ch 3.1: second order linear homogeneous …matysh/ma3220/chap3.pdfch 3.1: second order linear...

Ch 3.1: Second Order Linear Homogeneous Equations with Constant Coefficients

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) = 0 for all t, then the equation is called homogeneous. Otherwise the equation is nonhomogeneous.

),,( yytfy ′=′′

ytqytptgy )()()( −′−=′′

)()()()( tGytRytQytP =+′+′′

Homogeneous Equations, Initial ValuesIn Sections 3.6 and 3.7, 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 slope of solution at (t0, y0) is equal to y0'.

0=+′+′′ cyybya

( ) ( ) 0000 , ytyyty ′=′=

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.

0=−′′ yy

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

1)0(,3)0(,0 =′==−′′ yyyy

tt ececty −+= 21)(

1,21)0(3)0(

2121

21 ==⇒⎭⎬⎫

=−=′=+=

ccccyccy

tt eety −+= 2)(

Example 1: Solution Graphs (3 of 3)

Our initial value problem and solution are

Graphs of this solution are given below. The graph on the right suggests that both initial conditions are satisfied.

tt eetyyyyy −+=⇒=′==−′′ 2)(1)0(,3)0(,0

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.

,0=+′+′′ cyybya

02 =++ rtrtrt cebreear

0)( 2 =++ cbrarert

02 =++ cbrar

General SolutionUsing 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

trtr ececty 2121)( +=

aacbbr

242 −±−

=

Initial ConditionsFor 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.

,)(,)(,0 0000 ytyytycyybya ′=′==+′+′′

0201

0201

0201

21

0102

21

2001

02211

021 , trtrtrtr

trtr

erryryce

rrryyc

yercerc

yecec −−

−′−

=−−′

=⇒⎪⎭

⎪⎬⎫

′=+

=+

trtr ececty 2121)( +=

Example 2Consider the initial value problem

Assuming exponential soln leads to characteristic equation:

Factoring yields two solutions, r1 = -4 and r2 = 3The general solution has the form

Using the initial conditions:

Thus

1)0(,0)0(,012 =′==−′+′′ yyyyy

( )( ) 034012)( 2 =−+⇔=−+⇒= rrrrety rt

tt ececty 32

41)( += −

71,

71

1340

2121

21 =−

=⇒⎭⎬⎫

=+−=+

cccccc

tt eety 34

71

71)( +

−= −


Then

Factoring yields two solutions, r1 = 0 and r2 = -3/2The general solution has the form


Thus

( ) ( ) 30,10,032 =′==′+′′ yyyy

( ) 032032)( 2 =+⇔=+⇒= rrrrety rt

2/321

2/32

01)( ttt eccececty −− +=+=

2,33

23

1212

21

−==⇒⎪⎭

⎪⎬⎫

=−

=+ccc

cc

2/323)( tety −−=

Example 4: Initial Value Problem (1 of 2)

Consider the initial value problem

Then

Factoring yields two solutions, r1 = -2 and r2 = -3The general solution has the form

Using initial conditions:

Thus

( ) ( ) 30,20,065 =′==+′+′′ yyyyy

( )( ) 032065)( 2 =++⇔=++⇒= rrrrety rt

tt ececty 32

21)( −− +=

7,93322

2121

21 −==⇒⎭⎬⎫

=−−=+

cccccc

tt eety 32 79)( −− −=

Example 4: Find Maximum Value (2 of 2)

Find the maximum value attained by the solution.

204.21542.0

)6/7ln(6/7

7602118)(

79)(

32

32

32

≈≈==

=

=+−=′

−=

−−

−−

−−

ytt

eee

eety

eety

t

tt

settt

tt

Ch 3.2: Fundamental Solutions of Linear Homogeneous EquationsLet p, q be continuous functions on an interval I = (α, β), 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,

[ ] yqypyyL +′+′′=

[ ] )()()()()()( tytqtytptytyL +′+′′=

( )[ ] )sin(2)cos()sin()(

2,0),sin()(,)(,)(22

22

tettttyLIttyetqttp

t

t

++−=

==== π

Differential Operator NotationIn 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.

[ ]1000 )(,)(

0)()(ytyyty

ytqytpyyL=′=

=+′+′′=

Theorem 3.2.1Consider 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 y = φ(t) 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.

0000 )(,)()()()(

ytyytytgytqytpy

′=′==+′+′′

Example 1Consider the second order linear initial value problem

In Section 3.1, we showed that this initial value problem had the following solution:

Note that p(t) = 0, q(t) = -1, g(t) = 0 are each continuous on (-∞, ∞), and the solution y is defined and twice differentiable on (-∞, ∞).

tt eety −+= 2)(

( ) ( ) 10,30,0 =′==−′′ yyyy

1000 )(,)()()()(

ytyytytgytqytpy

=′==+′+′′

Example 2Consider 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.

( ) ( ) 00,00,0)()( =′==+′+′′ yyytqytpy

Example 3Determine the longest interval on which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution.

First put differential equation into standard form:

The longest interval containing the point t = 0 on which the coefficient functions are continuous is (-1, ∞).It follows from Theorem 3.2.1 that the longest interval on which this initial value problem is certain to have a twice differentiable solution is also (-1, ∞).

( ) ( ) ( ) 00,10,13)(cos1 =′==+′−′′+ yyyytyt

( ) ( ) 00,10,1

11

31

cos=′=

+=

++′

+−′′ yy

ty

ty

tty

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.

0)()(][ =+′+′′= ytqytpyyL

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:

0)()(][ =+′+′′= ytqytpyyL

0000 )(,)( ytyyty ′=′=

0022011

0022011

)()()()(

ytyctycytyctyc′=′+′

=+


Solving the equations, we obtain

In terms of determinants:

0022011

0022011

)()()()(

ytyctycytyctyc′=′+′

=+

)()()()()()(

)()()()()()(

02010201

0100102

02010201

0200201

tytytytytyytyyc

tytytytytyytyyc

′−′′+′−

=

′−′′−′

=

)()()()(

)()(

,

)()()()(

)()(

0201

0201

001

001

2

0201

0201

020

020

1

tytytyty

ytyyty

c

tytytyty

tyytyy

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

)()()()()()()()(

020102010201

0201 tytytytytytytyty

W ′−′=′′

=

Wytyyty

cW

tyytyy

c 001

001

2020

020

1

)()(

,)()(

′′=

′′=

( )( )021, tyyW

Theorem 3.2.3Suppose y1 and y2 are solutions to the equation

and that the Wronskian

is not zero at the point t0 where the initial conditions

are assigned. Then there is a choice of constants c1, c2 for which y = c1y1 + c2 y2 is a solution to the differential equation (1) and initial conditions (2).

