()( ytqytpy
)()()( tgytqytpy
Page 58
Theorem 3.5.1
• If Y1 and Y2 are solutions of the nonhomogeneous equation
then Y1 – Y2 is a solution of the homogeneous equation
• If, in addition, {y1, y2} forms a fundamental solution set of
the homogeneous equation, then there exist constants c1 and
c2 such that
)()()()( 221121 tyctyctYtY
)()()( tgytqytpy
0)()( ytqytpy
Page 59
Theorem 3.5.2 (General Solution)
• To solve the nonhomogeneous equation
we need to do three things:
1. Find the general solution c1y1(t) + c2y2(t) of the
corresponding homogeneous equation. This is called the
complementary solution and may be denoted by yc(t).
2. Find any solution Y(t) of the nonhomogeneous equation.
This is often referred to as a particular solution.
3. Form the sum of the functions found in steps 1 and 2.
)()()()( 2211 tYtyctycty
)()()( tgytqytpy
Page 60
Method of Undetermined Coefficients
• Recall the nonhomogeneous equation
with general solution
• In this section we use the method of undetermined
coefficients to find a particular solution Y to the
nonhomogeneous equation, assuming we can find solutions
y1, y2 for the homogeneous case.
• The method of undetermined coefficients is usually limited
to when p and q are constant, and g(t) is a polynomial,
exponential, sine or cosine function.
)()()()( 2211 tYtyctycty
)()()( tgytqytpy
Page 61
Example 1: Exponential g(t)
• Consider the nonhomogeneous equation
• We seek Y satisfying this equation. Since exponentials
replicate through differentiation, a good start for Y is:
• Substituting these derivatives into the differential equation,
• Thus a particular solution to the nonhomogeneous ODE is
teyyy 2343
ttt AetYAetYAetY 222 4)(,2)()(
2/136
3464
22
2222
AeAe
eAeAeAe
tt
tttt
tetY 2
2
1)(
Page 62
Example 2: Sine g(t), First Attempt (1 of 2)
• Consider the nonhomogeneous equation
• We seek Y satisfying this equation. Since sines replicate through differentiation, a good start for Y is:
• Substituting these derivatives into the differential equation,
• Since sin(x) and cos(x) are not multiples of each other, we must have c1= c2 = 0, and hence 2 + 5A = 3A = 0, which is
impossible.
tyyy sin243
tAtYtAtYtAtY sin)(,cos)(sin)(
0cossin
0cos3sin52
sin2sin4cos3sin
21
tctc
tAtA
ttAtAtA
Page 63
Example 2: Sine g(t), Particular Solution (2 of 2)
• Our next attempt at finding a Y is
• Substituting these derivatives into ODE, we obtain
• Thus a particular solution to the nonhomogeneous ODE is
tBtAtYtBtAtY
tBtAtY
cossin)(,sincos)(
cossin)(
-Asin t - Bcos t( ) - 3 Acos t - Bsin t( ) - 4 Asin t + Bcos t( ) = 2sin t
Û -5A + 3B( )sin t + -3A - 5B( )cost = 2sin t
Û - 5A + 3B = 2, - 3A - 5B = 0
Û A = -5
17, B =
3
17
tyyy sin243
tttY cos17
3sin
17
5)(
Page 64
Example 3: Product g(t)
• Consider the nonhomogeneous equation
• We seek Y satisfying this equation, as follows:
• Substituting these into the ODE and solving for A and B:
¢¢y - 3 ¢y - 4y = -8et cos(2t)
Y (t) = Aet cos(2t) + Bet sin(2t)
¢Y (t) = Aet cos(2t) - 2Aet sin(2t) + Bet sin(2t) +2Bet cos(2t)
= A +2B( )et cos(2t) + -2A + B( )et sin(2t)
¢¢Y (t) = A +2B( )et cos(2t) - 2 A +2B( )et sin(2t) + -2A + B( )et sin(2t)
+ 2 -2A + B( )et cos(2t)
= -3A + 4B( )et cos(2t) + -4A - 3B( )et sin(2t)
A =10
13, B =
2
13ÞY (t) =
10
13et cos(2t)+
2
13et sin(2t)
Page 65
Discussion: Sum g(t)
• Consider again our general nonhomogeneous equation
• Suppose that g(t) is sum of functions:
• If Y1, Y2 are solutions of
respectively, then Y1 + Y2 is a solution of the
nonhomogeneous equation above.
)()()( tgytqytpy
)()()( 21 tgtgtg
)()()(
)()()(
2
1
tgytqytpy
tgytqytpy
Page 66
Example 4: Sum g(t)
• Consider the equation
• Our equations to solve individually are
• Our particular solution is then
teteyyy tt 2cos8sin2343 2
tetettetY ttt 2sin13
22cos
13
10sin
17
5cos
17
3
2
1)( 2
teyyy
tyyy
eyyy
t
t
2cos843
sin243
343 2
Page 67
Example 5: First Attempt (1 of 3)
• Consider the nonhomogeneous equation
• We seek Y satisfying this equation. We begin with
• Substituting these derivatives into differential equation,
• Since the left side of the above equation is always 0, no
value of A can be found to make a solution to the
nonhomogeneous equation.
