non-homogeneous equations_undetermined coefficients

7/29/2019 Non-Homogeneous Equations_Undetermined Coefficients

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Copyright . Ren Barrientos Page 1

NON-HOMOGENEOUS EQUATIONS

Introduction

Consider the non-homogeneous equation Where, as usual,

; The general solution of (1) is found by adding itscomplementary solution, found by solving 0, andany particular solution. We recall that a particular solution is any function free of arbitrary constantswhich satisfies . The goal of these notes is to introduce you to one of several methods used tofind particular solutions the method of undetermined coefficient.

But first some preliminaries that will be needed in the course of our work.

Box 1

Box 2

Box 3

Example 1 Find the general solution of the equation

5

6 3 0 given that

5 2 5 /6is a

particular solution.

SolutionFirst we find by solving 5 6 0. The auxiliary equation is 5 6 0.

Therefore, we obtain . Box tells us that desired general solution is 525/6

The Superposition Principle

Suppose that is a particular solution of and is a particular solution of .Then

is a particular solution of

.

Proof

Therefore, is a solution of

; , , , ,

Existence and Uniqueness

Let have continuous coefficients on an interval in which 0 and suppose is alsocontinuous in. Let . Then there is an interval centered at and a unique function defined in such thatfor any values of, , .

General Solutions (Non-Homogeneous Equations)

Let be ann-parameter family of solutions of 0 and be a particular solution of .Then is the general solution of .


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Example 2 To find the general solution of 3 we can invoke the superposition principle andsolve two separate but easier non-homogeneous equations: 3 . .

.

.

Once solutions and are obtained for each equation, is a particularsolution of the original equation. This is the essence of the superposition principle for non-homogeneousequations, but fortunately we will not have to go through this tedious separation procedure as we shall see inthe examples that follow.

Example 3 The equation 2 5 5 2 75 has a particular solution given by 27 3[verify]. Find its general solution.

SolutionThe auxiliary equation is 2 5 0Therefore,

cos5and

sin5are the two linearly independent solutions of the

corresponding homogeneous equation and the complementary solution is Thus, the non-homogeneous equations general solution is given by

Thus, the key to solving non-homogeneous equations lies in finding their particular solutions. We turn ourattention to that task.

Undetermined Coefficients

The idea behind this method is to assume that the particular solution is of the same form as the forcing

function. However, certain restrictions apply and we will address them in due course.

Suppose that theforcing function is a linear combination of functions of the form :

where , , and are constants and is a polynomial of degree . At first, this may look like a verystringent limitation, but as we shall see many applications fall under this category and the method ofundetermined coefficients allows us to find particular solutions fairly quickly. For more general cases, themethods of variationof parameters or annihilators(which use the properties of certain operators) may beused.

Here is an example of the method of undetermined coefficients (sometimes calledjudicious guessing) at work:

Example 4 Find a particular solution of the equation 1 0 Solution

Since the forcing function is 10 which a polynomial of first degree, weassume a particularsolution of the same form: a polynomial of first degree. So let Notice that we have chosen the most general representative of the family of polynomials of firstdegree. It is possible the we will find that 0, but we cannot know that ahead of time.


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Substituting in the equation, 10And now we proceed to identify the coefficients; hence the nameundetermined coefficients.

Differentiating on the left-hand side gives us

1 0 Which must hold for all . Equating coefficients: 1 0 0Solving the system gives us 10, 10. Therefore, A quick check shows that this is in fact a particular solution:1010 1010 1010 10

10 1010 10The last equation is an identity. Therefore, 1 0 1 0 is a particular solution.Example 5 Find a particular solution of the equation Solution

Since the forcing function is which an exponential function, we assume a particularsolution of the same form: an exponential function. Thus, let Substituting in the equation in order to determine the undetermined coefficient: Performing the required operations:

4

or5 This equation is true for all if and only if5 1 or 1/5. Hence,

which you should verify.

Once again, notice thatparticular solutions do not involve arbitrary constants.

