boyce/diprima/meade 11th nded, ch 3.1: 2 order...

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)

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

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)(

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

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

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

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

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)(

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)(

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)(

Example 4: 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

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)(

Boyce/DiPrima/Meade 11th ed, Ch 3.2: Fundamental Solutions of

Linear Homogeneous Equations


• 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)

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

Theorem 3.2.1 (Existence and Uniqueness)


• 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)

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

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

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

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


• 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


• 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

Theorem 3.2.3


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

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

Theorem 3.2.4 (Fundamental Solutions)


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.

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

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

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

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

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

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

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

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

Theorem 3.2.7 (Abel’s Theorem)


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

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

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

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

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

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

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

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

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

Example 2


• 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

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


Repeated Roots; Reduction of Order


• 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)

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)

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

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)

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

Example 1 (1 of 2)


• 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

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

Example 2 (1 of 2)


• 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

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

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

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

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


• 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

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

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

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

Method of Undetermined Coefficients


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.


)()()( tgytqytpy

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)(

Example 2: Sine g(t), First Attempt (1 of 2)


• 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

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)(

Example 3: Product g(t)


• 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)

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

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

Example 5: First Attempt (1 of 3)


• 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 )(

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)(

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

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

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


Variation of Parameters


& Sons, Inc


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

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)

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)

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

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

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


• 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)

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

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


)2(0)()(

)1()()()(

ytqytpy

tgytqytpy

Boyce/DiPrima/Meade 11th ed, Ch 3.7: Mechanical &

Electrical Vibrations


& 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.

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

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)

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

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)

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

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

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

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

Spring Model:


• 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

Spring Model:


• 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

Spring Model:


• 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

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

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)(

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

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

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/)(

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

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

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


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


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


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

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

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


• 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


• 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

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


Forced Periodic Vibrations


& Sons, Inc.

• We continue the discussion of the last section, and now

consider the presence of a periodic external force:

tFtuktutum cos)()()( 0

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)

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

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

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)

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

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

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

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

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

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

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

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

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

Example 2:

Forced Vibrations with Damping (1 of 4)


• 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

Example 2:


• Graphs of the solution, along with the graph of the

ratios Rk/F vs. for . w /w0 w = 0.3

Example 2:



ratios Rk/F vs. for . w /w0 w = 1

Example 2:



ratios Rk/F vs. for . w /w0 w = 2

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

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

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

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

Example 3: Undamped Equation,

Mass Initially at Rest (1 of 2)


• 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

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

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

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

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

boyce/diprima/meade 11th nded, ch 3.1: 2 order...

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