• To understand why this happens, we will look at the solution
of the corresponding homogeneous differential equation
teyyy 243
ttt AetYAetYAetY )(,)()(
tt eeAAA 2)43(
tAetY )(
Page 68
Example 5: Homogeneous Solution (2 of 3)
• To solve the corresponding homogeneous equation:
• We use the techniques from Section 3.1 and get
• Thus our assumed particular solution solves
the homogeneous equation instead of the nonhomogeneous
equation.
• So we need another form for Y(t) to arrive at the general
solution of the form:
)()( 421 tYececty tt
043 yyy
tAetY )(
tt etyety 4
21 )(and)(
Page 69
Example 5: Particular Solution (3 of 3)
• Our next attempt at finding a Y(t) is:
• Substituting these into the ODE,
• So the general solution to the IVP is
0.0 0.2 0.4 0.6 0.8 1.0t
10
20
30
40
y t
ttttt
tt
t
AeAteAteAeAetY
AteAetY
AtetY
2)(
)(
)(
t
tttt
tttttt
tetY
AeAteAteAte
eAteAteAeAeAte
5
2)(
5/22550
24332
teyyy 24'3
te
y(t) = c1e- t + c2e
4 t -2
5t e - t
ttt eteety 5/24)( 4
Page 70
Summary – Undetermined Coefficients (1 of 2)
• For the differential equation
where a, b, and c are constants, if g(t) belongs to the class
of functions discussed in this section (involves nothing
more than exponential functions, sines, cosines,
polynomials, or sums or products of these), the method of
undetermined coefficients may be used to find a particular
solution to the nonhomogeneous equation.
• The first step is to find the general solution for the
corresponding homogeneous equation with g(t) = 0.
)(tgcyybya
)()()( 2211 tyctyctyC
Page 71
Summary – Undetermined Coefficients (2 of 2)
• The second step is to select an appropriate form for the
particular solution, Y(t), to the nonhomogeneous equation and
determine the derivatives of that function.
• After substituting Y(t), Y’(t), and Y”(t) into the nonhomo-
geneous differential equation, if the form for Y(t) is correct, all
the coefficients in Y(t) can be determined.
• Finally, the general solution to the nonhomogeneous
differential equation can be written as
)()()()()()( 2211 tYtyctyctYtyty Cgen
Page 72
Boyce/DiPrima/Meade 11th ed, Ch 3.6:
Variation of Parameters
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley
& Sons, Inc
• Recall the nonhomogeneous equation
where p, q, g are continuous functions on an open interval I.
• The associated homogeneous equation is
• In this section we will learn the variation of parameters method to solve the nonhomogeneous equation. As with the method of undetermined coefficients, this procedure relies on knowing solutions to the homogeneous equation.
• Variation of parameters is a general method, and requires no detailed assumptions about solution form. However, certain integrals need to be evaluated, and this can present difficulties.
0)()( ytqytpy
)()()( tgytqytpy
Page 73
Example 1: Variation of Parameters (1 of 6)
• We seek a particular solution to the equation below.
• We cannot use the undetermined coefficients method since g(t) is a quotient of sin t or cos t, instead of a sum or product.
• Recall that the solution to the homogeneous equation is
• To find a particular solution to the nonhomogeneous equation, we begin with the form
• Then
• or
¢¢y + 4y = 8 tant, -p / 2 < t < p / 2
yC (t) = c1 cos(2t)+ c2 sin(2t)
y(t) = u1(t)cos(2t)+u2(t)sin(2t)
¢y (t) = ¢u1(t)cos(2t)- 2u1(t)sin(2t)+ ¢u2(t)sin(2t)+ 2u2(t)cos(2t)
¢y (t) = -2u1(t)sin(2t)+ 2u2(t)cos(2t)+ ¢u1(t)cos(2t)+ ¢u2(t)sin(2t)
Page 74
Example 1: Derivatives, 2nd Equation (2 of 6)
• From the previous slide,
• Note that we need two equations to solve for u1 and u2. The
first equation is the differential equation. To get a second
equation, we will require
• Then
• Next,
¢y (t) = -2u1(t)sin(2t)+ 2u2(t)cos(2t)+ ¢u1(t)cos(2t)+ ¢u2(t)sin(2t)
¢u1(t)cos(2t)+ ¢u2(t)sin(2t) = 0
¢y (t) = -2u1(t)sin(2t)+ 2u2(t)cos(2t)
¢¢y (t) = -2 ¢u1(t)sin(2t)- 4u1(t)cos(2t)+ 2 ¢u2(t)cos(2t)- 4u2(t)sin(2t)