Example 6 Find the general solution of the equation 4 Solution

We need to do a bit more work here. Recall that the general solution is found by adding thecomplementary solution to a particular solution . We first find the complementary solution bysolving 4 0

This is just a homogeneous equation with constant coefficients and it has two linearly independentsolutions: sin2 and cos2. Hence, s i n2 cos2


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Now we seek a particular solution of the form , which is the most general form ofthe forcing function . We substitute in the original equation in order to determine thecoefficients, , and: 4 24 4 4

or4 4 2 4 For this equation to hold for all , we must have

4 14 02 4 0Solving the system, 1/4, 0, 1 / 8. Thus,

14 18Finally, the general solution is given by

:

Notice that general solutions do have arbitrary constants. These can become known only if additional initialconditions are given.

So the method seems to be pretty straight forward. However, there are complications as the following exampledemonstrates.

Example 7 Find a particular solution of the equation 2 Solution

As before, we assume a particular solution of the form . Substituting,

2

or 2 we have a problem: The last equation implies that0 which is absurd. What went wrong here isthat the forcing function happens to have the same form as one of the solutions of the correspondinghomogeneous equation 2 0. Obviously, any particular solution of the form willfail to work.

We address this problem by assuming a particular solution of the form instead. Onceagain, substituting, 2

or

2 2 Combining line terms: 3 Hence, 1/3 and a particular solution is


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The moral of the story here is this: If any term of the forcing function is equal to a member of the fundamentalset of solutions of the corresponding homogeneous equation, an adjustment must be made to the tentativeparticular solution.

It is therefore important to have a systematic way of identifying the particular solution and we proceed tooutline one.

Solving non-homogeneous equation with Undetermined Coefficients

Step 1 Find the fundamental set corresponding to 0 and identify thecomplementary solution .Step 2 Identify the form of the particular solution using the guidelines described in the following table1:

Solving Form of Auxiliary Equation Form of Tentative Particul ar Solution

is not a root

is a simple root has multiplicity 2

or is not a root

cos sin is root cos

Step 3 Determine the undetermined coefficients by substituting in .Step 4 The general solution is given by the sum of the complementary and particular solutions.

Step 5 Finally apply initial conditions if given.

Step 2 says that we should multiply the tentative particular solution by enough powers of the independentvariable until none of its terms resembles any of the members of the fundamental set of solutions of thecorresponding homogeneous equation. Furthermore, we should begin with the most general form of the forcingfunction.

Here are some more examples when there are no complications:

Differential Equation f

Fundamental set of the

associated homogeneousequation Tentative Form of t he Particular Soluti on

2 2

1We use the superposition principle when is a combination of these.


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4 cos2,sin2 sin sin cos 32 , sin 32 cos sin

sin

sin cos 32 , sin 32 sin cos cos,sin

Example 8 What would be the form of the particular solution of the equation 4 4 for each ofthe following cases? [note: (b), (d) and (e) usetas the independent variable].

(a) 6; (b) 4; (c) ; (d) 10; (e) sin4Solution

all these have the same characteristic polynomial so step 1 will be the same for all of them.Step 1 The associated homogeneous equation has characteristic polynomial 4 4 2 2 so the fundamental set in each of the cases is , . We now proceed to identify a tentativeparticular solution for each case.

Step 2

(a) 6 has the form with 6 , 3 , 0. That is, it is a polynomial of degree3 without an exponential factor. Accordingly, we set That is, the most general form of a third degree polynomial.

(b) 4 has the form with 4 , 0 , 2. This is an exponential function sowe begin by setting However, notice that is a root of multiplicity two of the auxiliary equation. Therefore, both and are elements of the fundamental set. We must modify our tentative particular solutionby multiplying by a power oftuntil we no longer have a function that is part of the fundamental set.Multiplying once, No good, this is still in the fundamental set. Multiplying again,

Good! this in no longer in the fundamental set so we may use it as our tentative solution.(c) has the form with 1 , 2 , 1 . This is a second degreepolynomial exponential function. Thus, let does not involve any of the function of the fundamental set. Therefore there are no adjustmentsto be made. may be used as our tentative particular solution.