)1(0)()(][ =+′+′′= ytqytpyyL

2121 yyyyW ′−′=

)2()(,)( 0000 ytyyty ′=′=

Example 4Recall the following initial value problem and its solution:

Note that the two functions below are solutions to the differential equation:

The Wronskian of y1 and y2 is

Since W ≠ 0 for all t, linear combinations of y1 and y2 can be used to construct solutions of the IVP for any initial value t0.

tt eyey −== 21 ,

22 02121

21

21 −=−=−−=′−′=′′

= −− eeeeeyyyyyyyy

W tttt

( ) ( ) tt eetyyyyy −+=⇒=′==−′′ 2)(10,30,0

2211 ycycy +=

Theorem 3.2.4 (Fundamental Solutions)Suppose y1 and y2 are solutions to the equation

If there is a point t0 such that W(y1,y2)(t0) ≠ 0, then the family of solutions y = c1y1 + c2 y2 with arbitrary coefficients c1, c2includes every solution to the differential equation.

The expression y = c1y1 + c2 y2 is called the general solutionof the differential equation above, and in this case y1 and y2are said to form a fundamental set of solutions to the differential equation.

.0)()(][ =+′+′′= ytqytpyyL

Example 5Recall the equation below, with the two solutions indicated:


Thus y1 and y2 form a fundamental set of solutions to the differential equation above, and can be used to construct all of its solutions. The general solution is

tt eyeyyy −===−′′ 21 ,,0

. allfor 022 0

21

21 teeeeeyyyy

W tttt ≠−=−=−−=′′

= −−

tt ececy −+= 21

Example 6Consider 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 rreyey trtr ≠==

( ) ( ) . allfor 021

21

21

122121

21 terrerer

eeyyyy

W trrtrtr

trtr

≠−==′′

= +

0)()( =+′+′′ ytqytpy

trtr ececy 2121 +=

Example 7: 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:

12

2/11 , −== tyty

0,032 2 >=−′+′′ tyytyt

0123

21

23

42 2/12/1

2/12/32 =⎟

⎠⎞

⎜⎝⎛ −+−=−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ − −−

tttttt

( ) ( ) ( ) 0134322 11232 =−−=−−+ −−−− tttttt

Example 7: 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, y2 form a fundamental set of solutions for the differential equation

3

2/32/32/322/1

12/1

21

21

23

23

21

21

tttttt

tt

yyyy

W −=−=−−=−=′′

= −−−−−

−

12

2/11 , −== tyty

0,032 2 >=−′+′′ tyytyt

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, y2 form a fundamental set of solutions to the given differential equation.

0)()(][ =+′+′′= ytqytpyyL

0)(,1)( 0101 =′= tyty

1)(,0)( 0202 =′= tyty

Example 7: Theorem 3.2.5 (1 of 3)

Find the fundamental set specified by Theorem 3.2.5 for the differential equation and initial point

We showed previously that

were fundamental solutions, since W(y1, y2)(t0) = -2 ≠ 0.But these two solutions don’t 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 ,

0,0 0 ==−′′ tyy

1)0(,0)0(;0)0(,1)0( 4433 =′==′= yyyy

Example 7: General Solution (2 of 3)

Since y1 and y2 form a fundamental set of solutions,

Solving each equation, we obtain


Thus y3, y4 forms 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

yyececytt

tt

)sinh(21

21)(),cosh(

21

21)( 43 teetyteety tttt =−==+= −−

01sinhcoshcoshsinhsinhcosh 22

21

21 ≠=−==′′

= tttttt

yyyy

W

)sinh()cosh()( 21 tktkty +=

Example 7: 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 == −

0,0 0 ==−′′ tyy

SummaryTo 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.

βα <<=+′+′′ tytqytpy ,0)()(

Ch 3.3: Linear Independence and the Wronskian

Two functions f and g are linearly dependent if there exist constants c1 and c2, not both zero, such that

for all t in I. Note that this reduces to determining whether f and g are multiples of each other. If the only solution to this equation is c1 = c2 = 0, then f and g are linearly independent. For example, let f(x) = sin2x and g(x) = sinxcosx, and consider the linear combination

This equation is satisfied if we choose c1 = 1, c2 = -2, and hence f and g are linearly dependent.

0)()( 21 =+ tgctfc

0cossin2sin 21 =+ xxcxc

Solutions of 2 x 2 Systems of EquationsWhen solving

for c1 and c2, it can be shown that

Note that if a = b = 0, then the only solution to this system of equations is c1 = c2 = 0, provided D ≠ 0.

bycycaxcxc

=+=+

2211

2211

21

2111

2121

112

22

2121

221

where,

,

yyxx

DD

bxayxyyx

bxayc

Dbxay

xyyxbxayc

=+−

=−+−

=

−=

−−

=

Example 1: Linear Independence (1 of 2)

Show that the following two functions are linearly independent on any interval:

Let c1 and c2 be scalars, and suppose

for all t in an arbitrary interval (α, β). We want to show c1 = c2 = 0. Since the equation holds for all t in (α, β), choose t0 and t1 in (α, β), where t0 ≠ t1. Then

tt etgetf −== )(,)(

0)()( 21 =+ tgctfc

0

011

00

21

21

=+

=+−

−

tt

tt

ecec

ecec

Example 1: Linear Independence (2 of 2)

The solution to our system of equations

will be c1 = c2 = 0, provided the determinant D is nonzero:

Then

Since t0 ≠ t1, it follows that D ≠ 0, and therefore f and g are linearly independent.

01101010

11

00tttttttt

tt

tt

eeeeeeeeee

D −−−−−

−

−=−==

( )10

2

1

11 0

10

10

10

100110

tte

ee

eeeD

tt

tttt

tttttt

=⇔=⇔

=⇔=⇔=⇔=

−

−−

−−−

0

011

00

21

21

=+

=+−

−

tt

tt

ecec

ecec

Theorem 3.3.1If f and g are differentiable functions on an open interval Iand if W(f, g)(t0) ≠ 0 for some point t0 in I, then f and g are linearly independent on I. Moreover, if f and g are linearly dependent on I, then W(f, g)(t) = 0 for all t in I.

Proof (outline): Let c1 and c2 be scalars, and suppose

for all t in I. In particular, when t = t0 we have

Since W(f, g)(t0) ≠ 0, it follows that c1 = c2 = 0, and hence f and g are linearly independent.

0)()( 21 =+ tgctfc

0)()(0)()(

0201

0201

=′+′=+

tgctfctgctfc

Theorem 3.3.2 (Abel’s Theorem)Suppose y1 and y2 are solutions to the equation

where p and q are continuous on some open interval I. Then 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).