Page 75
Example 1: Two Equations (3 of 6)
• Recall that our differential equation is
• Substituting y'' and y into this equation, we obtain
• This equation simplifies to
• Thus, to solve for u1 and u2, we have the two equations:
-2 ¢u1(t)sin(2t)- 4u1(t)cos(2t)+ 2 ¢u2 (t)cos(2t)- 4u2(t)sin(2t)
+ 4 u1(t)cos(2t)+ u2(t)sin(2t)( ) = 8 tant
-2 ¢u1(t)sin(2t)+ 2 ¢u2 (t)cos(2t) = 8 tan t
¢u1(t)cos(2t)+ ¢u2 (t)sin(2t) = 0
-2 ¢u1(t)sin(2t)+ 2 ¢u2(t)cos(2t) = 8tant
¢¢y + 4y = 8 tant
Page 76
Example 1: Solve for u1' (4 of 6)
• To find u1 and u2 , we first need to solve for
• From second equation,
• Substituting this into the first equation,
t
ttutu
2sin
2cos)()( 12
-2 ¢u1(t)sin(2t) + 2 - ¢u1(t)cos(2t)
sin(2t)
é
ëê
ù
ûúcos(2t) = 8 tan t
-2 ¢u1(t)sin2 2t( ) - 2 ¢u1(t)cos2 2t( ) = 8 tan t sin(2t)
-2 ¢u1(t) sin2 2t( ) + cos2 2t( )éë ùû = 82sin2 t cos t
cos t
é
ëê
ù
ûú
¢u1(t) = -8sin2 t
-2 ¢u1(t)sin(2t)+ 2 ¢u2 (t)cos(2t) = 8 tan t
¢u1(t)cos(2t)+ ¢u2 (t)sin(2t) = 0
21 and uu
Page 77
Example 1: Solve for u1 and u2 (5 of 6)
• From the previous slide,
• Then
• Thus
¢u2(t) = 8sin2 tcos(2t)
sin(2t)= 4
sint(2cos2 t -1)
cost= 4sint 2cost -
1
cos t
æ
èçö
ø÷
u1(t) = ¢u1(t)dt =ò 4sin t cos t - 4t + c1
u2(t) = ¢u2 (t)dt =ò 4 ln(cost)- 4 cos2 t + c2
¢u1(t) = -8sin2 t, ¢u2(t) = - ¢u1(t)cos2t
sin2t
Page 78
Example 1: General Solution (6 of 6)
• Recall our equation and homogeneous solution yC:
• Using the expressions for u1 and u2 on the previous slide, the
general solution to the differential equation is
y(t) = u1(t)cos2t + u2 (t)sin2t + yC (t)
= (4sin t cost)cos(2t)+ (4 ln(cost)- 4 cos2 t)sin(2t)+ c1 cos(2t)+ c2 sin(2t)
= -2sin(2t)- 4t cos(2t)+ 4 ln(cos t)sin(2t)+ c1 cos(2t)+ c2 sin(2t)
¢¢y + 4y = 8 tant, yC (t) = c1 cos(2t)+ c2 sin(2t)
Page 79
Summary
• Suppose y1, y2 are fundamental solutions to the homogeneous
equation associated with the nonhomogeneous equation
above, where we note that the coefficient on y'' is 1.
• To find u1 and u2, we need to solve the equations
• Doing so, and using the Wronskian, we obtain
• Thus
)()()()()(
0)()()()(
2211
2211
tgtytutytu
tytutytu
)()()()()(
)()()(
2211 tytutytuty
tgytqytpy
)(,
)()()(,
)(,
)()()(
21
12
21
21
tyyW
tgtytu
tyyW
tgtytu
2
21
121
21
21
)(,
)()()(,
)(,
)()()( cdt
tyyW
tgtytucdt
tyyW
tgtytu
Page 80
Theorem 3.6.1
• Consider the equations
• If the functions p, q and g are continuous on an open interval I,
and if y1 and y2 are fundamental solutions to Eq. (2), then a
particular solution of Eq. (1) is
and the general solution is
dttyyW
tgtytydt
tyyW
tgtytytY
)(,
)()()(
)(,
)()()()(
21
12
21
21
)()()()( 2211 tYtyctycty
)2(0)()(
)1()()()(
ytqytpy
tgytqytpy
Page 81
Boyce/DiPrima/Meade 11th ed, Ch 3.7: Mechanical &
Electrical Vibrations
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley
& Sons, Inc.
• Two important areas of application for second order linear
equations with constant coefficients are in modeling
mechanical and electrical oscillations.
• We will study the motion of a mass on a spring in detail.
• An understanding of the behavior of this simple system is the
first step in investigation of more complex vibrating systems.
Page 82
Spring – Mass System
• Suppose a mass m hangs from a vertical spring of original
length l. The mass causes an elongation L of the spring.
• The force FG of gravity pulls the mass down. This force has
magnitude mg, where g is acceleration due to gravity.
• The force FS of the spring stiffness pulls the mass up. For
small elongations L, this force is proportional to L.
That is, Fs = kL (Hooke’s Law).
• When the mass is in equilibrium, the forces balance each
other: kLmg
Page 83
Spring Model
• We will study the motion of a mass when it is acted on by an
external force (forcing function) and/or is initially displaced.
• Let u(t) denote the displacement of the mass from its
equilibrium position at time t, measured downward.
