Second-Order Linear Equations - Christian Brothers Universityfacstaff.cbu.edu/~lbecker/Math231/DEQBook/Chapter9.pdf · Second-Order Linear Equations Pedagogically speaking, a good

Chapter 9

Second-Order Linear Equations

Pedagogically speaking, a good share of physics and mathematics was—and is—writing differential equations on a blackboard and showing students how to solvethem. Differential equations represent reality as a continuum, changing smoothlyfrom place to place and from time to time, not broken in discrete grid points ortime steps.

James Gleick1

9.1 Horizontal Mass-Spring System

Consider the simple apparatus depicted in Figure 9.1. A body of mass m sits on atrack and is attached to a spring. The other end of the spring is fastened to a stationarysupport. This is an example of a horizontal mass-spring system.2 We will use the letterm to refer either to the body or to its mass. It should be clear from the context in whichsense m is used.

Our objective is to model the back-and-forth motion of m when it is pulled to theright or pushed to the left and then released. Suppose the mass-spring system has thefollowing features:

1. The mass of the spring is negligible compared to m. In effect, the spring isessentially massless.

2. The track on which m is free to slide is like the surface of an air hockey table.There are many small perforations in the track through which air is pumped tosignificantly reduce the frictional force exerted on m by the track. Consequently,friction will have a negligible effect on the motion of m.

1See [27, p. 67]. James Gleick helped popularize the mathematics and science of chaos through hisnewspaper column in the New York Times and in his best-selling nonfiction book: Chaos: Making a NewScience [27]. After that, the general public began hearing of chaos elsewhere, even in the movies, such asSpielberg’s 1993 film Jurassic Park. Jeff Goldblum played the part of a mathematician with expertise inchaos theory.

2Here is another meaning of “mass”: it refers to the body attached to the spring.

241

242 CHAPTER 9. SECOND-ORDER LINEAR EQUATIONS

Fig. 9.1: Horizontal mass-spring system

3. When the natural length of the spring is changed by stretching or compressingit, a force acts to restore it back to its natural length. Assume that the springobeys Hooke’s law, namely that the magnitude of the restoring force at any giveninstant is proportional to the difference between the length of the spring at thatinstant and its natural length. This is the case for many common springs providedthe elastic limit of the spring is not exceeded. Such springs are said to be linear(or ideal) springs.

To specify the horizontal position of m at time t , let

x.t/ WD [length of the spring at time t] � [natural length of the spring].

Thus, when x > 0, the spring is stretched and when x < 0, it is compressed. SeeFigure 9.2, where P marks the end of the spring attached to m.

Fig. 9.2: Comparison of a deformed spring to its natural length

When the spring is stretched, the restoring force is directed to the left. So in thiscase the force is a negative quantity. It follows from the statement of Hooke’s law thatthis restoring force F is given by the formula

F D �kx; (9.1.1)

9.1. HORIZONTAL MASS-SPRING SYSTEM 243

where k > 0. The negative sign is included so that the constant of proportionalityk , called the spring constant or stiffness of the spring, is positive. When the springis compressed, the restoring force is directed to the right: so F > 0. Also, �kx > 0

when x < 0 as k > 0. And from (9.1.1) we get F D 0 when x D 0, as we should whenthe spring is in its natural state. We conclude that (9.1.1) is the correct formulation ofHooke’s law.

Since the force �kx is the only horizontal force acting on m, we obtain fromNewton’s second law of motion the differential equation

m Rx D �kx: (9.1.2)

Letting ! WD pk=m, we can rewrite this as

Rx C !2x D 0: (9.1.3)

The parameter ! is positive since both k and m are positive.A physical system whose dynamics are modeled by (9.1.3), such as the horizontal

mass-spring system, is called a simple (linear) harmonic oscillator. The motion ofthe mass m as you would expect is oscillatory and is called simple (linear) harmonicmotion. Besides the mass-spring system, there are other phenomena that are modeledby (9.1.3), such as the current in an LC circuit: an electrical circuit consisting of aninductor and a charged capacitor but no voltage source and which has no resistance. Ofcourse, in an actual LC circuit there is always some resistance just as there is alwayssome friction present in a mechanical system.

The horizontal mass-spring system that we have described is idealistic because thefrictional force between the track and the body m cannot be totally neglected. Evenif we could somehow pull off an engineering feat by eliminating friction altogetherthere is still a damping (retarding) force acting on m as it moves through the air. Thisdamping force would be even more pronounced if the mass moved through a moreviscous medium such as oil. In fact, the suspension system for each wheel of a motorvehicle consists of a mass-spring system connected to a shock absorber which damps orsmooths out the constant up-and-down motion of the vehicle’s frame. A shock absorberis basically a cylinder filled with a fluid, such as oil or pressurized air, with a slidingpiston inside it.3

Experimental evidence indicates the damping force described above is more or lessproportional to the velocity of the body m; that is, the force is �r Px, where r > 0 is theconstant of proportionality. The negative sign arises from the opposing directions ofthe damping force and m’s velocity. If we still ignore friction but now add the dampingforce, (9.1.2) becomes

m Rx D �r Px � kx (9.1.4)

orRx C b Px C !2x D 0; (9.1.5)

where b D r=m and ! D pk=m.

Equations 9.1.2, 9.1.3, 9.1.4, and 9.1.5 are members of the family of differentialequations known as second-order homogeneous linear differential equations. The ob-jective of this chapter is to learn to recognize such equations and how to solve them.

3The next time you see a parked motorcycle look at its rear shock absorber and spring.


9.2 Second-Order Homogeneous Linear Equations

Equations 9.1.3 and 9.1.5 are examples of second-order homogeneous linear equationswith constant coefficients. The order of a differential equation is two when the equationhas a second-order derivative but no derivatives of higher order. Members of this familyare easy to recognize. Any equation that looks like equation (9.2.1) below, or whichcan be algebraically manipulated into that form, is a member of this family.

Second-Order Homogeneous Linear Equations

Definition 9.1. A differential equation expressible in the form

ad2x

dt2C b

dx

dtC cx D 0; (9.2.1)

where a, b, c are constants and a ¤ 0 is called a second-order homogeneouslinear differential equation with constant coefficients.

It goes without saying that x 00 must be present in order for (9.2.1) to be considereda second-order equation, hence, the proviso a ¤ 0. On the other hand, the other termsmay or may not be present. That depends on the values of b and c. For example, fora D 1, b D �1, and c D 0, (9.2.1) is

d2x

dt2� dx

dtD 0: (9.2.2)

The word homogeneous means the right-hand side of (9.2.1) is always equal tozero, regardless of the value of the independent variable t . If we replace “0” with afunction g that is not always equal to zero, then (9.2.1) becomes

ad2x

dt2C b

dx

dtC cx D g.t/

and is called a nonhomogeneous second-order differential equation.4

9.3 Characteristic Roots and Solutions

We begin our search for a method to solve equations of the form (9.2.1). First, considerthe following simple examples which will help us get some experience with dealingwith these equations and a feel for the nature of their solutions. Unless it is otherwiseindicated, t is the independent variable and the prime symbol ( 0 ) denotes differentia-tion with respect to t .

4The term inhomogeneous differential equation is also used.

9.3. CHARACTERISTIC ROOTS AND SOLUTIONS 245

Example 9.1. Find solutions of the first-order equation x 0 � x D 0.

Solution. By inspection (or by separating variables or the integrating factor method),

x.t/ D cet

is a solution on the interval .�1;1/ for any constant value of c.

Example 9.2. Find solutions of the second-order equation x 00 � x0 D 0.

Solution. Solutions of this equation are easy to find due to the missing x term: simplywrite

.x0/0 � x0 D 0

and then replace x0 with y. In that way the second-order equation reduces to the first-order equation

y0 � y D 0:

By the previous example, it has the general solution y.t/ D ce t . And so x 0.t/ D cet .Integrating, we see that any function of the form

x.t/ D cet C k;

where c and k are constants, is a solution on .�1;1/. Notice that we could view thissolution as a combination of two exponential functions

x.t/ D cet C ke0t :

Example 9.3. Verify that e2t and e�3t are solutions of x 00 C x0 � 6x D 0.

Solution. For x.t/ D e2t , x0.t/ D 2e2t and x00.t/ D 4e2t . And so

x00.t/ C x0.t/ � 6x.t/ D 4e2t C 2e2t � 6e2t D 0;

which shows that x.t/ D e2t is a solution. It is left to the reader to show x.t/ D e �3t

is also a solution. In fact,x.t/ D c1e2t C c2e�3t

is a solution on .�1;1/ for any values of the constants c1 and c2.

The second-order equations in the previous examples have solutions of the forme�t , where � is a constant. So it seems reasonable to conjecture all second-order equa-tions of the form (9.2.1) have exponential solutions. Since we are concerned only withthese kinds of equations from now on, it will be understood that a ¤ 0 throughout therest of this chapter.