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(d) 10 has the form with 1 0 , 1 , which is a linear polynomial an exponential function. Thus our first trial is But observe this tentative solution involves both and . Therefore we need to introduce apower of

until this is no longer the case:

is not good because we still have a term, namely, . Therefore, we multiply by again: Now none of the terms of this tentative solution appears in the fundamental set and we are ready toproceed with a tentative particular solution of the from (e) sin which is a product of a polynomial of degree2, an exponential function, and asine function. Let

s i n

cosbe the tentative particular solution. Since none of the terms in the tentative solution appear in thefundamental set. Therefore, we do not need to make any adjustment. The tentative particularsolution is

Example 9 Find a tentative form for the particular solution of the equation4 10Solution

Step 1: identify the fundamental set of the corresponding homogeneous equation by solving theauxiliary equations of4 0: , / . Hence the fundamental set is

1,/The forcing function

10 has the form

with

, and

. Since

0is a

root of the characteristic polynomial, we need to adjust the tentative solution

by multiplying once by :

or Example 10 Find a tentative form for the particular solution of the equation 2 3 s i n Solution

The roots of the characteristic polynomial are . Therefore, the fundamental set canbe taken to by /, /.Since

has a trigonometric form, we assume that its particular solution also has a trigonometric

form. Thus tentatively we set which is the most general form of a trigonometric solution involving the sine and cosine functions.No adjustments are needed.

Example 11 Find a tentative form for the particular solution of the equation 2 5 /


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SolutionThe roots of the characteristic polynomial are 1 , /. Thus a fundamental set is, /. Notice that one of the elements of the fundamental set appears as a term in the forcingfunction. Hence, our initial guess of /must be modified by multiplying by

:

/This will do and we may proceed to determine.Example 12 Find the form of a particular solution for 3 .Solution

The auxiliary equation is 0. Factoring and solving: 1 0 0 mult 2, 1 , and 1Thus, is the complementary solution. Since 3, our first guessfor is However, both terms in appear in already. Therefore, we must multiply by a power of until this is no longer the case. The smallest power that will accomplish this is. Thus, is the correct trial function. Notice that none of its term appears in.

Example 13 Find a tentative particular solution of 9 2 sin3.Solution

The roots of the auxiliary equation are 0 3 and 0 3 . Therefore, the associatedhomogeneous equation hascos3,sin3 as its fundamental set.Since 2 sin3 has form with 2 , 2 , 0 , and 3 . Our firstguess for the particular solution is

cos3 sin3However, the terms cos3 and sin3 of the particular solution involve the two members of thefundamental set. Accordingly, we modify our guess by multiplying once by:

cos3 sin3Our particular solution has the form Let us now move beyond identifying a trial solution and find particular and general solutions.

Example 14 Find a particular solution of 5 6 2 0 SolutionStep 1 We begin by finding the complementary solution 5 6 0 Step 2 20. Thus, our first guess is


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But as you can see,this term already appears in the complementary solution. Therefore, we modifyour guess by multiplying it by : If a term of the form was present in the complementary solution, we would modify it again bymultiplying one more time by t, and so on. Since this is not the case, we proceed todetermine

.

Step 3 Substituting, 5 6 Performing the indicated operations,6 9 5 15 6 20Cancelling and equating coefficients: 20

Thus, Example 15 Find a particular solution of 4 3 c o s 2 Solution

Step 1 We first solve 4 0. Thus, 4 0 .Accordingly, cos2,sin2 is the fundamental set and the complementary solution is given by c os 2 sin2Step 2 3cos2 has the form with 3 , 0 , 0 , and 2. Begin byassuming a solution of the form c os 2 s i n2 However, we see that these are exactly the functions in the fundamental set. Thus, we need to makean adjustment by multiplying byx: cos2sin2

cos2sin2None of the terms of appear in the fundamental set and we may proceed to step3.Step 3 Substitute in the differential equation in order to determine its coefficients. We need tocompute some derivatives first: cos22sin2sin22cos2 2sin22sin22cos2 2 c os 2 2cos22sin2Cleaning this expression up, 4sin24cos2 4cos24sin2

Thus, we must have

4sin24cos2 4cos24sin2 4cos2sin23cos2Combining like terms,44 44 4 4 3cos2or443

Equating coefficients, we obtain , /. Note : there is no sine term in the right-hand side.