• Let f be the net force acting on the mass. We will use
Newton’s 2nd Law:
• In determining f, there are four separate forces to consider:
– Weight: w = mg (downward force)
– Spring force: Fs = – k(L+ u) (up or down force, see next slide)
– Damping force: (up or down, see following slide)
– External force: F (t) (up or down force, see text)
)()( tftum
Fd (t) = -g u '(t)
Page 84
Spring Model:
Spring Force Details
• The spring force Fs acts to restore a spring to the natural
position, and is proportional to L + u. If L + u > 0, then the
spring is extended and the spring force acts upward. In this
case
• If L + u < 0, then spring is compressed a distance of |L + u|,
and the spring force acts downward. In this case
• In either case,
)( uLkFs
uLkuLkuLkFs
)( uLkFs
Page 85
Spring Model:
Damping Force Details
• The damping or resistive force Fd acts in the opposite direction as
the motion of the mass. This can be complicated to model. Fd may
be due to air resistance, internal energy dissipation due to action of
spring, friction between the mass and guides, or a mechanical
device (dashpot) imparting a resistive force to the mass.
• We simplify this and assume Fd is proportional to the velocity.
• In particular, we find that
– If u’ > 0, then u is increasing, so the mass is moving downward.
Thus Fd acts upward and hence .
– If u’ < 0, then u is decreasing, so the mass is moving upward.
Thus Fd acts downward and hence
– In either case, 0),()( tutFd
Fd (t) = -g ¢u (t)
Fd (t) = -g ¢u (t)
Page 86
Spring Model:
Differential Equation
• Taking into account these forces, Newton’s Law becomes:
• Recalling that mg = kL, this equation reduces to
where the constants m, , and k are positive.
• We can prescribe initial conditions also:
• It follows from Theorem 3.2.1 that there is a unique solution to
this initial value problem. Physically, if the mass is set in motion
with a given initial displacement and velocity, then its position is
uniquely determined at all future times.
)()()(
)()()()(
tFtutuLkmg
tFtFtFmgtum ds
)()()()( tFtkututum
00 )0(,)0( vuuu
g
Page 87
Example 1:
Find Coefficients (1 of 2)
• A 4 lb mass stretches a spring 2". The mass is displaced an
additional 6" and then released; and is in a medium that exerts a
viscous resistance of 6 lb when the mass has a velocity of 3 ft/sec.
Formulate the IVP that governs the motion of this mass:
• Find m:
• Find :
• Find k:
ft
seclb
8
1
sec/ft32
lb4 2
2 mm
g
wmmgw
ft
seclb2
sec/ft3
lb6lb6 u
ft
lb24
ft6/1
lb4
in2
lb4 kkkLkFs
00 )0(,)0(),()()()( vuuutFtkututum
g
Page 88
Example 1:
Find IVP (2 of 2)
• Thus our differential equation becomes
and hence the initial value problem can be written as
• This problem can be solved using the
methods of Chapter 3.3 and yields
the solution
0)(24)(2)(8
1 tututu
0)0(,2
1)0(
0)(192)(16)(
uu
tututu
))28sin(2)28cos(2(4
1)( 8 ttetu t 0.2 0.4 0.6 0.8
t
0.2
0.2
0.4
0.6
0.8
u t
))28sin(2)28cos(2(4
1)( 8 ttetu t
Page 89
Spring Model:
Undamped Free Vibrations (1 of 4)
• Recall our differential equation for spring motion:
• Suppose there is no external driving force and no damping.
Then F(t) = 0 and = 0, and our equation becomes
• The general solution to this equation is
0)()( tkutum
)()()()( tFtkututum
mk
tBtAtu
/
where
,sincos)(
2
0
00
g
Page 90
Spring Model:
Undamped Free Vibrations (2 of 4)
• Using trigonometric identities, the solution
can be rewritten as follows:
where
• Note that in finding , we must be careful to choose the
correct quadrant. This is done using the signs of cos and
sin .
u(t) = Acosw0t + Bsinw0t, w 0
2 =k
m
,sinsincoscos)(
cos)(sincos)(
00
000
tRtRtu
tRtutBtAtu
A
BBARRBRA tan,sin,cos 22
d
dd
Page 91
Spring Model:
Undamped Free Vibrations (3 of 4)
• Thus our solution is
where
• The solution is a shifted cosine (or sine) curve, that describes simple
harmonic motion, with period
• The circular frequency (radians/time) is the natural frequency of
the vibration, R is the amplitude of the maximum displacement of
mass from equilibrium, and is the phase or phase angle
(dimensionless).
tRtBtAtu 000 cos sincos)(
k
mT
2
2
0
mk /0
w 0
d
Page 92
Spring Model:
Undamped Free Vibrations (4 of 4)
• Note that our solution
is a shifted cosine (or sine) curve with period
• Initial conditions determine A & B, hence also the amplitude R.
• The system always vibrates with the same frequency ,
regardless of the initial conditions.