To find out if our conjecture is correct, let us substitute e �t for x in (9.2.1) to see ifit is a solution for some value of �:

ad2

dt2e�t C b

d

dte�t C ce�t D .a�2 C b�C c/e�t D 0:


From this, it follows that e �t is a solution if the factor a�2 Cb�Cc is equal to zero.Moreover, it is a solution only if a� 2 C b�C c D 0 since e�t is never zero regardlessof the value of �. In other words, e�t is a solution of (9.2.1) if � is a solution of thequadratic equation

a�2 C b�C c D 0; (9.3.1)

but not otherwise. In the context of solving a differential equation, such a quadraticequation is known as a characteristic equation. So (9.3.1) is the characteristic equationfor the differential equation (9.2.1). The polynomial

P .�/ WD a�2 C b�C c (9.3.2)

is called the characteristic polynomial. By the quadratic formula, (9.3.1) has the roots(solutions)

� D �b ˙p

b2 � 4ac

2a: (9.3.3)

Roots of a characteristic equation are called characteristic roots. Let’s label them �1

and �2 as follows:

�1 D �b �p

b2 � 4ac

2aand �2 D �b C

pb2 � 4ac

2a: (9.3.4)

Observe that the characteristic roots are unequal, or distinct, if the so-called discrim-inant b2 � 4ac is nonzero. If b2 � 4ac > 0, these distinct roots are real numbers.However, if b2 � 4ac < 0, then they clearly are not real numbers, but rather complex(or imaginary) numbers. There is also the case: b2 � 4ac D 0. Then

�1 D �2 D � b

2a

We could say, as in an algebra course, that there is only one root. However, it will beadvantageous as we shall see shortly to view all characteristic equations as having tworoots. With this point of view, there are still two characteristic roots when b 2�4ac D 0,namely, the real and equal roots

�1 D � b

2aand �2 D � b

2a

We say that �b=2a is a repeated root or a double root or a root of multiplicity 2.

Up to this point, we have discovered that

Characteristic Roots and Exponential Solutions

e�t is a solution of ax 00 C bx0 C cx D 0 (9.3.5a)

if and only if

� is a root of a�2 C b�C c D 0: (9.3.5b)


It is clear that a multiple of e�t is also a solution of (9.2.1). And then there is thequestion as to whether other solutions exist that we have not yet discovered. We willanswer this next by searching for a general solution of (9.2.1)—the form in which allof its solutions can be written.

Our approach will be to rewrite (9.2.1) in terms of its characteristic roots and thento reduce it to a first-order equation similar to what was done in Example 9.2. To thisend, suppose x.t/ is a solution of (9.2.1), which means of course that it satisfies (9.2.1)on some interval J and so

ax00.t/ C bx0.t/C cx.t/ D 0 (9.3.6)

for t 2 J . Whether the characteristic roots are distinct or repeated, their respective sumand product are

�1 C �2 D �b

aand �1�2 D c

a: (9.3.7)

This is easily shown by adding and multiplying the values of � 1 and �2 given by (9.3.4).These relations make it possible to rewrite (9.3.6) in terms of its characteristic roots asfollows. First divide (9.3.6) by the nonzero coefficient a:

x00.t/ C b

ax0.t/ C c

ax.t/ D 0:

Now replace the above coefficients with their respective equivalents in (9.3.7):

x00.t/ � .�1 C �2/x0.t/ C �1�2x.t/ D 0: (9.3.8)

By rewriting this as �x0.t/ � �2x.t/

�0 � �1

�x0.t/ � �2x.t/

� D 0; (9.3.9)

we can simplify it by defining y.t/ by

y.t/ WD x0.t/ � �2x.t/: (9.3.10)

As a result, the second-order equation (9.3.8), which we do not know yet how to solvedirectly, has been changed to the first-order equation

y0.t/ � �1y.t/ D 0; (9.3.11)

which we do know how to solve. All of its solutions are of the form y.t/ D k 1e�1t ,where k1 denotes a constant. Consequently, by (9.3.10),

x0.t/ � �2x.t/ D k1e�1t :

Multiplying by the integrating factor e ��2 t , we get

d

dt

he��2t x.t/

iD k1e.�1��2/t : (9.3.12)


If the roots are distinct (�2 ¤ �1/, we can integrate (9.3.12) and solve for x.t/ obtain-ing

x.t/ D k1

�1 � �2

e�1t C k2e�2t ; (9.3.13)

where k2 is a constant. We conclude that if a function x.t/ is a solution of (9.2.1)and the characteristic roots are the distinct numbers �1 and �2, then x.t/ is given by(9.3.13) for some k1 and k2.

On second thought we need to rethink the very last statement. It is certainly correctwhen �1 and �2 are real numbers. But what if they are complex numbers? Then thereis a problem because e�1t and e�2t would make no sense to someone who has not dealtwith complex powers before. In point of fact, it does make sense but the explanationof this will have to wait until Section 9.4. As for now, we will only deal with realcharacteristic roots.

Note that (9.3.13) is undefined when �2 D �1. We will investigate what to do whenthis happens after we have disposed of the case of distinct real roots.

9.3.1 Distinct Real Characteristic Roots

Let us replace the constants k1=.�1 � �2/ and k2 in (9.3.13) by c1 and c2 respectively.Then (9.3.13) simplifies to

x.t/ D c1e�1t C c2e�2t : (9.3.14)

We discovered in the previous section that if the characteristic roots �1 and �2 are realand distinct, then any solution x.t/ of (9.2.1) can be expressed in the form (9.3.14)for some c1 and c2. Conversely, any function of the form (9.3.14) is a solution of(9.2.1) if �1 and �2 are roots of (9.3.1). This is easy to verify: take the first and secondderivatives of (9.3.14), substitute them and x.t/ into the left-hand side of (9.2.1), anduse that �1 and �2 are solutions of (9.3.1). The result is zero. We conclude that (9.3.14)is the form in which all solutions of (9.2.1) can be expressed when the characteristicroots are real and distinct—for that reason it is known as a general solution of (9.2.1).Finally, we note that (9.3.14) is a solution of (9.2.1) on the interval .�1;1/.

Assigning actual values to the coefficients c1 and c2 results in a particular solutionof (9.2.1). For example, setting c1 D 2 and c2 D �3=7 gives the particular solution

x.t/ D 2e�1t � 3

7e�2t :

The particular solutionx.t/ D e�1t

is obtained by setting c1 D 1 and c2 D 0. Similarly, setting c1 D 0 and c2 D 1 givesthe solution

x.t/ D e�2t :

An expression of the form

c1e�1t C c2e�2t ;


where c1 and c2 are constants, is said to be a linear combination of the functionse�1t and e�2t . Accordingly, it follows from (9.3.14) and our discussion of the notionof a general solution that every solution x.t/ of (9.1) is some linear combination ofthe solutions e�1t and e�2t whenever �1 and �2 are distinct real characteristic roots.For this reason, the pair e�1t and e�2t are said to be a fundamental set of solutionsof (9.1). Furthermore, e�1t and e�2t are completely different from each other in thesense of not being proportional to each other. The mathematical term for this is linearindependence. That is to say, two functions are said to be linearly independent on aninterval J if neither is equal to a constant times the other on the interval. As this is thecase for e�1t and e�2t for �1 < t < 1, they are linearly independent solutions of(9.1) on the interval .�1;1/.

Theorem 9.1 summarizes the case of distinct real roots.

Distinct Real Characteristic Roots & General Solution

Theorem 9.1. If the characteristic equation for

ax00 C bx0 C cx D 0 (9.3.15)

has the distinct real roots �1 and �2, then e�1t and e�2t are linearly independentsolutions of (9.3.15) on the interval .�1;1/. Furthermore,

x.t/ D c1e�1t C c2e�2t ; (9.3.16)

where c1 and c2 are arbitrary constants, is a general solution on .�1;1/.

A simple mneumonic for finding a general solution of (9.3.15) is to associate thefunctions e�1t and e�2t with the characteristic roots �1 and �2:

�1 ! e�1t

�2 ! e�2t :

Then their linear combination gives the general solution (9.3.16) of (9.3.15).

Example 9.4. Find a general solution of 3x 00 � 13x0 � 10x D 0.

Solution. The characteristic equation is 3�2 �13��10 D 0. By the quadratic formula,its roots are

� D �.�13/˙p.�13/2 � 4.3/.�10/

2.3/D 13 ˙ p

289

6

D 13 ˙ 17

6D

8<ˆ:

�4

6D �2

3

30

6D 5:


But of course all of this arithmetic could be avoided by simply factoring the character-istic equation:

.3�C 2/.� � 5/ D 0:

At any rate, the characteristic roots are �1 D �2=3 and �2 D 5. The particularsolutions associated with these roots are

�1 D �2

3! e�2t=3

�2 D 5 ! e5t :

By Theorem 9.1, the linear combination

x.t/ D c1e�2t=3 C c2e5t :

is a general solution of the differential equation on .�1;1/.