0)()(][ =+′+′′= ytqytpyyL

∫=− dttp

cetyyW)(

21 ))(,(

Example 2: Wronskian and Abel’s TheoremRecall the following equation and two of its solutions:

The Wronskian of y1and y2 is

Thus y1 and y2 are linearly independent on any interval I, by Theorem 3.3.1. Now compare W with Abel’s Theorem:

Choosing c = -2, we get the same W as above.

tt eyeyyy −===−′′ 21 ,,0

. allfor 022 0

21

21 teeeeeyyyy

W tttt ≠−=−=−−=′′

= −−

ccecetyyWdtdttp=∫=∫=

−− 0)(21 ))(,(

Theorem 3.3.3Suppose y1 and y2 are solutions to equation below, whose coefficients p and q are continuous on some open interval I:

Then y1 and y2 are linearly dependent on I iff W(y1, y2)(t) = 0 for all t in I. Also, y1 and y2 are linearly independent on I iff W(y1, y2)(t) ≠ 0 for all t in I.

0)()(][ =+′+′′= ytqytpyyL

SummaryLet y1 and y2 be solutions of

where p and q are continuous on an open interval I. Then the following statements are equivalent:

The functions y1 and y2 form a fundamental set of solutions on I.The functions y1 and y2 are linearly independent on I.W(y1,y2)(t0) ≠ 0 for some t0 in I.W(y1,y2)(t) ≠ 0 for all t in I.

0)()( =+′+′′ ytqytpy

Linear Algebra NoteLet V be the set

Then V is a vector space of dimension two, whose bases are given by any fundamental set of solutions y1 and y2. For example, the solution space V to the differential equation

has bases

with { } { }ttSeeS tt sinh,cosh,, 21 == −

( ){ }βα ,,0)()(: ∈=+′+′′= tytqytpyyV

0=−′′ yy

21 SpanSpan SSV ==

Ch 3.4: Complex Roots of Characteristic EquationRecall 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 & r2:

If b2 – 4ac < 0, then complex roots: r1 = λ + iμ, r2 = λ - iμThus


0)( 2 =++⇒= cbrarety rt

aacbbr

242 −±−

=

( ) ( )titi etyety μλμλ −+ == )(,)( 21

Euler’s Formula; Complex Valued SolutionsSubstituting it into Taylor series for et, we obtain Euler’s formula:

Generalizing Euler’s formula, we obtain

Then

Therefore

( ) ( ) titn

tin

tnite

n

nn

n

nn

n

nit sincos

!12)1(

!2)1(

!)(

1

121

0

2

0+=

−−

+−

== ∑∑∑∞

=

−−∞

=

∞

=

tite ti μμμ sincos +=

( ) [ ] tietetiteeee ttttitti μμμμ λλλμλμλ sincossincos +=+==+

( )

( ) tieteety

tieteetyttti

ttti

μμ

μμλλμλ

λλμλ

sincos)(

sincos)(

2

1

−==

+==−

+

Real Valued SolutionsOur 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

tietetytt

tt

μμ

μμλλ

λλ

sincos)(

sincos)(

2

1

−=

+=

tietyty

tetytyt

t

μ

μλ

λ

sin2)()(

cos2)()(

21

21

=−

=+

tetytety tt μμ λλ sin)(,cos)( 43 ==

Real Valued Solutions: The WronskianThus 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 μμ λλ sin)(,cos)( 43 ==

( ) ( )0

cossinsincossincos

2 ≠=

+−=

t

tt

tt

e

ttettetete

W

λ

λλ

λλ

μ

μμμλμμμλμμ

tectecty tt μμ λλ sincos)( 21 +=

Example 1Consider the equation

Then

Therefore

and thus the general solution is

( ) ( )2/3sin2/3cos)( 2/2

2/1 tectecty tt −− +=

0=+′+′′ yyy

iirrrety rt

23

21

231

241101)( 2 ±−=

±−=

−±−=⇔=++⇒=

2/3,2/1 =−= μλ


Then

Therefore

and thus the general solution is

04 =+′′ yy

irrety rt 204)( 2 ±=⇔=+⇒=

2,0 == μλ

( ) ( )tctcty 2sin2cos)( 21 +=


Then

Therefore the general solution is

023 =+′−′′ yyy

irrrety rt

32

31

612420123)( 2 ±=

−±=⇔=+−⇒=

( ) ( )3/2sin3/2cos)( 3/2

3/1 tectecty tt +=

Example 4: Part (a) (1 of 2)

For the initial value problem below, find (a) the solution u(t) and (b) the smallest time T for which |u(t)| ≤ 0.1

We know from Example 1 that the general solution is

Using the initial conditions, we obtain

Thus

( ) ( )2/3sin2/3cos)( 2/2

2/1 tectectu tt −− +=

1)0(,1)0(,0 =′==+′+′′ yyyyy

33

3,11

23

21

1

2121

1

===⇒⎪⎭

⎪⎬

⎫

=+−

=cc

cc

c

( ) ( )2/3sin32/3cos)( 2/2/ tetetu tt −− +=

Example 4: Part (b) (2 of 2)

Find the smallest time T for which |u(t)| ≤ 0.1 Our solution is

With the help of graphing calculator or computer algebra system, we find that T ≅ 2.79. See graph below.

( ) ( )2/3sin32/3cos)( 2/2/ tetetu tt −− +=

Ch 3.5: Repeated Roots; Reduction of OrderRecall our 2nd order linear homogeneous ODE

where a, b and c are constants. Assuming an exponential soln leads to characteristic equation:

Quadratic formula (or factoring) yields two solutions, r1 & r2:

When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives one solution:


0)( 2 =++⇒= cbrarety rt

aacbbr

242 −±−

=

atbcety 2/1 )( −=

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 =⇒

atbatb etvtyety 2/2

2/1 )()( try solution a )( −− =⇒=

atbatbatbatb

atbatb

atb

etvabetv

abetv

abetvty

etvabetvty

etvty

2/2

22/2/2/

2

2/2/2

2/2

)(4

)(2

)(2

)()(

)(2

)()(

)()(

−−−−

−−

−

+′−′−′′=′′

−′=′

=

Finding Multiplying Factor v(t)Substituting derivatives into ODE, we seek a formula for v:


43

2

222

22

22

2

22/

)(0)(

0)(4

4)(

0)(44

4)(0)(

44

42

4)(

0)(24

)(

0)()(2

)()(4

)()(

0)()(2

)()(4

)()(

ktktvtv

tva

acbtva

tvaac

abtvatv

aac

ab

abtva

tvca

ba

btva

tcvtva

btvbtva

btvbtva

tcvtvabtvbtv

abtv

abtvae atb

+=⇒=′′

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −−′′

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+′′⇔=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−+′′

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−+′′

=+−′++′−′′

=⎭⎬⎫

⎩⎨⎧

+⎥⎦⎤

⎢⎣⎡ −′+⎥

⎦

⎤⎢⎣

⎡+′−′′−

General SolutionTo find our general solution, we have:

Thus the general solution for repeated roots is

( )abtabt

abtabt

abtabt

tecec

ektkek

etvkekty

2/2

2/1

2/43

2/1

2/2

2/1 )()(

−−

−−

−−

+=

++=

+=

abtabt tececty 2/2

2/1)( −− +=

WronskianThe 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.

abtabt tececty 2/2

2/1)( −− +=

abtabt tetyety 2/2

2/1 )(,)( −− ==

tea

btea

bte

ea

bteab

teetyyW

abt

abtabt

abtabt

abtabt

allfor 022

1

21

2))(,(

/

//

2/2/

2/2/

21

≠=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −−=

−

−−

−−

−−



Thus the general solution is


Thus

( ) ( ) 10,10,02 =′==+′+′′ yyyyy

10)1(012)( 22 −=⇔=+⇔=++⇒= rrrrety rt

tt tececty −− += 21)(

2,111

2121

1 ==⇒⎭⎬⎫

=+−=

cccc

c

tt teety −− += 2)(





Thus

( ) ( ) 2/10,20,025.0 =′==+′−′′ yyyyy

2/10)2/1(025.0)( 22 =⇔=−⇔=+−⇒= rrrrety rt

2/2

2/1)( tt tececty +=

21,2

21

21

221

21

1−==⇒

⎪⎭

⎪⎬⎫

=+

=cccc

c

2/2/

212)( tt teety −=





Thus

( ) ( ) 2/30,20,025.0 =′==+′−′′ yyyyy

2/10)2/1(025.0)( 22 =⇔=−⇔=+−⇒= rrrrety rt

2/2

2/1)( tt tececty +=

21,2

23

21

221

21

1==⇒

⎪⎭

⎪⎬⎫

=+

=cccc

c

2/2/

212)( tt teety +=

Reduction of OrderThe 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

tytvtytvtytvtytytvtytvty

tytvty

′′+′′+′′=′′′+′=′

=

( ) ( ) 02 111111 =+′+′′+′+′+′′ vqyypyvpyyvy

( ) 02 111 =′+′+′′ vpyyvy

Example 4: 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 ODE and collecting terms,

,)(;0,03 11

2 −=>=+′+′′ ttytyytyt

3212

212

12

)(2 )(2 )()(

)( )()(

)()(

−−−

−−

−

+′−′′=′′

−′=′

=

ttvttvttvty

ttvttvty

ttvty

( ) ( )

)()( where,00

033220322

111

1213212

tvtuuutvvt

vtvtvvtvtvvtvttvtvttvtvt

′==+′⇔=′+′′⇔

=+−′++′−′′⇔

=+−′++′−′′−−−

−−−−−−

Example 4: Finding v(t) (2 of 3)

To solve

for u, we can use the separation of variables method:

Thus

and hence

.0 since,

lnln10

11 >=⇔=⇔

+−=⇔−=⇔=+

−−

∫∫tctuetu

Ctudttu

duudtdut

C

)()(,0 tvtuuut ′==+′

tcv =′

ktctv += ln)(

Example 4: General Solution (3 of 3)

We have

Thus

Recall

and hence we can neglect the second term of y2 to obtain

Hence the general solution to the differential equation is

( ) 1112 lnln)( −−− +=+= tktcttktcty

ktctv += ln)(

.ln)( 12 ttty −=

11 )( −= tty

ttctcty ln)( 12

11

−− +=

Ch 3.6: Nonhomogeneous Equations; Method of Undetermined Coefficients

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 homogeneous equation.

)()()( tgytqytpy =+′+′′

0)()( =+′+′′ ytqytpy

Theorem 3.6.1

If Y1, Y2 are solutions of nonhomogeneous equation

then Y1 - Y2 is a solution of the homogeneous equation

If y1, y2 form a fundamental solution set of homogeneous equation, then there exists constants c1, c2 such that

)()()()( 221121 tyctyctYtY +=−

)()()( tgytqytpy =+′+′′

0)()( =+′+′′ ytqytpy

Theorem 3.6.2 (General Solution)

The general solution of nonhomogeneous equation

can be written in the form

where y1, y2 form a fundamental solution set of homogeneous equation, c1, c2 are arbitrary constants and Y is a specific solution to the nonhomogeneous equation.

)()()()( 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 =+′+′′

)()()()( 2211 tYtyctycty ++=

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 differential equation,

Thus a particular solution to the nonhomogeneous ODE is

teyyy 2343 =−′−′′

ttt AetYAetYAetY 222 4)(,2)()( =′′=′⇒=

2/1363464

22

2222

−=⇔=−⇔

=−−

AeAeeAeAeAe

tt

tttt

tetY 2

21)( −=

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:


Since sin(x) and cos(x) are linearly independent (they 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

0cos3sin52sin2sin4cos3sin

21 =+⇔=++⇔=−−−

tctctAtA

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


tBtAtYtBtAtYtBtAtY

cossin)(,sincos)(cossin)(

−−=′′−=′⇒+=

( ) ( ) ( )( ) ( )

17/3 ,17/5053,235

sin2cos53sin35sin2cossin4sincos3cossin

=−=⇔=−−=+−⇔

=−−++−⇔=+−−−−−

BABABA

ttBAtBAttBtAtBtAtBtA

tyyy sin243 =−′−′′

tttY cos173sin

175)( +

−=

Example 3: Polynomial g(t)Consider the nonhomogeneous equation

We seek Y satisfying this equation. We begin with



1443 2 −=−′−′′ tyyy

AtYBAttYCBtAttY 2)(,2)()( 2 =′′+=′⇒++=

( ) ( )( ) ( )

8/11 ,2/3 ,11432,046,44

14432464144232

22

22

−==−=⇔−=−−=+=−⇔

−=−−++−−⇔

−=++−+−

CBACBABAA

tCBAtBAAttCBtAtBAtA

811

23)( 2 −+−= tttY

Example 4: Product g(t)Consider the nonhomogeneous equation

We seek Y satisfying this equation, as follows:

Substituting derivatives into ODE and solving for A and B:

teyyy t 2cos843 −=−′−′′

( ) ( )( ) ( ) ( )

( )( ) ( ) teBAteBA

teBAteBAteBAteBAtY

teBAteBAtBetBetAetAetY

tBetAetY

tt

t

ttt

tt

tttt

tt

2sin342cos432cos22

2sin22sin222cos2)(2sin22cos2

2cos22sin2sin22cos)(2sin2cos)(

−−++−=

+−+

+−++−+=′′

+−++=

++−=′

+=

tetetYBA tt 2sin1322cos

1310)(

132 ,

1310

+=⇒==

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

tgytqytpytgytqytpy

=+′+′′=+′+′′

Example 5: Sum g(t)Consider the equation

Our equations to solve individually are

Our particular solution is then

teteyyy tt 2cos8sin2343 2 −+=−′−′′

tetettetY ttt 2sin1322cos

1310sin

175cos

173

21)( 2 ++−+−=

teyyytyyy

eyyy

t

t

2cos843sin243

343 2

−=−′−′′

=−′−′′=−′−′′

Example 6: First Attempt (1 of 3)