• The period T increases as m increases, so larger masses vibrate
more slowly. However, T decreases as k increases, so stiffer
springs cause a system to vibrate more rapidly.
mktRtBtAtu /,cos sincos)( 0000
k
mT 2
w 0
Page 93
Example 2: Find IVP (1 of 3)
• A 10 lb mass stretches a spring 2". The mass is displaced an
additional 2" and then set in motion with an initial upward
velocity of 1 ft/sec. Determine the position of the mass at any
later time, and find the period, amplitude, and phase of the motion.
• Find m:
• Find k:
• Thus our IVP is
ft
seclb
16
5
sec/ft32
lb10 2
2 mm
g
wmmgw
ft
lb60
ft6/1
lb10
in2
lb10 kkkLkFs
5
16¢¢u (t)+ 60u(t) = 0, u(0) =
1
6, ¢u (t) = -1
00 )0(,)0(,0)()( vuuutkutum
Page 94
Example 2: Find Solution (2 of 3)
• Simplifying, we obtain
• To solve, use methods of Ch 3.3 to obtain
or
1)0(,6/1)0(,0)(192)( uututu
tttu 192sin192
1192cos
6
1)(
tttu 38sin38
138cos
6
1)(
Page 95
Example 2:
Find Period, Amplitude, Phase (3 of 3)
• The natural frequency is
• The period is
• The amplitude is
• Next, determine the phase :
tttu 38sin38
138cos
6
1)(
rad/sec 856.1338192/0 mk
sec 45345.0/2 0 T
ft 18162.022 BAR
rad 40864.04
3tan
4
3tantan 1
A
B
ABRBRA /tan,sin,cos
409.038cos182.0)( Thus ttu
d
Page 96
Spring Model: Damped Free Vibrations (1 of 8)
• Suppose there is damping but no external driving force F(t):
• What is effect of the damping coefficient on the system?
• The characteristic equation is
• Three cases for the solution:
0)()()( tkututum
2
2
21
411
22
4,
mk
mm
mkrr
term.damping
thefrom expected as ,0)(limcases, threeallIn :Note
.02
4,sincos)(:04
;02/where,)(:04
;0,0where,)(:04
22/2
2/2
21
2 21
tu
m
mktBtAetumk
meBtAtumk
rrBeAetumk
t
mt
mt
trtr
g
Page 97
Damped Free Vibrations: Small Damping (2 of 8)
• Of the cases for solution form, the last is most important,
which occurs when the damping is small:
• We examine this last case. Recall
• Then
and hence
(damped oscillation)
0,sincos)(:04
02/,)(:04
0,0,)(:04
2/2
2/2
21
2 21
tBtAetumk
meBtAtumk
rrBeAetumk
mt
mt
trtr
sin,cos RBRA
teRtu mt cos)( 2/
mteRtu 2/)(
Page 98
Damped Free Vibrations: Quasi Frequency (3 of 8)
• Thus we have damped oscillations:
• The amplitude R depends on the initial conditions, since
• Although the motion is not periodic, the parameter
determines the mass oscillation frequency.
• Thus is called the quasi frequency.
• Recall
sin,cos,sincos)( 2/ RBRAtBtAetu mt
mtmt eRtuteRtu 2/2/ )(cos)(
m
mk
2
4 2
m
m
Page 99
Damped Free Vibrations: Quasi Period (4 of 8)
• Compare with , the frequency of undamped motion:
• Thus, small damping reduces oscillation frequency slightly.
• Similarly, the quasi period is defined as . Then
• Thus, small damping increases quasi period.
kmkmmkkm
kmkm
km
mkm
km
mkm
km
81
81
6441
41
4
4
/4
4
/2
4
22
2
22
42
22
2
22
0
For small
kmkmkmT
Td
81
81
41
/2
/2 21
22/1
2
0
0
Td = 2p / m
g 2
4km
m w 0
Page 100
Damped Free Vibrations:
Neglecting Damping for Small (5 of 8)
• Consider again the comparisons between damped and
undamped frequency and period:
• Thus it turns out that a small is not as telling as a small
ratio .
• For small , we can neglect the effect of damping when
calculating the quasi frequency and quasi period of motion.
But if we want a detailed description of the motion of the
mass, then we cannot neglect the damping force, no matter
how small it is.
2/12
2/12
0 41,
41
kmT
T
km
d
gg 2
4km g 2
4km
g 2
4km
Page 101
Damped Free Vibrations:
Frequency, Period (6 of 8)
• Ratios of damped and undamped frequency, period:
• Thus
• The importance of the relationship between and 4km is
supported by our previous equations:
2/12
2/12
0 41,
41
kmT
T
km
d
dkmkm
T22
lim and 0lim
0,sincos)(:04
02/,)(:04
0,0,)(:04
2/2
2/2
21
2 21
tBtAetumk
meBtAtumk
rrBeAetumk
mt
mt
trtr
g 2
Page 102
Damped Free Vibrations:
Critical Damping Value (7 of 8)
• Thus the nature of the solution changes as passes through the value
• This value of is known as the critical damping value, and for larger values of the motion is said to be overdamped.
• Thus for the solutions given by these cases,
we see that the mass creeps back to its equilibrium position for solutions (1) and (2), but does not oscillate about it, as it does for small in solution (3).
• Solution (1) is overdamped and (2) is critically damped.