9.3.2 Repeated Real Characteristic Roots

Theorem 9.1 pertains to real characteristic roots �1 and �2 that are not equal. However,if it also were true when a characteristic root � is repeated, then, as �1 D �2 D �, thetheorem would state that a general solution of (9.3.15) is

x.t/ D c1e�t C c2e�t D .c1 C c2/e�t :

This would say that every solution of (9.3.15) is merely a multiple of e �t . But thisflies in the face of the distinct roots case where two linearly independent solutions areneeded to express all solutions. The discrepancy is resolved when we come to realizeTheorem 9.1 does not apply to repeated roots because the integration yielding (9.3.13)is not valid when �2 D �1. This means we have to backtrack to (9.3.12) when �1 and�2 have the same value �. In that case,

d

dt

he��tx.t/

iD k1:

Changing the name of the constant k1 to c2 and integrating, we get

e��tx.t/ D c2t C c1:

So,x.t/ D .c1 C c2t/e�t :

In short, if the characteristic equation for (9.3.15) has the repeated root �, then anysolution of (9.3.15) can be expressed in the form

x.t/ D c1e�t C c2te�t (9.3.17)

for some c1 and c2. Conversely, it is left as an exercise (cf. Problem 2 on page 308)to show that if � is a repeated root of the characteristic equation for (9.3.15), then


any function of the form (9.3.17) is a solution of (9.3.15). Clearly, it is a solution on.�1;1/.

Setting c1 D 1 and c2 D 0, we obtain the particular solution x.t/ D e �t . Similarly,setting c1 D 0 and c2 D 1 gives the particular solution

x.t/ D te�t :

Observe that te�t is not a multiple of e �t , and vice versa, since t is a variable, not aconstant. Or, in the new terminology, e�t and te�t are linearly independent solutionsof (9.3.15). Our findings are summarized in the next theorem.

Repeated Characteristic Roots & General Solution


ax00 C bx0 C cx D 0 (9.3.18)

has the real repeated root �, then e�t and te�t are linearly independent solutionsof (9.3.18) on .�1;1/. Furthermore,

x.t/ D c1e�t C c2te�t ; (9.3.19)

where c1 and c2 are arbitrary constants, is a general solution on .�1;1/.

This theorem is easy to apply if we merely remember that two linearly independentsolutions associated with the repeated root � are

� !8<: e�t

te�t

and that their linear combination is a general solution.

Example 9.5. Find a general solution of 9x 00 C 24x0 C 16x D 0.

Solution. The characteristic equation is 9�2 C 24� C 16 D 0, the left-hand side ofwhich is the perfect square .3�C4/2. Now if we think of this as .3�C4/.3�C4/D 0,we see that � D �4=3 is a repeated root. From

� D �4

3!

8<: e�4t=3

te�4t=3;

we obtain the general solution

x.t/ D c1e�4t=3 C c2te�4t=3

for t 2 .�1;1/.


9.3.3 Operators and the Superposition Principle

Theorems 9.1 and 9.2 state that a linear combination of the two solutions of (9.2.1)corresponding to the two characteristic roots is also a solution. This suggests that linearcombinations of any solutions, regardless of how they are obtained, are also solutions.Let us see if this is true.

Our investigation will proceed more smoothly if we first introduce some new termsand notation. We will focus on functions f .t/ that are defined on an open intervalJ D .˛; ˇ/ and whose second derivatives f 00.t/ exist at each t in J . The reason isthat these are the kinds of functions we must deal with when searching for solutionsof (9.2.1). Such functions are said to be twice differentiable on J . We will assign toeach such f the output af 00 C bf 0 C cf , itself a function, where a, b, and c are realconstants. A concise way to express this is as follows:

f 7! af 00 C bf 0 C cf: (9.3.20)

Notice this too is a rule—similar to what is typically called a function. However itsdomain consists of a set of twice-differentiable functions rather than a set of numbers.So it would be far too confusing to also call this a function since associated with theword “function” is a domain consisting of real numbers. Instead this kind of rule isknown as a differential operator. We denote it with the symbol LŒ � �, where the “dot”is a placeholder for a twice-differential function. That is, LŒf � represents the output ofapplying the differential operator to f , namely, the right-hand side of (9.3.20):

LŒf � WD af 00 C bf 0 C cf: (9.3.21)

It is common to shorten this even more by writing

L WD aD2 C bD C c;

where D and D2 denote the operations of taking the first and second derivatives, re-spectively, of the function placed after them. In other words, D and D 2 are merely

alternate symbols for the differentiation operatorsd

dtand

d2

dt2, respectively. So the

result of applying them to a twice-differentiable function f is

Df .t/ D d

dtf .t/ D f 0.t/ and D2f .t/ D d2

dt2f .t/ D f 00.t/:

The way to write that L operates on f is

LŒf � D �aD2 C bD C c

�Œf � D aD2f C bDf C cf D a

d2f

dt2C b

df

dtC cf:

Note that the result is (9.3.21). The point to be made is that we can now represent thedifferential equation (9.2.1) in the abbreviated form

LŒx� D 0:


The operator L has the important property of linearity. What this means is that forany two functions f and g, both of which are twice differentiable on an interval J , andfor any two constants and �, the result of L operating on the sum f C �g is

LŒf C �g� D LŒf �C �LŒg�; (9.3.22)

where equality holds on the interval J . The sum f C �g is known as a linearcombination of the functions f and g. Property (9.3.22) follows directly from thedefinition of L:

LŒf C �g� D ad2

dt2.f C �g/C b

d

dt.f C �g/C c.f C �g/

D .af 00 C bf 0 C cf /C �.ag00 C bg0 C cg/ D LŒf �C �LŒg�:

Because L is expressed in terms of the differential operators D and D2 and has prop-erty 9.3.22, it is said to be a linear differential operator.

It is common to see (9.3.22) recast in the following two equivalent statements:

LŒf � D LŒf � (9.3.23)

andLŒf C g� D LŒf �C LŒg�; (9.3.24)

on J , where f , g, are as before. The reader should verify that (9.3.23) and (9.3.24)are equivalent to (9.3.22). Note that L becomes D if a D c D 0 and b D 1. Then(9.3.23) and (9.3.24) are the familiar statements that “the derivative of a constant timesa function is the constant times the derivative of the function” and “the derivative ofthe sum is the sum of the derivatives,” respectively.

Besides L, there are other well-known operators, such as the Laplace transform (cf.Chapter 10), that are used extensively in the sciences and engineering. Linear operatorsby definition have properties (9.3.23) and (9.3.24). If an operator fails to satisfy eitherone of them, or both, then it is not a linear operator—but rather a nonlinear operator.

Finally we are ready to show that all linear combinations of solutions of (9.2.1) arethemselves solutions.

Superposition Principle

Theorem 9.3. If x1.t/ and x2.t/ are solutions of

ax00 C bx0 C cx D 0

on .�1;1/, then so isc1x1.t/ C c2x2.t/

regardless of the values of the constants c1 and c2.


Proof. Let x1.t/ and x2.t/ be any two solutions of LŒx� D 0 on .�1;1/, whereL D aD2 C bD C c. This means that

LŒx1.t/� D 0 and LŒx2.t/� D 0

for all �1 < t < 1. Then, by (9.3.22), for any constants c1 and c2, we have

LŒc1x1.t/C c2x2.t/� D c1LŒx1.t/� C c2LŒx2.t/� D c1 � 0 C c2 � 0 D 0

for all �1 < t < 1.

9.3.4 Complex Characteristic Roots

Introduction to Complex Numbers

Now let us determine the solutions of second-order linear equations when their charac-teristic equations have complex roots. This occurs when b2 � 4ac < 0. To see this, letus rewrite (9.3.3) as

� D � b

2a˙

pb2 � 4ac

2aD � b

2a˙

p�1 �p

4ac � b2

2a

and then introduce some new notation. Since 4ac � b2 > 0,p

4ac � b2 is a positivereal number. But

p�1 is different: it is not even a real number. Why? The answerhas to do with the meaning of

px. It is defined as the nonnegative real number whose

square is x. According to this then,p�1 should represent a nonnegative real number

whose square is �1. But that cannot be: the square of any real number is nonnegative.Nonetheless, we can still say it denotes a number, just not a real number—instead, it isa so-called imaginary number5 and is usually denoted by the letter i .6 Thus, i D p�1

and i2 D �1. By defining the real numbers ˛ and ! by

˛ WD � b

2aand ! WD

p4ac � b2

2a; (9.3.25)

the expressions for the two characteristic roots simplify to ˛C i! and ˛ � i!, therebymaking them less tedious to deal with in the future. Since a is any nonzero real numberand

p4ac � b2 > 0, it follows that ! is a nonzero real number.

Numbers that are not real can always be expressed in the form a C ib, where a

and b ¤ 0 are real numbers. These numbers are called complex numbers.7 It is easy

5Imaginary numbers are numbers of the form bi, where b is any real number. The derivation of theterm “imaginary” goes back to the 17th century when they were first being discovered and were not fullyunderstood. The term is a misnomer because there is nothing unreal about imaginary numbers if they arecorrectly interpreted.

6In some disciplines, such as electrical engineering, j is used because i is reserved for designating elec-trical current.

7Every imaginary number is a complex number: for example, 5i can also be written as 0 C 5i. Every realnumber is a complex number too: for example, 5 can also be written as 5 C 0i. But, of course, not everycomplex number is a real number. It boils down to this: The set of all complex numbers is larger than the setof real numbers in the sense that it includes all of the real numbers.


to show that complex numbers can be handled algebraically in the same way as realnumbers. It follows from (9.3.3) that one characteristic root cannot be complex andthe other one real. Either both roots are complex or both are real. Since characteristicroots always occur as an ˛ ˙ i! pair, they are said to be complex conjugates of eachone other. The word conjugate used in this sense means joined together in a pair.Ordinarily, a bar is placed over a complex number to denote its conjugate. With thisnotation, the two characteristic roots are

� D ˛ C i! and � D ˛ � i!: (9.3.26)

The real number ˛ is called the real part of the complex number �. It is convenient todesignate the real part of � like this:

Re � D ˛:

The real number ! next to i is called the imaginary part of � and we write

Im � D !:

Note that the imaginary part does not include i . Thus, Re � D ˛ and Im � D �!.