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A particular solution is Example 16 Find a particular solution of 9 2 0 2 7 .Solution

The forcing function is a sum of two functions so se can use the superposition principle and solve thefollowing two cases individually: 9 2 0 2 9 2 0 7 For both cases, the associated homogeneous equation has as solution and neither 2 nor 7 involve the functions that make up the complementary solution. Wecould proceed to find two particular solution and , but this isunnecessary; it suffices to write a tentative particular solution involving undetermined coefficientscorresponding to both cases: 2 and 7

Thus, let

. The rest of the procedure is that same and we must determine the

values of, , and.Substituting, 9 20 2 7 Thus, 9 20 2 7 Collecting like terms, 9 2 0 920 20 2 7 Equating coefficients: 3 0 7 ; 9 2 0 0 ; 2 0 2Solving for the undetermined coefficients: 9200 ; 110 ; 730A particular solution of 9 2 0 2 7 is

Example 17 Find the general solution of the equation 5 6 1 2 .Solution

Step 1 the associated homogeneous equation has, as fundamental set.Step 2 the form of the forcing function

12is

with

1 2 , 1 , 0. None

of the terms of

appear in the fundamental set. Thus, we may set

which is the most general form of a linear polynomial.

Step 3 The quantities A and B are the undetermined coefficients. In order to determine them, wesubstitute in the differential equation: 5 6 12Performing the required differentiations,


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Example 20 Find the solution of the IVP 3 c o s 2 ; 0 0 0 1Solution

Step 1 The corresponding homogeneous equation 0 has solutions c os s i nTherefore, the complementary solution is

c o s sinStep 2 The forcing function cos2 does not involve any of the terms in the fundamental set.Therefore, we may assume that PITFALL 1: Always assume the particular solution has the most general form of the forcing function. If the lattercontains the terms or , assume is a linear combination of both.It might very well be that turns out to be0, but we cannot know that ahead of time.Step 3 Determine the undetermined coefficientsAandB. Substitute in the original equation:cos2sin2 cos 2 sin 2 3cos2Performing the required differentiations,

4cos24sin2cos2sin23cos2Collecting like trigonometric terms,4 cos24sin23cos2

This equation is true for all t if and only if 4 34 0Solving the system: 1, 0.

Therefore, a particular solution is given by

Step 4 The equations general solution is PITFALL 2: Do not apply the initial conditi ons to the complementary solution. You must always first find thegeneral solut ion.

Step 5 Having obtained the general solution, we may now apply the initial conditions to c o s sincos20 0 cos0 sin0cos00Therefore, and the general solution reduces to

sincos2The second condition requires that we find a derivative: cos2sin20 1 cos02sin01Therefore, and the solution of the IVP is given by


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Example 21 Find the solution the equation 2 ; 0 1Solution

This is just a first order linear equation so there is no need for the method of undeterminedcoefficients we could very easily use an integrating factor to find a solution. However, let us seewhat our new method gives us.

Step 1 The corresponding homogeneous equation 2 0 has characteristic polynomial 2 0. Therefore, we only have one root: 2. Thus, where is the arbitrary constant.Step 2 has the formwith 1 , 1 , 1 which is a polynomial of degree1times an exponential function. Thus, let Once again, careful! we are looking at the most general form of the forcing function .It is possible thatthe coefficientBturns out to be0, but we cannot assume that ahead of time.Since the terms of the tentative particular solution are not part of we may proceed with step 3:Step 3 Determine the undetermined coefficients: 2 Differentiating 2 Sinceis never0, we can divide it out. Collecting like terms and grouping like powers of: 2 2

Thus, 0 1Solving the system, 1, 1.Substituting in

:

1 Step 4 The general solution is given by Step 5 Apply initial conditions to determine the arbitrary constant(s)0 1 0 1 1Solving for , 2. Thus, is the solution of the initial value problem.

An Application Resonance

When we studied mechanical systems, we found that they have a characteristic frequency (and sometimesmore than one) at which they prefer to oscillate. For example, the mass-spring system in which an object of

mass is attached to a secured spring of spring constant has a natural frequency of /. Whatwould happen if an external periodic force is applied to such a system?