Consider the equation

We seek Y satisfying this equation. We begin with

Substituting these derivatives into ODE:

Thus no particular solution exists of the form

tyy 2cos34 =+′′

tBtAtYtBtAtYtBtAtY

2cos42sin4)(,2sin22cos2)(2cos2sin)(

−−=′′−=′⇒+=

( ) ( )( ) ( )

tttBBtAAttBtAtBtA

2cos302cos32cos442sin442cos32cos2sin42cos42sin4

==+−++−=++−−

tBtAtY 2cos2sin)( +=

Example 6: Homogeneous Solution (2 of 3)

Thus no particular solution exists of the form

To help understand why, recall that we found the corresponding homogeneous solution in Section 3.4 notes:

Thus our assumed particular solution solves homogeneous equation

instead of the nonhomogeneous equation.

tBtAtY 2cos2sin)( +=

tctctyyy 2sin2cos)(04 21 +=⇒=+′′

tyy 2cos34 =+′′

04 =+′′ yy

Example 6: Particular Solution (3 of 3)

Our next attempt at finding a Y is:

Substituting derivatives into ODE,

tBttAttBtAtBttBtBtAttAtAtY

tBttBtAttAtYtBttAttY

2cos42sin42sin42cos42cos42sin22sin22sin42cos22cos2)(

2sin22cos2cos22sin)(2cos2sin)(

−−−=−−−−+=′′

−++=′+=

tttY

BAttBtA

2sin43)(

0,4/32cos32sin42cos4

=⇒

==⇒=−

tyy 2cos34 =+′′

Ch 3.7: Variation of Parameters


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 parametersmethod to solve the nonhomogeneous equation. As with the method of undetermined coefficients, this procedure relies on knowing solutions to 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.

)()()( tgytqytpy =+′+′′

0)()( =+′+′′ ytqytpy

Example: Variation of Parameters (1 of 6)

We seek a particular solution to the equation below.

We cannot use method of undetermined coefficients since g(t) is a quotient of sint 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

tyy csc34 =+′′

tctctyC 2sin2cos)( 21 +=

ttuttuty 2sin)(2cos)()( 21 +=

ttuttuttuttuty 2cos)(22sin)(2sin)(22cos)()( 2211 +′+−′=′

ttuttuttuttuty 2sin)(2cos)(2cos)(22sin)(2)( 2121 ′+′++−=′

Example: 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,

ttuttuttuttuty 2sin)(2cos)(2cos)(22sin)(2)( 2121 ′+′++−=′

02sin)(2cos)( 21 =′+′ ttuttu

ttuttuty 2cos)(22sin)(2)( 21 +−=′

ttuttuttuttuty 2sin)(42cos)(22cos)(42sin)(2)( 2211 −′+−′−=′′

Example: 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:

( ) tttuttuttuttuttuttu

csc32sin)(2cos)(42sin)(42cos)(22cos)(42sin)(2

21

2211

=++−′+−′−

02sin)(2cos)(csc32cos)(22sin)(2

21

21

=′+′=′+′−

ttuttutttuttu

tttuttu csc32cos)(22sin)(2 21 =′+′−

tyy csc34 =+′′

Example: Solve for u1' (4 of 6)

To find u1 and u2 , we need to solve the equations

From second equation,

Substituting this into the first equation,tttutu

2sin2cos)()( 12 ′−=′

( ) ( )

( ) ( )[ ]ttu

ttttttu

ttttuttu

tttttuttu

cos3)(sin

cossin232cos2sin)(2

2sincsc32cos)(22sin)(2

csc32cos2sin2cos)(22sin)(2

1

221

21

21

11

−=′

⎥⎦⎤

⎢⎣⎡=+′−

=′−′−

=⎥⎦⎤

⎢⎣⎡ ′−+′−

02sin)(2cos)(csc32cos)(22sin)(2

21

21

=′+′=′+′−

ttuttutttuttu

Example : Solve for u1 and u2 (5 of 6)

From the previous slide,

Then

Thus

tttt

t

tt

tttt

ttttu

sin3csc23

sin2sin2

sin213

sin2sin213

cossin2sin21cos3

2sin2coscos3)(

2

22

2

−=⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡ −=⎥

⎦

⎤⎢⎣

⎡ −=⎥⎦

⎤⎢⎣⎡=′

222

111

cos3cotcscln23sin3csc

23)()(

sin3cos3)()(

ctttdtttdttutu

cttdtdttutu

++−=⎟⎠⎞

⎜⎝⎛ −=′=

+−=−=′=

∫∫

∫∫

tttututtu

2sin2cos)()(,cos3)( 121 ′−=′−=′

Example: 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

[ ]

( )[ ]

tctctttt

tyttttttt

tyttttttt

tyttttttt

tyttuttuty

C

C

C

C

2sin2cos2sincotcscln23sin3

)(2sincotcscln231cos2sincossin23

)(2sincotcscln232cossin2sincos3

)(2sincos32sincotcscln232cossin3

)(2sin)(2cos)()(

21

22

21

++−+=

+−+−−=

+−+−=

++−+−=

++=

tctctytyy C 2sin2cos)(,csc34 21 +==+′′

SummarySuppose 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

tgtytutytutytutytu=′′+′′=′+′

)()()()()()()()(

2211 tytutytutytgytqytpy

+==+′+′′

( ) ( ) )(,)()()(,

)(,)()()(

21

12

21

21 tyyW

tgtytutyyW

tgtytu =′−=′

( ) ( )∫∫ +=+−= 221

121

21

21 )(,

)()()(,)(,

)()()( cdttyyW

tgtytucdttyyW

tgtytu

Theorem 3.7.1Consider 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

tgtytydttyyW

tgtytytY)(,

)()()()(,

)()()()(21

12

21

21

)()()()( 2211 tYtyctycty ++=

)2(0)()()1()()()(

=+′+′′=+′+′′

ytqytpytgytqytpy

Ch 3.8: Mechanical & Electrical Vibrations

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 SystemSuppose a mass m hangs from vertical spring of original length l. The mass causes an elongation L of the spring. The force FG of gravity pulls mass down. This force has magnitude mg, where g is acceleration due to gravity. The force FS of spring stiffness pulls mass up. For small elongations L, this force is proportional to L. That is, Fs = kL (Hooke’s Law). Since mass is in equilibrium, the forces balance each other:

kLmg =

Spring Model

We will study motion of mass when it is acted on by an external force (forcing function) 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 mass. 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: Fd(t) = - γ u′ (t) (up or down, see following slide)External force: F (t) (up or down force, see text)

)()( tftum =′′

Spring Model: Spring Force Details

The spring force Fs acts to restore spring to natural position, and is proportional to L + u. If L + u > 0, then 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 DetailsThe damping or resistive force Fd acts in opposite direction as motion of mass. Can be complicated to model.Fd may be due to air resistance, internal energy dissipation due to action of spring, friction between mass and guides, or a mechanical device (dashpot) imparting resistive force to mass. We keep it simple and assume Fd is proportional to velocity. In particular, we find that

If u′ > 0, then u is increasing, so mass is moving downward. Thus Fd acts upward and hence Fd = - γ u′, where γ > 0.If u′ < 0, then u is decreasing, so mass is moving upward. Thus Fd acts downward and hence Fd = - γ u′ , γ > 0.