Remark 9.1. Observe that we could replace ! with �! in (9.3.26) and nothing changes:we still have the same pair of complex numbers.

Simple Harmonic Oscillator and Solutions

When the characteristic roots are complex numbers, we expect the mathematics to bemore involved than when they are real numbers, because somehow we will have tofigure out how to use the complex roots to obtain real-valued solutions of (9.2.1). Atthis point, we have no idea where to even start. So instead of tackling the generaldifferential equation (9.2.1) head on, let us consider a specific example: let a D 1,b D 0, c D w2, where ! is any nonzero real number. With these values, (9.2.1)becomes

x00 C !2x D 0: (9.3.27)

The roots of (9.3.27) are complex numbers since b2 � 4ac D �4!2 < 0. Recall that(9.3.27) is the equation of motion of a simple harmonic oscillator when the independentvariable t denotes time and ! > 0.

We will prove a number of results for (9.3.27) because, as it turns out, these resultswill enable us to find the general solution of any second-order linear equation withcomplex characteristic roots.

Lemma 9.1. Let t0 and ! be given real numbers where ! ¤ 0. The initial valueproblem

x00 C !2x D 0 (9.3.28a)

x.t0/ D 0; x0.t0/ D 0 (9.3.28b)

has the unique solution x.t/ � 0 on .�1;1/.


Proof. Obviously the function x.t/ D 0 solves the differential equation (9.3.28a) for�1 < t < 1. It also satisfies the initial conditions (9.3.28b). So we only have toestablish that there is no other such function. To that end, suppose a function u.t/

solves (9.3.28a) for �1 < t < 1 and also satisfies (9.3.28b). In other words, wesuppose that

u00.t/ C !2u.t/ D 0 (9.3.29)

andu.t0/ D 0; u0.t0/ D 0: (9.3.30)

Now multiply (9.3.29) by 2u 0.t/ and then integrate the result from t0 to t :Z t

t0

2u0.s/�u00.s/C !2u.s/

�ds D

Z t

t0

0 ds: (9.3.31)

By the Fundamental Theorem of Calculus and (9.3.30),Z t

t0

�2u0.s/u00.s/C 2!2u.s/u0.s/

�ds D

nŒu0.s/�2 C !2Œu.s/�2

oˇtt0

D Œu0.t/�2 C !2Œu.t/�2 � Œu0.t0/�2 � !2Œu.t0�2

D Œu0.t/�2 C !2Œu.t/�2:

Since the right-hand side of (9.3.31) is equal to 0, we have

Œu0.t/�2 C !2Œu.t/�2 D 0: (9.3.32)

Since both terms on the left-hand side of (9.3.32) are nonnegative, each must in fact beequal to 0. Finally, as !2 > 0, we are forced to conclude that u.t/ � 0.

Remark 9.2. Lemma 9.1 has an obvious physical interpretation. If at time t D t0, themass of a horizontal mass-spring system is located at the origin (the position where thespring is neither stretched nor compressed) and it has no velocity, then it will continueto remain there unless a force is exerted on it at some future time.

Lemma 9.2. Let t0, �, �, ! be given real numbers where ! ¤ 0. If

x00 C !2x D 0

x.t0/ D �; x0.t0/ D �

has a solution on .�1;1/, then it is unique.

Proof. Suppose x.t/ and Qx.t/ are solutions of this initial value problem on the interval.�1;1/. Consider the difference u.t/ WD x.t/� Qx.t/. By the SuperpositionPrinciple,u is also a solution of x 00 C !2x D 0. Moreover,

u.t0/ D x.t0/ � Qx.t0/ D � � � D 0


andu0.t0/ D x0.t0/ � Qx0.t0/ D � � � D 0:

This makes u a solution of the initial value problem in Lemma 9.1. Consequently,u.t/ � 0. Therefore, x.t/ � Qx.t/ on .�1;1/.

We have just shown that there can be at most one solution of the initial value prob-lem in Lemma 9.2. However, aside from the zero solution in Lemma 9.1 when � D 0

and � D 0, we have not proved that it actually has a solution. Does it? It does in fact,regardless of the values of � and �, as we will now prove. But first let us point outthat in both of the lemmas, ! represents any nonzero real number. But in the followingtheorem, we state that ! is positive. As it turns out, there is no loss of generality indoing so as we will point after the proof of the theorem.

Simple Harmonic Oscillator and Solutions I

Theorem 9.4. Let t0, �, � be real numbers where ! > 0. There exists a uniquesolution of

x00 C !2x D 0 (9.3.33)

on .�1;1/ satisfying the initial conditions

x.t0/ D �; x0.t0/ D �: (9.3.34)

This solution can be expressed in the form

x.t/ D c1 cos!t C c2 sin!t ; (9.3.35)

where the constants c1 and c2 are uniquely determined by (9.3.34).

Proof. Recall that (9.3.33) models the motion of the mass of an undamped mass-springsystem when ! > 0. Because of the total absence of damping, our intuition suggeststhat the mass of such a system will move back and forth in the same repetitious mannerforever. This suggests that sine and cosine functions may be solutions. Let us confirmthis by substituting cos!t for x in t(9.3.33):

x00 C !2x D d2

dt2cos!t C !2 cos!t D �!2 cos!t C !2 cos!t D 0:

Likewise, it is easy to show that x D sin!t is also a solution of (9.3.33). It followsfrom the Superposition Principle that all linear combinations of cos!t and sin!t aresolutions.

So far we have established that all functions of the form (9.3.35) are solutions of(9.3.33). On the other hand, we have not ruled out the possibility that (9.3.33) may haveother kinds of solutions. But let us first determine if (9.3.33) always has a solution(regardless of the initial conditions). And if it has, can it be expressed in the form(9.3.35)? If so, this would rule out the existence of other kinds of solutions.


Since we have already determined that (9.3.35) is a solution of (9.3.33) for allvalues of c1 and c2, let us see if a pair of values for c1 and c2 exists that will alsosatisfy the initial conditions (9.3.34), where � and � are any given real numbers. Itboils down to this: Does the system of equations

c1 cos!t0 C c2 sin!t0 D �

�c1! sin!t0 C c2! cos!t0 D �

have a solution? The answer is yes. This can be seen by multiplying the first equationby ! cos!t0, the second one by � sin!t0, and adding the results. For c1, we obtainthe value

c1 D � cos!t0 � �

!sin!t0: (9.3.36)

Similarly,

c2 D � sin!t0 C �

!cos!t0: (9.3.37)

Therefore, x.t/ D c1 cos!t C c2 sin!t , with c1 and c2 given by (9.3.36) and (9.3.37),solves the differential equation (9.3.33), and it also satisfies the initial conditions (9.3.34).Finally, Lemma 9.2 asserts this is the only solution!

Remark 9.3. In Theorem 9.4, we state that ! > 0. But what if ! < 0? Then �! > 0.In that case, by (9.3.35), a general solution of (9.3.33) is

x.t/ D c1 cos.�!t/ C c2 sin.�!t/:

Recall that the cosine and sine functions are even and odd functions, respectively. Thuscos.�!t/ D cos.!t/ and sin.�!t/ D � sin.!t/. Letting k1 WD c1 and k2 WD �c1, wehave

x.t/ D c1 cos.!t/ � c2 sin.!t/ D k1 cos.!t/ C k2 sin.!t/:

The point to be made is that we still obtain the same general solution. So from now onwe will always choose ! > 0 to avoid the hassle of dealing with a negative sign.

Example 9.6. Find the solution of the initial value problem

d2x

dt2C 4x D 0I x.0/ D 2; x0.0/ D 4

p3:

Solution. Since !2 D 4, ! D 2 by Remark 9.3. By (9.3.35), the solution has the form

x.t/ D c1 cos 2t C c2 sin 2t :

Now we must find values of c1 and c2 to satisfy the initial conditions. To that end, sett equal to 0 in x.t/ and its derivative

x0.t/ D �2c1 sin 2t C 2c2 cos 2t


and then set them equal to 2 and 4p

3 respectively. The result is the system

c1 cos 0 C c2 sin 0 D 2

�2c1 sin 0 C 2c2 cos 0 D 4p

3;

which has the solution c1 D 2 and c2 D 2p

3. Therefore, the solution of the initialvalue problem is

x.t/ D 2 cos 2t C 2p

3 sin 2t :

There are other ways of expressing solutions of the undamped mass-spring systemthat are more suited for graphing and determining the frequency of the oscillation of themass m. Referring back to the form of the general solution (9.3.35), define a constantA by

A WDq

c21 C c2

2 (9.3.38)

and the angle ' byA cos' D c1 and A sin' D c2: (9.3.39)

Then, by one of the addition formulas of trigonometry, (9.3.35) becomes

x.t/ D A cos ' cos!t C A sin ' cos!t D A cos .!t � '/ :By (9.3.38), A � 0. As a result, we have the following alternative form of a generalsolution of a simple harmonic oscillator.