.2 km
)3(0,sincos)(:04
)2( 02/,)(:04
)1(0,0,)(:04
2/2
2/2
21
2 21
tBtAetumk
meBtAtumk
rrBeAetumk
mt
mt
trtr
g
g = 2 kmg
g
Page 103
Damped Free Vibrations:
Characterization of Vibration (8 of 8)
• The mass creeps back to the equilibrium position for
solutions (1) & (2), but does not oscillate about it, as it does
for small in solution (3).
• Solution (1) is overdamped and
• Solution (2) is critically damped.
• Solution (3) is underdamped
)3( (Blue)sincos)(:04
)2( Black) (Red,02/,)(:04
)1((Green)0,0,)(:04
2/2
2/2
21
2 21
tBtAetumk
meBtAtumk
rrBeAetumk
mt
mt
trtr
g
Page 104
Example 3: Initial Value Problem (1 of 4)
• Suppose that the motion of a spring-mass system is governed by the initial value problem
• Find the following:
(a) quasi frequency and quasi period;
(b) time at which mass passes through equilibrium position;
(c) time such that |u(t)| < 0.1 for all t > .
• For Part (a), using methods of this chapter we obtain:
where
¢¢u +1
8¢u + u = 0, u(0) = 2, ¢u (0) = 0
tettetu tt
16
255cos
255
32
16
255sin
255
2
16
255cos2)( 16/16/
)sin,cos (recall 06254.0255
1tan RBRA
t t
Page 105
Example 3: Quasi Frequency & Period (2 of 4)
• The solution to the initial value problem is:
• The graph of this solution, along with solution to the
corresponding undamped problem, is given below.
• The quasi frequency is
and quasi period is
• For the undamped case:
tettetu tt
16
255cos
255
32
16
255sin
255
2
16
255cos2)( 16/16/
998.016/255
295.6/2 dT
283.62,10 T
Page 106
Example 3: Quasi Frequency & Period (3 of 4)
• The damping coefficient is = 0.125 = 1/8, and this is 1/16
of the critical value
• Thus damping is small relative to mass and spring stiffness.
Nevertheless the oscillation amplitude diminishes quickly.
• Using a solver, we find that |u(t)| < 0.1 for t > sec
22 km
g
t @ 47.5149
Page 107
Example 3: Quasi Frequency & Period (4 of 4)
• To find the time at which the mass first passes through the
equilibrium position, we must solve
• Or more simply, solve
016
255cos
255
32)( 16/
tetu t
sec 637.12255
16
216
255
t
t
Page 108
Electric Circuits
• The flow of current in certain basic electrical circuits is
modeled by second order linear ODEs with constant
coefficients:
• It is interesting that the flow of current in this circuit is
mathematically equivalent to motion of spring-mass system.
• For more details, see text.
00 )0(,)0(
)()(1
)()(
IIII
tEtIC
tIRtIL
Page 109
Boyce/DiPrima/Meade 11th ed, Ch 3.8:
Forced Periodic Vibrations
Elementary Differential Equations and Boundary Value Problems, 11th edition, by William E. Boyce, Richard C. DiPrima, and Doug Meade ©2017 by John Wiley
& Sons, Inc.
• We continue the discussion of the last section, and now
consider the presence of a periodic external force:
tFtuktutum cos)()()( 0
Page 110
Forced Vibrations with Damping
• Consider the equation below for damped motion and external
forcing function .
• The general solution of this equation has the form
where the general solution of the homogeneous equation is
and the particular solution of the nonhomogeneous equation is
tFtkututum cos)()()( 0
)()(sincos)()()( 2211 tUtutBtAtuctuctu C
)()()( 2211 tuctuctuC
tBtAtU sincos)(
F0 cos(wt)
Page 111
Homogeneous Solution
• The homogeneous solutions u1 and u2 depend on the roots r1
and r2 of the characteristic equation:
• Since m, , and k are are all positive constants, it follows that
r1 and r2 are either real and negative, or complex conjugates
with negative real part. In the first case,
while in the second case
• Thus in either case,
m
mkrkrrmr
2
40
22
0)(lim
tuCt
,0lim)(lim 21
21
trtr
tC
tecectu
.0sincoslim)(lim 21
tectectu tt
tC
t
g
Page 112
Transient and Steady-State Solutions
• Thus for the following equation and its general solution,
we have
• Thus uC(t) is called the transient solution. Note however that
is a steady oscillation with same frequency as forcing function.
• For this reason, U(t) is called the steady-state solution, or
forced response.
tBtAtU sincos)(
0)()(lim)(lim 2211
tuctuctut
Ct
,sincos)()()(
cos)()()(
)()(
2211
0
tUtu
tBtAtuctuctu
tFtkututum
C
Page 113
Transient Solution and Initial Conditions
• For the following equation and its general solution,
the transient solution uC(t) enables us to satisfy whatever initial
conditions might be imposed.
• With increasing time, the energy put into system by initial
displacement and velocity is dissipated through damping force.
The motion then becomes the response U(t) of the system to
the external force .