In either case,0),()( >′−= γγ tutFd

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, γ, 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 mass is set in motion with a given initial displacement and velocity, then its position is uniquely determined at all future times.

[ ] )()()()()()()(

tFtutuLkmgtFtFtFmgtum ds

+′−+−=+++=′′

γ

)()()()( tFtkututum =+′+′′ γ

00 )0(,)0( vuuu =′=

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 velocity of mass is 3 ft/sec. Formulate the IVP that governs motion of this mass:

Find m:

Find γ :

Find k:

ftseclb

81

sec/ft32lb4 2

2 =⇒=⇒=⇒= mmgwmmgw

ftseclb2

sec/ft3lb6lb6 =⇒=⇒=′ γγγ u

ftlb24

ft6/1lb4

in2lb4

=⇒=⇒=⇒−= kkkLkFs

00 )0(,)0(),()()()( vuuutFtkututum =′==+′+′′ γ

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 methods of Chapter 3.4. Given on right is the graph of solution.

0)(24)(2)(81

=+′+′′ tututu

0)0(,21)0(

0)(192)(16)(

=′=

=+′+′′

uu

tututu

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 γ = 0, and our equation becomes

The general solution to this equation is0)()( =+′′ tkutum

)()()()( tFtkututum =+′+′′ γ

mk

tBtAtu

/where

,sincos)(

20

00

=

+=

ω

ωω


Using trigonometric identities, the solution

can be rewritten as follows:

where

Note that in finding δ, we must be careful to choose correct quadrant. This is done using the signs of cosδ and sinδ.

mktBtAtu /,sincos)( 2000 =+= ωωω

( ),sinsincoscos)(cos)(sincos)(

00

000

tRtRtutRtutBtAtu

ωδωδδωωω

+=⇔−=⇔+=

ABBARRBRA =+=⇒== δδδ tan,sin,cos 22


Thus our solution is

where

The solution is a shifted cosine (or sine) curve, that describes simple harmonic motion, with period

The circular frequency ω0 (radians/time) is natural frequencyof the vibration, R is the amplitude of max displacement of mass from equilibrium, and δ is the phase (dimensionless).

( )δωωω −=+= tRtBtAtu 000 cos sincos)(

kmT π

ωπ 22

0

==

mk /0 =ω


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 same frequency ω0 , regardless of 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 system to vibrate more rapidly.

( ) mktRtBtAtu /,cos sincos)( 0000 =−=+= ωδωωω

kmT π2=

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 initial upward velocity of 1 ft/sec. Determine position of mass at any later time. Also find period, amplitude, and phase of the motion.

Find m:

Find k:

Thus our IVP is

ftseclb

165

sec/ft32lb10 2

2 =⇒=⇒=⇒= mmgwmmgw

ftlb60

ft6/1lb10

in2lb10

=⇒=⇒=⇒−= kkkLkFs

1)(,6/1)0(,0)(60)(16/5 −=′==+′′ tuututu

00 )0(,)0(,0)()( vuuutkutum =′==+′′

Example 2: Find Solution (2 of 3)

Simplifying, we obtain

To solve, use methods of Ch 3.4 to obtain

or

1)0(,6/1)0(,0)(192)( −=′==+′′ uututu

tttu 192sin1921192cos

61)( −=

tttu 38sin38

138cos61)( −=

Example 2: Find Period, Amplitude, Phase (3 of 3)

The natural frequency is

The period is

The amplitude is

Next, determine the phase δ :

tttu 38sin38

138cos61)( −=

rad/sec 856.1338192/0 ≅=== mkω

sec 45345.0/2 0 ≅= ωπT

ft 18162.022 ≅+= BAR

rad 40864.04

3tan4

3tantan 1 −≅⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⇒

−=⇒= −δδδ

AB

ABRBRA /tan,sin,cos === δδδ

( )409.038cos182.0)( Thus += ttu

Spring Model: Damped Free Vibrations (1 of 8)

Suppose there is damping but no external driving force F(t):

What is effect of damping coefficient γ on system? The characteristic equation is

Three cases for the solution:

0)()()( =+′+′′ tkututum γ

⎥⎦

⎤⎢⎣

⎡−±−=

−±−= 2

2

21411

224

,γ

γγγ mkmm

mkrr

( )

( )

term.damping from expected as ,0)(limcases, threeallIn :Note

.02

4,sincos)(:04

;02/where,)(:04

;0,0where,)(:04

22/2

2/221

2 21

=

>−

=+=<−

>+==−

<<+=>−

∞→

−

−

tu m

mktBtAetumk

meBtAtumk

rrBeAetumk

t

mt

mt

trtr

γμμμγ

γγ

γ

γ

γ

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

2/221

2 21

>+=<−

>+==−

<<+=>−

−

−

μμμγ

γγ

γ

γ

γ

tBtAetumkmeBtAtumk

rrBeAetumk

mt

mt

trtr

δδ sin,cos RBRA ==

( )δμγ −= − teRtu mt cos)( 2/

mteRtu 2/)( γ−≤

Damped Free Vibrations: Quasi Frequency (3 of 8)

Thus we have damped oscillations:

Amplitude R depends on the initial conditions, since

Although the motion is not periodic, the parameter μdetermines mass oscillation frequency.Thus μ is called the quasi frequency.Recall

( ) δδμμγ sin,cos,sincos)( 2/ RBRAtBtAetu mt ==+= −

( ) mtmt eRtuteRtu 2/2/ )(cos)( γγ δμ −− ≤⇒−=

mmk2

4 2γμ

−=

Damped Free Vibrations: Quasi Period (4 of 8)

Compare μ with ω0 , the frequency of undamped motion:

Thus, small damping reduces oscillation frequency slightly. Similarly, quasi period is defined as Td = 2π/μ. Then

Thus, small damping increases quasi period.

kmkmmkkm

kmkmkm

mkm

kmmkm

km

81

81

6441

41

44

/4

4/2

4

222

22

42

22

2

22

0

γγγγ

γγγγωμ

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=+−≅

−=−

=−

=−

=

For small γ

kmkmkmTTd

81

81

41

/2/2 2122/12

0

0

γγγμω

ωπμπ

+≅⎟⎟⎠

⎞⎜⎜⎝

⎛−≅⎟⎟

⎠

⎞⎜⎜⎝

⎛−===

−−

Damped Free Vibrations: Neglecting Damping for Small γ 2/4km (5 of 8)

Consider again the comparisons between damped and undamped frequency and period:

Thus it turns out that a small γ is not as telling as a small ratio γ 2/4km. For small γ 2/4km, we can neglect effect of damping when calculating quasi frequency and quasi period of motion. But if we want a detailed description of motion of mass, then we cannot neglect damping force, no matter how small.