Simple Harmonic Oscillator and Solutions II

Theorem 9.5. Let ! > 0. A general solution of

x00 C !2x D 0

on .�1;1/ isx.t/ D A cos .!t � '/; (9.3.40)

where A � 0 and ' are arbitrary constants.

Remark 9.4. If we define ' by

A sin' D c1 and A cos ' D c2

rather than by (9.3.39), then we obtain

x.t/ D A sin .!t C '/

as the general solution. Moreover, as ' may be positive or negative,

A sin .!t � '/ and A cos .!t C '/

are also general solutions. The one that is used is a matter of personal preference.


Example 9.7. Find the solution of the initial-value problem

x00 C 4x D 0I x.0/ D 2; x0.0/ D 4p

3:

Solution. By (9.3.40),x.t/ D A cos .2t � '/

is a general solution of x 00 C 4x D 0. Its derivative is x 0.t/ D �2A sin .2t � '/.Because of the initial conditions,

A cos .�'/ D A cos ' D 2 and � 2A sin .�'/ D 2A sin' D 4p

3:

Thus,

cos' D 2

Aand sin' D 2

p3

A:

As cos2 ' C sin2 ' D 1, �2

A

�2

C

2p

3

A

!2

D 1:

Solving, we get A D 4. Hence,

cos ' D 1

2and sin' D

p3

2:

This implies ' D �=3. Therefore, a general solution is

x.t/ D 4 cos2t � �

3

:

Example 9.8. For the horizontal mass-spring system in Fig. 9.1, determine the signif-icance of the parameters A and ' in (9.3.40). Also, determine the significance of !other than being defined in terms of the mass m and spring constant k by

pk=m.

Solution. Since the cosine function takes on values only from �1 to 1, the minimumand maximum values of (9.3.40) are �A and A, respectively. Hence, A is the maximumdisplacement of m from its equilibrium position and is called the amplitude of themotion.

As for ', compare the graphs of x.t/ D A cos .!t � '/ and A cos!t . Since

A cos .!t � '/ D A cos .!.t � '=!//;we see that the graph of x.t/ can be obtained from the graph of A cos!t by shifting itto the right by '=! units. The quantity !t � ' is called the phase of the motion andthe constant ' is called the phase angle.

Since the cosine is a periodic function with period 2� ,

x.t/ D A cos .!t � '/ D A cos .!t � ' C 2�/

D A cos

�!

�t C 2�

!

�� '

�

D x

�t C 2�

!

�:


Therefore,

T WD 2�

!

is the period of the motion, the time it takes for m to make one complete oscillation.Related to this is the frequency of motion, which is defined as the number of completeoscillations per unit of time. Denoting it by f , we have

f D 1

TD !

2�D 1

2�

rk

m: (9.3.41)

Example 9.9. Suppose the mass-spring system in Example 9.8 is modified by attachinga second massless spring as shown in Fig. 9.3. If the spring constants of the left and

Fig. 9.3: Two springs and mass system

right springs are k1 and k2, respectively, determine the position of m at time t and itsfrequency of oscillation.

Solution. Suppose at some point in time the mass m is located x units to the rightof its equilibrium position (the position where both springs are neither stretched norcompressed). Then the left spring must be stretched beyond its natural length, say byx1 units. Consequently the right spring is stretched by x 2 units, where x2 D x � x1.By Hooke’s law, it exerts a force �k2x2 on m. It follows from Newton’s third lawof motion that m exerts an equal but oppositely directed force F on the right spring. 8

Thus,F D k2x2:

The right spring also exerts a force k2x2 on the left spring which in turn exerts a force�k1x1 on the right spring. Applying Newton’s third law to this pair of forces, we have

k1x1 D k2x2:

8Isaac Newton’s laws of motion first appeared in a three-volume work written in Latin by him and pub-lished in 1687 called the Philosophiæ Naturalis Principia Mathematica (Latin for “mathematical principlesof natural philosophy”). A typical translation of the third law is: “To every action there is always opposed anequal reaction.” Action and reaction refer to the pair of forces that two interacting bodies exert on each other.In present-day vernacular, Newton’s third law of motion states that if a body A exerts a force on a body B,then body B exerts an equal but oppositely directed force on body A.


Hence,

x D x1 C x2 D F

k1

C F

k2

D F

�k2 C k1

k1k2

�:

Solving for F , we get

F D�

k1k2

k1 C k2

�x: (9.3.42)

It is left to the reader to show that (9.3.42) is also the result when the mass m is locatedx units to the left of its equilibrium position.

Equation (9.3.42) implies that for a given set of initial conditions the motion of amass attached to two springs with spring constants k1 and k2 is identical to the motionof the same mass attached to a single spring with spring constant

k WD k1k2

k1 C k2

:

Consequently, by (9.3.41), m oscillates with the frequency

f D 1

2�

rk

mD 1

2�

sk1k2

m.k1 C k2/:

Thus, its position x.t/ is given by (9.3.40), where the so-called angular frequency is

! D 2�f D 2�

T

and the amplitude A and phase angle ' are determined by the initial position and ve-locity of m.

The General Case

The importance of Theorem 9.4 is that it will enable us to find a general solution of

ax00 C bx0 C cx D 0 (9.3.43)

when the characteristic roots are complex numbers. To this end, let � D ˛ C i! and� D ˛ � i! denote these roots, where ˛ and ! are given by (9.3.25). We may supposethat ! > 0 by Remark 9.3. Now replace the coefficients a, b, c in (9.3.43) with ˛ and!. To do this, first divide (9.3.43) by a:

x00 C b

ax0 C c

ax D 0: (9.3.44)

By (9.3.7), the coefficients of (9.3.44) are

b

aD �.�C �/ D �2˛

andc

aD � � � D .˛ C i!/.˛ � i!/ D ˛2 C !2:


Consequently, in terms of ˛ and !, (9.3.44) becomes

x00 � 2˛x0 C .˛2 C !2/x D 0: (9.3.45)

In short, if the characteristic equation for (9.3.43) has the complex roots ˛ ˙ i!, thenequations (9.3.43) and (9.3.45) are equivalent in that they have the same solutions. Itfollows that a general solution of (9.3.43) is a general solution of (9.3.45) and con-versely.

So let us find a general solution of (9.3.45). First notice that the characteristicequation for

x00 � 2˛x0 C ˛2x D 0 (9.3.46)

has the repeated real root ˛. By Theorem 9.4, a general solution of (9.3.46) is

x.t/ D c1e˛t C c2te˛t :

Thus, when ˛ ¤ 0, e˛t is a solution of (9.3.46) but not of (9.3.45). This suggests thatsomething may come of expressing a solution x.t/ of (9.3.45) in the form

x.t/ D e˛ty.t/

for some twice-differentiable function y.t/. Replacing x in the left-hand side of (9.3.45)with e˛t y, we obtain

x00�2˛x0 C .˛2 C !2/x D .˛2 C !2/.e˛t y/ � 2˛d

dt.e˛t y/ C d2

dt2.e˛ty/

D ˛2e˛ty C !2e˛ty � 2˛.e˛ty0 C ˛e˛ty/C e˛ty00 C 2˛e˛ty0 C ˛2e˛t y

D e˛t.y00 C !2y/:

From this, it follows:

If y.t/ is a solution ofy00 C !2y D 0; (9.3.47)

then x.t/ D e˛t y.t/ is a solution of (9.3.45). Conversely, if x.t/ is asolution of (9.3.45), then y.t/ D e �˛t x.t/ is a solution of (9.3.47).

By Theorem 9.4, a general solution of (9.3.47) is

y.t/ D c1 cos!t C c2 sin!t :

Consequently, as (9.3.45) is equivalent to (9.3.43), we conclude:

If x.t/ is a solution of (9.3.43), then

x.t/ D e˛t .c1 cos!t C c2 sin!t/

for some constants c1 and c2.


Setting c1 D 1 and c2 D 0 yields the particular solution e ˛t cos!t . Likewise, c1 D 0

and c2 D 1 yields the particular solution e ˛t sin!t . They are clearly linearly indepen-dent, i.e., not multiples of each other.

By Theorem 9.5, x.t/ can also be expressed in the form

x.t/ D Ae˛t cos .!t � '/;for some constants A � 0 and '.

In conclusion, we have found two forms of a general solution of (9.3.45)—andthereby of (9.3.43) in the foregoing discussion. Theorem 9.6 summarizes the foregoingresults.

Complex Characteristic Roots & General Solution


ax00 C bx0 C cx D 0 (9.3.48)

has the complex roots ˛ ˙ i!, where ! > 0, then e˛t cos!t and e˛t sin!t arelinearly independent solutions of (9.3.48) on .�1;1/. A general solution on.�1;1/ is

x.t/ D c1e˛t cos!t C c2e˛t sin!t ; (9.3.49)

where c1 and c2 are arbitrary constants. An alternative form is

x.t/ D Ae˛t cos .!t � '/; (9.3.50)

where A � 0 and ' are arbitrary constants.

A mnemonic is to associate the two particular solutions e ˛t cos!t and e˛t sin!t

with the characteristic roots � D ˛ C i! and � D ˛ � i!:

˛ ˙ i! !8<:e˛t cos!t

e˛t sin!t :

A general solution of (9.3.48) is a linear combination of these two particular solutions.