The following example illustrates the important concept ofresonance. Resonance is the phenomenon in wh icha system tends to oscill ate with greater amplitude at some frequencies more than others.


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Suppose we force a spring-mass system to oscillate by applying an external force. Assume that drag and otherlosses can be ignored so that the friction coefficient 0. We are interested in studying the situation inwhich is itself a periodic function, say cos. Thus, consider the non-homogeneous equation

, Where / is the systems characteristic ornatural angular frequency.What happens when the input frequency is equal to ? Intuitively, we are forcing a system to vibrate withits natural frequency; we are adding energy at just the precise moment thereby exacerbating the vibration. Ifin addition there are no damping forces, the systems energy can only increase resulting in a larger and largeramplitude. In order to quantify this observation, we must solve equation (3). However, we will assume thatthe system begins in a state of rest at equilibrium. Thus,

, From our study of non-homogeneous equations, we find the general solution of (4) by adding itscomplementary solution and a particular solution . The solution of the initial value problem is thenobtained by applying the initial conditions. The complementary solution is given by

cos sinThe methods ofvariation of parameters or undetermined coefficients can be used to find but we mustconsider the two cases and separately.Case I: . Let c os s i n be a particular solution. Then substituting in (3), cos sin cossin cos

Thus,

cos

sin

cos sin cos

Combining like terms: cos sin cosThis equation is true for all t if and only if and 0Solving forA andB: / 0A particular solution is /

cos

The general solution of equation (3) is

cos sin / cosApplying the initial conditions, 0 0 / 0 0 0Since 0, 0. The solution is


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/ cos / cosFactoring and simplifying,

/ coscosFrom trigonometry we know that

cos cos 2 sin 2 s i n 2 Applying2this identity with and ,

/ The graphs below illustrate for some specific values of and:

, .

, .

, . 2There are a couple of steps that you need to fill.

20 40 60 80 100 120

-1.0

-0.5

0.5

1.0

1.5

20 40 60 80 100 120

-3

-2

-1

1

2

3

20 40 60 80 100 120

-6

-4

-2

2

4

6


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The behavior of the system is very different for different input frequencies , but something interestinghappens as : we start to see the appearance ofbeats. These are characterized by thewave packetsseen in the figure above.

Case II: . The equation we want to solve is

,

The complementary solution is the same as inCase I. However, the particular solution must be modified.The forcing function now has exactly the same form as one of the members of the fundamental set of theassociated homogeneous equation. Therefore, our tentative particular solution must be Substituting in equation (4), cos s i n cos s i n cos

This requires some work. Let us find the derivatives first:

c os

s i n

sin

cos

sin cos sin cos cos sin 2 sin 2 cos cos s i n Hence equation (5) becomes

2 sin 2 cos cos s i n cos s i n cosCancelling and simplifying, 2 sin 2 cos cos

This equation is satisfied for all t if and only if

2 0and2 Therefore, , /and a particular solution of equation (4) is

0 c o s 2 sin 2 s i n

Hence , the general solution is

Applying the initial conditions 0 0, 0 0 we obtain . Thus,

As expected, we have a highly destructive situation: the factor in the solution makes the wave oscillate withever-increasing amplitude:


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When the input frequency is equal to , the system oscillates with increasingly higher amplitude, that is,resonancesets in and is characterized by the graph illustrated above.

Needless to say, this can be a very good thing or a very bad one you will encounter these phenomena in your

engineering and science courses.

Example 22 A 20 lb weight stretches a spring 10 inches. Once the mass is in equilibrium, a force given by10cos83/5 acts on the springs support. Find the springs position function.Solution

First let us compute: since2 0 , 2 4 lb/ft. Now we can calculate: 2420/32 2 4 3 220 83/5

The forcing function has the same angular frequency. Therefore, the position function is given by

2 s i n

Where

10. Thus,

10163/5 s i n83/5 / Example 23 With what frequency, in Hz., should a spring-mass system with 2 kg and 36 N/m beforced to oscillate in order to exhibit resonance?