2/122/12

0 41,

41

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

kmTT

kmd γγ

ωμ

Damped Free Vibrations: Frequency, Period (6 of 8)

Ratios of damped and undamped frequency, period:

Thus

The importance of the relationship between γ2 and 4km is supported by our previous equations:

2/122/12

0 41,

41

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

kmTT

kmd γγ

ωμ

∞==→→

dkmkmT

22lim and 0lim

γγμ

( )( ) 0,sincos)(:04

02/,)(:04

0,0,)(:04

2/2

2/221

2 21

>+=<−

>+==−

<<+=>−

−

−

μμμγ

γγ

γ

γ

γ

tBtAetumkmeBtAtumk

rrBeAetumk

mt

mt

trtr

Damped Free Vibrations: Critical Damping Value (7 of 8)

Thus the nature of the solution changes as γ passes through the value This value of γ is known as the critical damping value, and for larger values of γ 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 for small γ in solution (3). Soln (1) is overdamped and soln (2) is critically damped.

.2 km

( )( ) )3(0,sincos)(:04

)2( 02/,)(:04

)1(0,0,)(:04

2/2

2/221

2 21

>+=<−

>+==−

<<+=>−

−

−

μμμγ

γγ

γ

γ

γ

tBtAetumkmeBtAtumk

rrBeAetumk

mt

mt

trtr

Damped Free Vibrations: Characterization of Vibration (8 of 8)

Mass creeps back to equilibrium position for solns (1) & (2), but does not oscillate about it, as for small γ in solution (3).

Soln (1) is overdamped and soln (2) is critically damped.

( )( ) )3( (Blue)sincos)(:04

)2( Black) (Red,02/,)(:04

)1((Green)0,0,)(:04

2/2

2/221

2 21

tBtAetumkmeBtAtumk

rrBeAetumk

mt

mt

trtr

μμγ

γγ

γ

γ

γ

+=<−

>+==−

<<+=>−

−

−

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 τ such that |u(t)| < 0.1 for all t > τ.

For Part (a), using methods of this chapter we obtain:

where

0)0(,2)0(,0125.0 =′==+′+′′ uuuuu

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+= −− δtettetu tt

16255cos

25532

16255sin

2552

16255cos2)( 16/16/

)sin,cos (recall 06254.02551tan δδδδ RBRA ==≅⇒=

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

For undamped case:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+= −− δtettetu tt

16255cos

25532

16255sin

2552

16255cos2)( 16/16/

998.016/255 ≅=μ

295.6/2 ≅= μπdT

283.62,10 ≅== πω T


The damping coefficient is γ = 0.125 = 1/8, and this is 1/16 of the critical valueThus 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 > τ ≈ 47.515 sec

22 =km


To find the time at which the mass first passes through the equilibrium position, we must solve

Or more simply, solve

016255cos

25532)( 16/ =⎟⎟

⎠

⎞⎜⎜⎝

⎛−= − δtetu t

sec 637.12255

16216

255

≅⎟⎠⎞

⎜⎝⎛ +=⇒

=−

δπ

πδ

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

′=′=

′=+′+′′

Ch 3.9: Forced Vibrations

We continue the discussion of the last section, and now consider the presence of a periodic external force:

tFtuktutum ωγ cos)()()( 0=+′+′′

Forced Vibrations with DampingConsider the equation below for damped motion and external forcing funcion F0cosωt.

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 ωγ cos)()()( 0=+′+′′

( ) ( ) )()(sincos)()()( 2211 tUtutBtAtuctuctu C +=+++= ωω

)()()( 2211 tuctuctuC +=

( ) ( )tBtAtU ωω sincos)( +=

Homogeneous SolutionThe homogeneous solutions u1 and u2 depend on the roots r1and r2 of the characteristic equation:

Since m, γ, 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,

mmk

rkrrmr2

40

22 −±−

=⇒=++γγ

γ

0)(lim =∞→

tuCt

( ) ,0lim)(lim 2121 =+=

∞→∞→

trtr

tCtecectu

( ) .0sincoslim)(lim 21 =+=∞→∞→

tectectu tt

tCtμμ λλ

Transient and Steady-State SolutionsThus 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.


( ) 0)()(lim)(lim 2211 =+=∞→∞→

tuctuctutCt

( ) ( ),sincos)()()(cos)()()(

)()(

2211

0

444 3444 2144 344 21tUtu

tBtAtuctuctutFtkututum

C

ωωωγ

+++==+′+′′

Transient Solution and Initial ConditionsFor 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 F0cosωt. Without damping, the effect of the initial conditions would persist for all time.

( ) ( )444 3444 2144 344 21

)()(

2211

0

sincos)()()(cos)()()(

tUtu

tBtAtuctuctutFtkututum

C

ωωωγ

+++==+′+′′

Rewriting Forced ResponseUsing trigonometric identities, it can be shown that

can be rewritten as

It can also be shown that

where

( )δω −= tRtU cos)(


222220

2222220

2

220

222220

20

)(sin,

)()(cos

,)(

ωγωωωγδ

ωγωωωωδ

ωγωω

+−=

+−

−=

+−=

mmm

mFR

mk /20 =ω

Amplitude Analysis of Forced ResponseThe amplitude R of the steady state solution

depends on the driving frequency ω. For low-frequency excitation we have

where we recall (ω0)2 = k /m. Note that F0 /k is the static displacement of the spring produced by force F0. For high frequency excitation,

,)( 22222

02

0

ωγωω +−=

mFR

kF

mF

mFR 0

20

022222

02

0

00 )(limlim ==

+−=

→→ ωωγωωωω

0)(

limlim22222

02

0 =+−

=∞→∞→ ωγωωωω m

FR

Maximum Amplitude of Forced ResponseThus

At an intermediate value of ω, the amplitude R may have a maximum value. To find this frequency ω, differentiate R and set the result equal to zero. Solving for ωmax, we obtain

where (ω0)2 = k /m. Note ωmax < ω0, and ωmax is close to ω0for small γ. The maximum value of R is

)4(1 20

0max

mkFRγγω −

=

0lim,lim 00==

∞→→RkFR

ωω

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

mkm 21

2

2202

220

2max

γωγωω

Maximum Amplitude for Imaginary ωmax

We have

and

where the last expression is an approximation for small γ. If γ 2 /(mk) > 2, then ωmax is imaginary. In this case, Rmax= F0 /k,which occurs at ω = 0, and R is a monotone decreasing function of ω. Recall from Section 3.8 that critical damping occurs when γ 2 /(mk) = 4.