Example 9.10. Find a general solution of 2x 00 � 2x0 C 5x D 0.

Solution. The characteristic equation is 2�2 � 2�C 5 D 0. By the quadratic formula,

� D 2 ˙p.�2/2 � 4.2/.5/

2.2/D 2 ˙ p�36

4D 2 ˙ 6i

4D 1

2˙ 3

2i:

An alternative to the quadratic formula is to divide the characteristic equation by thecoefficient of �2 and then to complete the square as follows:

�2 � �C 1

4D �5

2C 1

4)

�� 1

2

�2

D �9

4:


Since � � 12

D ˙ 32i , the characteristic roots are � D 1

2˙ 3

2i . From

1

2˙ 3

2i !

8<ˆ:

et=2 cos3t

2

et=2 sin3t

2

and (9.3.49), a general solution is

x.t/ D c1et=2 cos3t

2C c2et=2 sin

3t

2:

Alternatively,

x.t/ D Aet=2 cos

�3t

2� '

�by (9.3.50).


x00 � 2x0 C 10x D 0I x.0/ D 0; x0.0/ D 6:

Solution. The characteristic equation is �2 � 2� C 10 D 0, which can be written as.� � 1/2 D �9 by completing the square. Consequently, the characteristic roots are

� D 1 ˙ 3i:

By (9.3.49), a general solution is

x.t/ D c1et cos 3t C c2et sin 3t :

As x.0/ D c1, it follows that c1 D 0 from the initial condition x.0/ D 0. Thus,

x.t/ D c2et sin 3t :

Differentiation of x.t/ with respect to t yields

x0.t/ D 3c2et cos 3t C c2et sin 3t :

As x0.0/ D 3c2 and x0.0/ D 6, c2 D 2. Therefore,

x.t/ D 2et sin 3t : (9.3.51)

It is left as an exercise to show that (9.3.50) gives the alternative form

x.t/ D 2et cos3t � �

2

and that it is equal to (9.3.51).


9.4 Complex Exponential Functions

If the characteristic equation for a linear second-order differential equation has a realroot �, one of its solutions is the exponential function e �t . What if � is a complexnumber? Is it still correct to say e�t is a solution? But there is a more basic question:Does it make sense to raise e to a complex power? To answer this, we consider thefollowing two examples.


x00 C x D 0I x.0/ D 1; x0.0/ D 0: (9.4.1)

Solution. The characteristic equation is

�2 C 1 D 0:

Thus, � D ˙i are the characteristic roots. Now for a new idea—well new to us butreally an old one. Suppose we try to obtain solutions corresponding to these complexcharacteristic roots just as we did from real roots. If this could be justified somehow,then it seems plausible that a general solution of the differential equation in (9.4.1) is

x.t/ D c1eit C c2e�it : (9.4.2)

But how do we define e it and e�it so that (9.4.2) makes sense and is really a solution?To answer this, let us begin by defining the derivatives of e it and e�it by the formulas

d

dteit D ieit and

d

dte�it D �ie�it

in order to be consonant with the differentiation formula

d

dte�t D �e�t

when � is a real number. Then the derivative of (9.4.2) is

x0.t/ D ic1eit � ic2e�it : (9.4.3)

From (9.4.2), (9.4.3), and the initial conditions, we obtain the system

c1 C c2 D 1 (9.4.4)

ic1 � ic2 D 0:

Its solution is c1 D c2 D 1=2. With these values (9.4.2) becomes

x.t/ D 1

2eit C 1

2e�it : (9.4.5)

But is (9.4.5) a legitimate way of expressing the solution? And if it is, how do weinterpret it? The answer is found by comparing (9.4.5) to what we truly know is thesolution. By inspection, x.t/ D cos t is a solution of (9.4.1). By Theorem 9.4, it is

9.4. COMPLEX EXPONENTIAL FUNCTIONS 267

the only solution of (9.4.1)! It follows that if we want to view (9.4.5) as another formof this solution, we have to say that it is equal to cos t . In short, any definition of e it

would have to be compatible with the formula

cos t D eit C e�it

2: (9.4.6)


x00 C x D 0I x.0/ D 0; x0.0/ D 1: (9.4.7)

Solution. The only differences between (9.4.1) and (9.4.7) are the initial values. Chang-ing (9.4.4) to reflect these new values, we have

c1 C c2 D 0 (9.4.8)

ic1 � ic2 D 1:

which has the solution c1 D 1=2i , c2 D �1=2i . Thus,

x.t/ D 1

2ieit � 1

2ie�it : (9.4.9)

Again, as was the case with (9.4.5), how is this to be interpreted? By inspection andTheorem 9.4, x.t/ D sin t is the unique solution of (9.4.7). So, if we want to be ableto say that (9.4.9) is another way of expressing this solution, we have to interpret it asbeing equal to sin t :

sin t D eit � e�it

2i: (9.4.10)

Now consider the linear combination cos t C i sin t . If we replace cos t and sin t

with the right-hand sides of (9.4.6) and (9.4.10), respectively, we obtain the well-knownand very useful formula

cos t C i sin t D eit C e�it

2C i

eit � e�it

2iD eit :

This is how we define eit since it is compatible with both (9.4.6) and (9.4.10). It is awell-known and very useful formula named after Leonhard Euler.

Euler’s Formula

Definition 9.2. For any real number t , the complex exponential e it is defined by

eit WD cos t C i sin t: (9.4.11)

A few noteworthy observations are:


1. Replacing t with �t , we get e i.�t/ D cos.�t/C i sin.�t/. Since cos t is an evenfunction and sin t is an odd function, this simplifies to

e�it WD cos t � i sin t: (9.4.12)

2. Differentiating (9.4.11) and (9.4.12) with respect to t , we obtain the formulas(see Problem 34)

d

dteit D ieit and

d

dte�it D �ie�it : (9.4.13)

3. Setting t D � , we obtain e i� D cos� C i sin� D �1 C i0, which simplifies to

ei� D �1: (9.4.14)

This equation is remarkable in that it relates three important constants innate todifferent branches of mathematics: � from geometry, Euler’s number e fromcalculus and real analysis, and i from complex analysis and abstract algebra.

4. Let us extend the laws of exponents for real numbers to complex numbers so that

ea˙ib D ea � e˙ib :

As a result, we have the following generalized versions of (9.4.11) and (9.4.12):

ea˙ib WD ea.cos b ˙ i sin b/: (9.4.15)

Now that we have managed to give meaning to complex exponential functions, areview of the work in Section 9.3 that led to (9.3.13) and (9.3.14) and which appearedat the time to be valid only for real characteristic roots is in fact valid for complexcharacteristic roots as well. So allowing for complex-valued solutions, we have thefollowing extension of Theorem 9.1.

Distinct Characteristic Roots & General Solution


ax00 C bx0 C cx D 0 (9.4.16)

has the distinct (real or complex) roots �1 and �2, then

x.t/ D c1e�1t C c2e�2t ; (9.4.17)

where c1 and c2 are arbitrary constants, is a general (complex-valued) solutionon .�1;1/.

9.4. COMPLEX EXPONENTIAL FUNCTIONS 269

Example 9.14. Use Theorem 9.7 to find the solution of the initial value problem

x00 C 4x D 0I x.0/ D 1; x0.0/ D 2i:

Solution. The characteristic roots are � D ˙2i . It follows from (9.4.17) that a generalsolution of the differential equation is

x.t/ D c1e2it C c2e�2it :

Differentiating with respect to t , we get

x0.t/ D 2ic1e2it � 2ic2e�2it :

From x.0/, x0.0/, and the initial conditions, we obtain the system of equations

c1 C c2 D 1

2ic1 � 2ic2 D 2i:

By inspection,c1 D 1 and c2 D 0:

Therefore, the solution of the initial value problem is the complex-valued function

x.t/ D e2it D cos 2t C i sin 2t:


x00 � 2x0 C 10x D 0I x.0/ D 0; x0.0/ D 6:

Solution. The characteristic roots are � D 1 ˙ 3i . By (9.4.17), a general solution ofthe differential equation is

x.t/ D c1e.1C3i/t C c2e.1�3i/t :

Differentiation of x.t/ with respect to t yields

x0.t/ D .1 C 3i/c1e.1C3i/t C .1 � 3i/c2e.1�3i/t :

Setting t D 0 and using the initial conditions, we get

c1 C c2 D 0

.1 C 3i/c1 C .1 � 3i/c2 D 6;

which has the solutionc1 D �i and c2 D i:

Thus,x.t/ D �ie.1C3i/t C ie.1�3i/t D �iet.e3it � e�3it /:

At first this solution appears to be complex-valued; however, by (9.4.10), we see that itsimplifies to

x.t/ D �iet

�e3it � e�3it

2i

�.2i/ D 2et sin 3t :

and is real-valued. Note that we obtained this very same result earlier in Example 9.11.