• Without damping, the effect of the initial conditions would
persist for all time.
)()(
2211
0
sincos)()()(
cos)()()(
tUtu
tBtAtuctuctu
tFtkututum
C
F0 cos(wt)
Page 114
Example 1 (1 of 2)
• Consider a spring-mass system satisfying the differential
equation and initial condition
• Begin by finding the solution to the homogeneous equation
• The methods of Chapter 3.3 yield the solution
• A particular solution to the nonhomogeneous equation will
have the form U(t) = A cos t + B sin t and substitution gives
. So
¢¢u + ¢u +5
4u = 3cost, u(0) = 2, ¢u (0) = 3
tectectu ttC sincos)( 2/
22/
1
U(t) =12
17cost +
48
17sint
A =12
17 and B =
48
17
Page 115
Example 1 (2 of 2)
• The general solution for the nonhomogeneous equation is
• Applying the initial conditions yields
• Therefore, the solution to the IVP is
• The graph breaks the solution into its steady state (U(t)) and transient ( ) components
3)0(,2)0(
025.1
uu
uuu
u(t) = c1e- t /2 cost + c2e
- t /2 sint +12
17cost +
48
17sin t
u(0) = c1 +12
17= 2
u '(0) = -1
2c1 + c2 +
48
17= 3
ü
ýïï
þïï
Þ c1 =22
17, c2 =
14
17
u(t) =22
17e- t /2 cost +
14
17e-t /2 sin t +
12
17cost +
48
17sint
5 10 15t
3
2
1
1
2
3
4
u t
solutionfull
statesteady
transient
)(tuC
Page 116
Rewriting Forced Response
• Using trigonometric identities, it can be shown that
can be rewritten as
• It can also be shown that
where
tRtU cos)(
tBtAtU sincos)(
22222
0
222222
0
2
22
0
22222
0
2
0
)(sin,
)(
)(cos
,)(
mm
m
m
FR
w 0
2 =k
m
Page 117
Amplitude Analysis of Forced Response
• The amplitude R of the steady state solution
depends on the driving frequency . For low-frequency
excitation we have
where we recall ( 0)2 = k /m. Note that F0 /k is the static
displacement of the spring produced by force F0.
• For high frequency excitation,
,)( 22222
0
2
0
m
FR
k
F
m
F
m
FR 0
2
0
0
22222
0
2
0
00 )(limlim
0)(
limlim22222
0
2
0
m
FR
w
w
Page 118
Maximum Amplitude of Forced Response
• Thus
• At an intermediate value of , the amplitude R may have a
maximum value. To find this frequency , differentiate R and
set the result equal to zero. Solving for max, we obtain
where ( 0)2 = k /m. Note max < 0, and max is close to
0 for small . The maximum value of R is
)4(1 2
0
0max
mk
FR
0lim,lim 00
RkFR
mkm 21
2
22
02
22
0
2
max
www
w w w w
w g
Page 119
Maximum Amplitude for Imaginary max
• We have
and
where the last expression is an approximation for small . If
2 /(mk) > 2, then max is imaginary.
• In this case the maximum value of R occurs for = 0, and R is
a monotone decreasing function of .
• Recall that critical damping occurs when .
mk
F
mk
FR
81
)4(1
2
0
0
2
0
0max
mk21
22
0
2
max
w
g
g w
ww
g 2
mk= 4
Page 120
Resonance
• From the expression
we see that Rmax for small .
• Thus for lightly damped systems, the amplitude R of the forced
response is large for near 0.
• This is true even for relatively small external forces, and the
smaller the the greater the effect.
• This phenomena is known as resonance. Resonance can be
either good or bad, depending on circumstances; for example,
when building bridges or designing seismographs.
mk
F
mk
FR
81
)4(1
2
0
0
2
0
0max
g@F0
gw 0
w w
g
Page 121
Graphical Analysis of Quantities
• To get a better understanding of the quantities we have been
examining, we graph the ratios for several
values of , as shown below.
• Note that the peaks tend to get higher as damping decreases.
• As damping decreases to zero, the values of Rk/F0 become
asymptotic to = 0.
• The graph corresponding to = 0.015625
is included because it appears in the next
example.
Rk
F0
versus w
w0G =g 2
mk
w w
G
Page 122
Analysis of Phase Angle
• Recall that the phase angle given in the forced response
is characterized by the equations
• For near zero, , and they rise and fall
together. Assuming their maxima and minima nearly together.
• For , , so and response
lags behind the excitation.
• For very large , , and the response is out of phase. That
is the response is a minimum when excitation is a maximum.
22222
0
222222
0
2
22
0
)(sin,
)(
)(cos
mm
m
tRtU cos)(
d
w cosd @1 and sind @ 0
w = w0cosd = 0 and sind =1 d =
p
2
w d @ p
Page 123
Example 2:
Forced Vibrations with Damping (1 of 4)
• Consider the initial value problem
• Then 0 = 1, F0/k = 3, and
• The unforced motion of this system was discussed in Ch 3.7,
with the graph of the solution on the next slide, along with the
graph of the ratios Rk/F vs. for different values of .