⎟⎟⎠

⎞⎜⎜⎝

⎛+≅

−=

mkF

mkFR

81

)4(1

2

0

02

0

0max

γγωγγω

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

mk21

220

2max

γωω

ResonanceFrom the expression

we see that Rmax≅ F0 /(γ ω0) for small γ. Thus for lightly damped systems, the amplitude R of the forced response is large for ω near ω0, since ωmax ≅ ω0 for small γ. This is true even for relatively small external forces, and the smaller the γ 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.

⎟⎟⎠

⎞⎜⎜⎝

⎛+≅

−=

mkF

mkFR

81

)4(1

2

0

02

0

0max

γγωγγω

Graphical Analysis of QuantitiesTo get a better understanding of the quantities we have been examining, we graph the ratios R/(F0/k) vs. ω/ω0 for several values of Γ = γ 2 /(mk), as shown below.Note that the peaks tend to get higher as damping decreases. As damping decreases to zero, the values of R/(F0/k) become asymptotic to ω = ω0. Also, if γ 2 /(mk) > 2, then Rmax= F0 /k,which occurs at ω = 0.

Analysis of Phase AngleRecall that the phase angle δ given in the forced response

is characterized by the equations

If ω ≅ 0, then cosδ ≅ 1, sinδ ≅ 0, and hence δ ≅ 0. Thus the response is nearly in phase with the excitation. If ω = ω0, then cosδ = 0, sinδ = 1, and hence δ ≅ π /2. Thus response lags behind excitation by nearly π /2 radians.If ω large, then cosδ ≅ -1, sinδ = 0, and hence δ ≅ π . Thus response lags behind excitation by nearly π radians, and hence they are nearly out of phase with each other.

222220

2222220

2

220

)(sin,

)()(cos

ωγωωωγδ

ωγωωωωδ

+−=

+−

−=

mmm

( )δω −= tRtU cos)(

Example 1: Forced Vibrations with Damping (1 of 4)


Then ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The unforced motion of this system was discussed in Ch 3.8, with the graph of the solution given below, along with the graph of the ratios R/(F0/k) vs. ω/ω0 for different values of Γ.

0)0(,2)0(,2cos3)()(125.0)( =′==+′+′′ uuttututu


Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The solution for the low frequency case ω = 0.3 is graphed below, along with the forcing function. After the transient response is substantially damped out, the steady-state response is essentially in phase with excitation, and response amplitude is larger than static displacement.Specifically, R ≅ 3.2939 > F0/k = 3, and δ ≅ 0.041185.


Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The solution for the resonant case ω = 1 is graphed below, along with the forcing function. The steady-state response amplitude is eight times the static displacement, and the response lags excitation by π /2 radians, as predicted. Specifically, R = 24 > F0/k = 3, and δ = π /2.


Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The solution for the relatively high frequency case ω = 2 is graphed below, along with the forcing function. The steady-state response is out of phase with excitation, and response amplitude is about one third the static displacement.Specifically, R ≅ 0.99655 ≅ F0/k = 3, and δ ≅ 3.0585 ≅ π.

Undamped Equation: General Solution for the Case ω0 ≠ ω

Suppose there is no damping term. Then our equation is

Assuming ω0 ≠ ω, then the method of undetermined coefficients can be use to show that the general solution is

tFtkutum ωcos)()( 0=+′′

tm

Ftctctu ωωω

ωω cos)(

sincos)( 220

00201 −++=

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 ω

0,)( 222

0

01 =

−−= c

mFc

ωω

( )ttm

Ftu 0220

0 coscos)(

)( ωωωω

−−

=

tm

Ftctctu ωωω

ωω cos)(

sincos)( 220

00201 −++=

Undamped Equation: Solution to Initial Value Problem (2 of 3)

Thus our solution is

To simplify the solution even further, let A = (ω0 + ω)/2 and B = (ω0 - ω)/2. Then A + B = ω0t and A - B = ωt. Using the trigonometric identity

it follows that

and hence

( )ttm

Ftu 0220

0 coscos)(

)( ωωωω

−−

=

,sinsincoscos)cos( BABABA m=±

BABAtBABAt

sinsincoscoscossinsincoscoscos

0 −=+=

ωω

BAtt sinsin2coscos 0 =− ωω

Undamped Equation: Beats (3 of 3)

Using the results of the previous slide, it follows that

When |ω0 - ω| ≅ 0, ω0 + ω is much larger than ω0 - ω, andsin[(ω0 + ω)t/2] oscillates more rapidly than sin[(ω0 - ω)t/2]. Thus motion is a rapid oscillation with frequency (ω0 + ω)/2, 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 022

0

0 tm

F ωωωω

−−

( ) ( )2

sin2

sin)(

2)( 0022

0

0 ttm

Ftu ωωωωωω

+⎥⎦

⎤⎢⎣

⎡ −−

=

Example 2: Undamped Equation,Mass Initially at Rest (1 of 2)


Then ω0 = 1, ω = 0.8, and F0 = 0.5, and hence the solution is

The displacement of the spring–mass system oscillates with a frequency of 0.9, slightly less than natural frequency ω0 = 1.The amplitude variation has a slow frequency of 0.1 and period of 20π. A half-period of 10π corresponds to a single cycle of increasing and then decreasing amplitude.

0)0(,0)0(,8.0cos5.0)()( =′==+′′ uuttutu

( )( )tttu 9.0sin1.0sin77778.2)( =

Example 2: Increased Frequency (2 of 2)

Recall our initial value problem

If driving frequency ω is increased to ω = 0.9, then the slow frequency is halved to 0.05 with half-period doubled to 20π. 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

Undamped Equation: General Solution for the Case ω0 = ω (1 of 2)

Recall our equation for the undamped case:

If forcing frequency equals natural frequency of system, i.e., ω = ω0 , then nonhomogeneous term F0cosωt is a solution of homogeneous equation. It can then be shown that

Thus solution u becomes unbounded as t →∞. Note: Model invalid when u getslarge, since we assume small oscillations u.

ttmFtctctu 0

0

00201 sin

2sincos)( ω

ωωω ++=

tFtkutum ωcos)()( 0=+′′

Undamped Equation: Resonance (2 of 2)

If forcing frequency equals natural frequency of system, i.e., ω = ω0 , then our solution is

Motion u remains bounded if damping present. However, response u to input F0cosωt may be large if damping is small and |ω0 - ω| ≅ 0, in which case we have resonance.

ttmFtctctu 0

0

00201 sin

2sincos)( ω

ωωω ++=

ch 3.1: second order linear homogeneous …matysh/ma3220/chap3.pdfch 3.1: second order linear...

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