9.5 Cauchy-Euler Equations

Even though the procedure for finding solutions of a homogeneous linear equation

ad2x

dt2C b

dx

dtC cx D 0

with constant coefficients a, b, c is relatively simple, it is quite another story if anyof the coefficients vary with t . Unfortunately the characteristic roots method does notwork for

p.t/d2x

dt2C q.t/

dx

dtC r .t/x D 0 (9.5.1)

if any of p, q, r vary with t . In fact, solutions of (9.5.1) generally cannot be expressedin terms of finitely many elementary functions. Other means such as numerical tech-niques or power series methods are generally needed to handle such equations. How-ever, if the coefficients have the special form

p.t/ D at2; q.t/ D bt; and r .t/ D c

where a, b, and c are constants, then expressions consisting of elementary functionsfor the solutions of(9.5.1) can be found. The name associated with such equations isCauchy-Euler, or sometimes just Euler.

Homogeneous Cauchy-Euler Equations

Definition 9.3. A second-order homogeneous Cauchy-Euler equation is anequation of the form

at2 d2x

dt2C bt

dx

dtC cx D 0 (9.5.2)

where a, b, c are constants.

Fortunately, we can change (9.5.2) to a linear homogeneous equation with constantcoefficients. One way to achieve this is to change the independent variable t to anothervariable, say u. If we are looking for a solutionon an interval where t is always positive,we can define u by u WD ln t or equivalently by

t D eu: (9.5.3)

First let us find the relation between the derivatives x 0.t/ and x0.u/. By the chain rule,it is

dx

duD dx

dt� dt

du: (9.5.4)

Asdt

duD d

dueu D eu D t;

9.5. CAUCHY-EULER EQUATIONS 271

this simplifies to

tdx

dtD dx

du: (9.5.5)

Next let us find the relation between the second-order derivatives x 00.t/ and x00.u/.By (9.5.5) and the product and chain rules, we obtain

d2x

du2D d

du

�tdx

dt

�D t

d

du

�dx

dt

�C dx

dt� dt

duD t

d

dt

�dx

dt

�� dt

duC dx

du:

As dt=du D t , this simplifies to

d2x

du2D t2 d2x

dt2C dx

du

and so

t2 d2x

dt2D d2x

du2� dx

du: (9.5.6)

Note that with the change in variables we have managed to transform the variablecoefficient terms to constant coefficient terms. Substituting (9.5.5) and (9.5.6) into(9.5.2), we obtain the equation

a

�d2u

du2� dx

du

�C b

dx

duC cx D 0;

which simplifies to

ad2x

du2C .b � a/

dx

duC cx D 0: (9.5.7)

The characteristic equation for (9.5.7) is

a�2 C .b � a/�C c D 0: (9.5.8)

Because of its connection to the Cauchy-Euler equation (9.5.2), we call (9.5.8) theCauchy-Euler characteristic equation.

Now we will show how to find a general solution of (9.5.2) by first finding a generalsolution of (9.5.7). Since the characteristic roots �1 and �2 of (9.5.8) may be real anddistinct numbers, real and equal numbers, or complex numbers, there are three separatecases to consider.

Case 1. The roots �1, �2 are real and �1 ¤ �2.

The general solution of (9.5.7) is

x.u/ D c1e�1u C c2e�2u:

Use (9.5.3) to revert back to the original variable t : e �1u D .eu/�1 D t�1 . Similarly,e�2u D t�2 . Thus, a general solution of (9.5.2) for this case is

x.t/ D c1t�1 C c2t�2 : (9.5.9)


Case 2. The roots �1, �2 are real and �1 D �2.

This is the repeated roots case. Let � WD �i for i D 1; 2. A general solution of (9.5.7)is

x.u/ D c1e�u C c2ue�u;

From (9.5.3), u D ln t and so a general solution of (9.5.2) for this case is

x.t/ D c1t� C c2t� ln t: (9.5.10)

Case 3. The roots �1, �2 are complex numbers.

Let �1 D ˛C i! and �2 D ˛� i!, where ! > 0, denote the two complex roots. Thena general solution of (9.5.7) is

x.u/ D c1e˛u cos!u C c2e˛u sin!u:

As u D ln t , a general solution of (9.5.2) for the case of complex roots is

x.t/ D c1t˛ cos.! ln t/C c2t˛ sin.! ln t/: (9.5.11)

If we seek solutions on intervals where t is always negative, then the definition ofu given in (9.5.3) must be changed to �t D eu. Then

dx

duD dx

dt� dt

duD dx

dt� d

du

� � eu� D t

dx

dt;

which is (9.5.5) again. Likewise (9.5.6) remains the same. Consequently the charac-teristic equation is still given by (9.5.8). The only change occurs in reverting back tothe original variable t : u D ln.�t/ rather than u D ln t .

There is no need to memorize the general solutions for all three cases. All we needto remember is the following method.

Method for Solving Homogeneous Cauchy-Euler Equations

To find a general solution of

at2 d2x

dt2C bt

dx

dtC cx D 0 (9.5.12)

on an interval that excludes t D 0, complete the following steps:

1. Find a general solution of the associated equation

ad2x

du2C .b � a/

dx

duC cx D 0: (9.5.13)

2. After completing Step 1, replace u with ln jt j and eu with jt j.

9.5. CAUCHY-EULER EQUATIONS 273

Example 9.16. Find a general solution of

t2x00 C tx0 � 4x D 0 (9.5.14)

on the interval .0;1/.

Solution. Comparing (9.5.14) to (9.5.12), we see that a D 1, b D 1, and c D �4. So,b � a D 0 and the associated equation is

d2x

du2� 4x D 0;

where u D ln jt j D ln t . The characteristic roots are � D ˙2. A general solution ofthe associated equation is

x.u/ D c1e�2u C c2e2u:

Since eu D t ,e�2u D .eu/�2 D t�2:

Similarly, e2u D t2. Therefore a general solution of (9.5.14) is

x.t/ D c1

t2C c2t2:


4t2x00 C x D 0I x.1/ D 2; x0.1/ D 0: (9.5.15)

Solution. Since a D 4, b D 0, c D 1, and b � a D �4, the associated equation is

4d2x

du2� 4

dx

duC x D 0:

The solution we seek is defined on .0;1/ since the initial conditions are given att D 1, Thus, the relation between u and t is u D ln t . The characteristic equationis .2� � 1/2 D 0. So, � D 1=2 is a repeated root. Hence, a general solution of theassociated equation is

x.u/ D c1eu=2 C c2ueu=2

Since u D ln t and eu D t , a general solution of (9.5.15) is

x.t/ D c1

pt C c2

pt ln t:

Finally, we need values for c1 and c2 that will satisfy the initial conditions. Since

x.1/ D c1 C c2 � 0;

we find c1 D 2: Differentiating, we have

x0.t/ D 2d

dtt1=2 C c2

d

dtt1=2 ln t D 1p

tC c2

�1pt

C ln t

2p

t

�:

Letting t D 1, we obtain 1 C c2 D 0. So c2 D �1. Therefore, the solution of (9.5.15)is

x.t/ D 2p

t � pt ln t:



t2x00 C 3tx0 C 2x D 0 (9.5.16)

on .�1; 0/.

Solution. Since a D 1, b D 3, c D 2, and b � a D 2, the associated equation is

d2x

du2C 2

dx

duC 2x D 0;

where u D ln.�t/ since t 2 .�1; 0/. The Cauchy-Euler characteristic equation is�2 C 2�C 2 D 0, which has the complex roots � D �1 ˙ i . A general solution of theassociated equation is

x.u/ D c1e�u cos u C c2e�u sin u:

Since

c1.eu/�1 cos u C c2.e

u/�1 sin u D c1.�t/�1 cos�

ln.�t/� C c2.�t/�1 sin

�ln.�t/

�;

by letting k1 WD �c1 and k2 WD �c2, we obtain the general solution

x.t/ D k1

tcos

�ln.�t/

� C k2

tsin�

ln.�t/�

on .�1; 0/.

9.6 Homogeneous Linear Equations of Higher Order

A homogeneous linear differential equation of order n with constant coefficients isan equation of the form

a0

dnx

dtnC a1

dn�1x

dtn�1C � � � C an�1

dx

dtC anx D 0; (9.6.1)

where a0; a1; � � � ; an denote the constant coefficients and a0 ¤ 0. Now that we knowhow to find a general solution of the second-order homogenous linear equation

a0

d2x

dt2C a1

dx

dtC a2x D 0 (9.6.2)

we will show how to obtain a general solution of (9.6.1) when n D 3. In other words,we will find a general solution of the homogenous third-order linear equation

a0

d3x

dt3C a1

d2x

dt2C a2

dx

dtC a3x D 0: (9.6.3)

We begin by dividing both sides of (9.6.3) by a 0:

d3x

dt3C b1

d2x

dt2C b2

dx

dtC b3x D 0; (9.6.4)

9.6. HOMOGENEOUS LINEAR EQUATIONS OF HIGHER ORDER 275

where bi D ai=a0 for i D 1; 2; 3. It is easy to verify that e�t is a solution of (9.6.4) ifand only if � is a root of

�3 C b1�2 C b2�C b3 D 0: (9.6.5)

Of course, (9.6.5) is the characteristic equation for (9.6.4). It has exactly 3 roots asdoes any nonconstant cubic polynomial when both real and complex roots are counted.Now recall the Remainder Theorem of algebra: If a polynomial P .�/ is divided by� � c, then the remainder is P .c/. It follows that � � c is a factor of P .�/ if and onlyif P .c/ D 0. Consequently if �1; �2; �3 are the three roots, then we can write (9.6.5)in the factored form