Solution

The natural angular frequency of this system is 3 6 /2 32 rad/sec. Therefore, if thesystem is forced to oscillate with this frequency it will exhibit resonance. Since we want to expressthis in cycles per second, we use 2 :

2

322 0.675 Hz.

Illustration

It is interesting to analyze graphically the relation between the forcing function and the resulting vibration.Suppose we have 2cos1.15 and this force is used on a system whose natural frequency is1 rad/sec[as we did above]. Here are the graphs of the input function [that is, the forcing function] and the inducedvibration:

20 40 60 80 100 120

-100

-50

50

100


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Example 24 A 30 cm long spring is attached to a rotating disk of radius8 cm which rotates with frequency3Hz. The springs opposite end is attached to a5 Kg mass which rests on a smooth surface and is confined tomove only in one dimension (see figure below). If the spring constant is

1 8 0 Nt/m determine if

resonance will take place.

SolutionWe need to make a simplifying assumption: as the wheel rotates, the spring remains essentiallyhorizontal. What is the input function? It is some kind of periodic function whose frequency is 3 Hz.First we convert 3 Hz to radians using the formula 2 : 2 3 6 rad/sec.A 0.08 m radius wheel rotating with this frequency can be modeled by the function

0.08cos6where is a phase angle. Thus, the equation of motion for the mass is5 1800.08cos6; 0 0, 0 0Resonance conditions will take place if the systems natural frequency is equal to the inputfrequency :

1805 6Since , we have resonance.

Example 25 A front-loading washing machine of mass M sits on a rubber pad that behaves like a spring.

When the machine is still, the pad is compressed1/2 cm and when it is on, the rotorspins with angular frequency rad/sec and it exerts an upward force cos.At what rotor speed, in revolutions per minute, will resonance occur? Neglectfriction.

SolutionThe spring constant is given by the condition0.005 sothat /0.005.

2 40 60 80

-2

-1

1

20 40 60 80

-10

-5

5

1

2cos1.15Input function

411.15sin1 1 . 1 52 sin11.152 Output

0.08

5 Kg


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If represents the position of the center of the machine, then the equation of motion is cosResonance occurs when / /. Simplifying the radical,

9.8/0.005 44.27 rad/secTo convert to revolutions per minute, we multiply by the appropriate conversion factors:44.27 radsec rev2 rad 60 secmin . rev/min

Example 26 (Damped Forced Oscillation) Solve the equation 4 4 1 6 9 c os 3 ; 0 0 , 0 0.Solution

The auxiliary equation is 4 4 0 whose roots are 2 and 2. Therefore, thecomplementary solution is We seek a particular solution of the form

c os 3 s i n3 Differentiating: 3sin33cos3 9cos39sin3Substituting,9cos39sin343sin33cos3 4 cos 3 sin 3 169cos3Collecting like terms,9124 cos3 9124 sin3169cos3Equating coefficients,

512169

1 2 5 0Thus, 5 and 1 2. Therefore, 5cos312sin3The general solution is 5cos312sin3Applying the initial conditions 0 0

The second condition requires that we compute:2 2 15sin336cos3Thus,

0 0 2 36Since 5, solving for : . The solution is therefore This example illustrates an interesting phenomenon: As , the exponential terms die out and only thetrigonometric ones survive. We call the former transient termsand denote by that part of the solutionwhich contains them. Thus,


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We denote by that part of the solution which governs the systems long term behavior. In the presentexample that is Writing in terms of a single cosine function:

cos3From basic trigonometry we know that 5 12 andc o s 513 s i n 1213

Therefore, is asecond quadrant angle and we may usecos to find it: c o s 1 . 9. Hence, 13cos31.97Thus, the solution may be written more compactly as .The graph below illustrates the influence of the transient terms and the long-term behavior of the solution.

The forcing function takes over very quickly, as you can see. However, transient terms can dominate thebehavior of a system for some time.

Exercise Solvethe equation

, Exercise Analyze the case by letting in the solution / coscos

2 8 10

-1

-5

5

1

non-homogeneous equations_undetermined coefficients

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