¢¢u (t)+1
8¢u (t)+ u(t) = 3cosw t, u(0) = 2, ¢u (0) = 0
w G =1/64 = 0.015625
w /w0 w
Page 124
Example 2:
Forced Vibrations with Damping (2 of 4)
• Graphs of the solution, along with the graph of the
ratios Rk/F vs. for . w /w0 w = 0.3
Page 125
Example 2:
Forced Vibrations with Damping (3 of 4)
• Graphs of the solution, along with the graph of the
ratios Rk/F vs. for . w /w0 w = 1
Page 126
Example 2:
Forced Vibrations with Damping (4 of 4)
• Graphs of the solution, along with the graph of the
ratios Rk/F vs. for . w /w0 w = 2
Page 127
Undamped Equation:
General Solution for the Case
• Suppose there is no damping term. Then our equation is
• Assuming , then the method of undetermined
coefficients can be use to show that the general solution is
m ¢¢u (t)+ ku(t) = F0 cos(w t)
u(t) = c1 cos(w0t)+ c2 sin(w0t)+F0
m(w0
2 -w 2 )cos(w t)
g = 0
w ¹ w0
Page 128
Undamped Equation:
Mass Initially at Rest (1 of 3)
• If the mass is initially at rest, then the corresponding initial
value problem is
• Recall that the general solution to the differential equation is
• Using the initial conditions to solve for c1 and c2, we obtain
• Hence
0)0(,0)0(,cos)()( 0 uutFtkutum
0,)(
222
0
01
c
m
Fc
ttm
Ftu 022
0
0 coscos)(
)(
tm
Ftctctu
cos
)(sincos)(
22
0
00201
Page 129
Undamped Equation:
Solution to Initial Value Problem (2 of 3)
• Thus our solution is
• To simplify the solution even further, let
and . . Then .
Using the trigonometric identity
it follows that
and hence
ttm
Ftu 022
0
0 coscos)(
)(
,sinsincoscos)cos( BABABA
BABAt
BABAt
sinsincoscoscos
sinsincoscoscos
0
BAtt sinsin2coscos 0
A =1
2(w 0 +w)t
B =1
2(w 0 -w)t A+ B = w0t and A- B = wt
Page 130
Undamped Equation: Beats (3 of 3)
• Using the results of the previous slide, it follows that
• When
• Thus motion is a rapid oscillation with frequency , but with slowly varying sinusoidal amplitude given by
• This phenomena is called a beat.
• Beats occur with two tuning forks of
nearly equal frequency.
2
sin2 0
22
0
0 t
m
F
2
sin2
sin)(
2)( 00
22
0
0 tt
m
Ftu
w0 -w @ 0, then w 0 +w is much greater than w 0 +w .
So sin(1
2(w0 +w)t) is oscillating more rapidly than sin(
1
2(w0 -w)t).
w 0 +w
2
Page 131
Example 3: Undamped Equation,
Mass Initially at Rest (1 of 2)
• Consider the initial value problem
• Then , and hence the solution is
• The displacement of the spring–mass system oscillates with a
frequency of 0.9, slightly less than natural frequency = 1.
• The amplitude variation has a slow
frequency of 0.1 and period of 20 .
• A half-period of 10 corresponds to
a single cycle of increasing and then
decreasing amplitude.
¢¢u (t)+u(t) = 0.5cos0.8t, u(0) = 0, ¢u (0) = 0
w0 = 1, w = 0.8, and F0 =1
2
w 0
p
p
Page 132
Example 3: Increased Frequency (2 of 2)
• Recall our initial value problem
• If driving frequency is increased to 0.9, then the slow
frequency is halved to 0.05 with half-period doubled to 20 .
• The multiplier 2.77778 is increased to 5.2632, and the fast
frequency only marginally increased, to 0.095.
0)0(,0)0(,8.0cos5.0)()( uuttutu
p
wp
Page 133
Undamped Equation:
General Solution for the Case (1 of 2)
• Recall our equation for the undamped case:
• If forcing frequency equals natural frequency of system, i.e.,
, then nonhomogeneous term is a solution of
homogeneous equation. It can then be shown that
• Thus solution u becomes unbounded.
• Note: Model invalid when u gets
large, since we assume small
oscillations u.
ttm
Ftctctu 0
0
00201 sin
2sincos)(
tFtkutum cos)()( 0
w0 = w
w = w0 F0 cosw t
Page 134
Undamped Equation: Resonance (2 of 2)
• If forcing frequency equals natural frequency of system, i.e.,
, then our solution is
• Motion u remains bounded if damping present. However,
response u to input may be large if damping is
small and , in which case we have resonance.
ttm
Ftctctu 0
0
00201 sin
2sincos)(
w = w0
F0 cosw tw @ w0
Page 135
Example 4
• Solve the initial value problem
And plot the graph of the solution.
The general solution of the differential equation is
And the initial conditions require that . Thus the
solution of the given initial value problem is
¢¢u + u =1
2cost, u(0) = 0, ¢u (0) = 0
u = c1 cost + c2 sint +1
4t sint
021 cc
u =t
4sin t