.� � �1/.� � �2/.� � �3/ D 0: (9.6.6)

Multiplying out the left-hand side of (9.6.6), we obtain

�3 � .�1 C �2 C �3/�2 C .�1�2 C �1�3 C �2�3/� � �1�2�3 D 0: (9.6.7)

Comparing (9.6.7) to (9.6.5), we see that

b1 D �.�1 C �2 C �3/

b2 D �1�2 C �1�3 C �2�3

b3 D ��1�2�3:

Hence, in terms of its characteristic roots, (9.6.4) is

x000 � .�1 C �2 C �3/x00 C .�1�2 C �1�3 C �2�3/x

0 � �1�2�3x D 0:

Now rearrange its terms as follows:

.x0 � �3x/00 � �2.x0 � �3x/0 � �1.x

0 � �3x/0 C �1�2.x0 � �3x/ D 0: (9.6.8)

Then by lettingy WD x0 � �3x (9.6.9)

equation (9.6.8) becomes

y00 � .�1 C �2/y0 C �1�2y D 0: (9.6.10)

Since a general solution of (9.6.4) depends on the characteristic roots—whetherthey are all distinct or some are equal and whether they are all real or some are complex—let us take a look at all of the possibilities. We know from algebra that complex rootsof polynomials must occur in conjugate pairs. That is, if z D ˛C i! is a complex rootof a polynomial P .�/, then its conjugate z D ˛ � i! is also a root. This implies that apolynomial P .�/ cannot have an odd number of complex roots. Thus the characteristicroots of (9.6.5) are either all real or two of them are complex conjugate pairs while theother is real. There are four distinct cases to consider.

Case 1. All three characteristic roots are real and distinct.


Since the characteristic roots associated with (9.6.10) are �1 and �2, its general solutionis

y.t/ D k1e�1t C k2e�2t ; (9.6.11)

where k1 and k2 are constants. An integrating factor for (9.6.9) is e ��3t . Hence,

d

dt

e��3t x

D e��3ty D k1e.�1��3/t C k2e.�2��3/t :

An integration yields

x.t/ D k1

�1 � �3

e�1t C k2

�2 � �3

e�2t C k3e�3t ; (9.6.12)

where k3 is another constant. This is a solution regardless of the values of k1; k2, andk3. Hence, we can replace the coefficients in (9.6.12) with c1; c2, and c3. In short, forthis case a general solution of (9.6.3) is

x.t/ D c1e�1t C c2e�2t C c3e�3t : (9.6.13)

Case 2. The characteristic roots are real and two of them are equal.

Let us label the roots as follows: �1 D �2 D m1 and �3 D m2, where m2 ¤ m1. Then(9.6.10) becomes

y00 � 2m1y0 C m21y D 0 (9.6.14)

with a general solutiony.t/ D k1em1t C k2tem1 t : (9.6.15)

From (9.6.9), we have

x0 � m2x D k1em1t C k2tem1t ;

which has the solution

x.t/ D k1

m1 � m2

em1t C k2

m1 � m2

tem1t C k3em2t :

Renaming the coefficients, we have the general solution

x.t/ D c1em1t C c2tem1t C c3em2t : (9.6.16)

Case 3. The three characteristic roots are real and equal.

Let � denote their common value. This makes � a root of multiplicity 3. Then, as inCase 2 (cf. (9.6.15)),

y.t/ D k1e�t C k2te�t :

Hence,x0 � �x D k1e�t C k2te�t :

Integrating and renaming constants, we obtain the general solution

x.t/ D c1e�t C c2te�t C c3t2e�t : (9.6.17)


Case 4. The roots are the complex conjugate pair ˛ ˙ i! and the real number �.

In (9.6.10) let �1 D ˛ C i! and �2 D ˛ � i!. And let �3 D � in (9.6.9). Then, as�1 C �2 D 2˛ and �1�2 D ˛2 C !2, we have

y00 � 2˛y0 C .˛2 C !2/y D 0 (9.6.18)

wherey D x0 � �x: (9.6.19)

In Section 9.3.4 we determined that a general solution of (9.6.18) is

y.t/ D k1e˛t cos!t C k2e˛t sin!t :

Consequently if x.t/ is a solution of (9.6.4), it must be one of the solutions of

x0 � �x D k1e˛t cos!t C k2e˛t sin!t :

Using the integrating factor e ��t , we find that

x.t/ D e�tk1

Ze.˛��/t cos!t dt C e�tk2

Ze.˛��/t sin!t dt C k3:

It is left as an exercise to verify that when these integrations are carried out that

x.t/ D k1b � k2!

b2 C !2e˛t cos!t C k1! C k2b

b2 C !2e˛t cos!t C k3e�t :

Finally, after replacing the above coefficients with c1; c2, and c3, we obtain

x.t/ D c1e˛t cos!t C c2e˛t sin!t C c3e�t :

Example 9.19. Find a general solution ofd3x

dt3� 3

dx

dtC 2x D 0.

Solution. In terms of the operator D D d=dt , the differential equation is

.D3 � 3D C 2/x D 0:

Hence, the characteristic equation is

�3 � 3�C 2 D 0:

By inspection � D 1 is a root. Consequently, the characteristic polynomial � 3 �3�C2

has the factor � � 1. To obtain another factor, divide it by � � 1 using the method ofsynthetic division:

1 1 0 �3 21 1 -2

1 1 �2 0


The numbers in the last row represent the factor �2 C �� 2. Hence,

�3 � 3�C 2 D .� � 1/.�2 C � � 2/ D .� � 1/2.�C 2/:

It follows that the characteristic roots are � D �2 and the double root � D 1. It followsfrom Case 2 that a general solution of the differential equation is

x.t/ D c1e�2t C c2et C c3tet :

Example 9.20. Find the solution of

4x000 � 4x00 � 3x0 C 5x D 0

satisfying the initial conditions

x.0/ D �17; x0.0/ D 0; x00.0/ D 34:

Solution. The corresponding characteristic equation is

4�3 � 4�2 � 3�C 5 D 0:

It is easy to see that � D �1 is a root; so a factor of the characteristic polynomial is� � .�1/ D �C 1. Synthetic division gives

�1 4 �4 �3 5�4 8 �5

4 �8 5 0

Hence, another factor is the quotient 4�2 � 8�C 5. Solutions of

4�2 � 8�C 5 D 0

are � D 1 ˙ 12i . Since the characteristic roots are �1; 1 ˙ 1

2i , it follows from Case 4

that a general solution is

x.t/ D c1e�t C c2et sint

2C c3et cos

t

2:

Now we need to determine the values of the constants c1; c2; c3 so that the initial con-ditions are also satisfied. The first and second derivatives of x.t/ are

x0.t/ D �c1e�t C c2et

�sin

t

2C 1

2cos

t

2

�C c3et

�cos

t

2� 1

2sin

t

2

�

x00.t/ D c1e�t C c2et

�3

4sin

t

2C cos

t

2

�C c3et

�3

4cos

t

2� sin

t

2

�:


Setting t D 0, we see that the initial conditions require that the constants satisfy thesystem of equations

c1 C c3 D �17

�c1 C 1

2c2 C c3 D 0

c1 C c2 C 3

4c3 D 34:

The solution of this system is

c1 D 3; c2 D 46; c3 D �20:

Therefore, the solution of the initial value problem is

x.t/ D 3e�t C 46et sint

2� 20et cos

t

2:

We could find solutions of fourth-order homogeneous linear equations in a way thatis similar to our approach for third-order equations. But that would be tedious. There isa better way but that requires a knowledge of linear algebra, something not assumed inthis book. So we sidestep this problem and simply state how to find a general solutionof an nth order homogeneous linear differential equation without giving a proof.

Solutions of Homogeneous Linear Equations of nth Order

Theorem 9.8. A general solution x.t/ of

a0

dnx

dtnC a1

dn�1x

dtn�1C � � � C an�1

dx

dtC anx D 0 (9.6.20)

is

x.t/ D c1x1.t/ C c2x2.t/ C � � � C cnxn.t/ DnX

iD1

cixi.t/

where the xi.t/ denote the following functions:

(a) If � is a real characteristic root of multiplicity j , then j of the x i.t/ are:e�t , te�t , t2e�t , : : :, tj�1e�t .

(b) If ˛ C i! is a characteristic root of multiplicity k , then 2k of the x i.t/

are: e˛t cos!t , e˛t sin!t , te˛t cos!t , te˛t sin!t , : : : , tk�1e˛t cos!t ,tk�1e˛t sin!t .

Example 9.21. Find a general solution of the 4th order linear equation

x.4/ C 8 Rx C 16x D 0: (9.6.21)


Solution. In operator notation, (9.6.21) is

.D4 C 8D2 C 16/x D 0:

Consequently, the characteristic equation is �4 C 8�2 C 16 D 0 or

.�2 C 4/2 D 0:

Thus, 2i is a characteristic root of multiplicity 2. By Theorem 9.8, a general solutionof (9.6.21) is

x.t/ D c1 cos 2t C c2 sin 2t C c3t cos 2t C c4t sin 2t :

9.7 Nonhomogeneous Linear Equations

A second-order nonhomogeneous9 linear differential equation with constant coeffi-cients has the form

ad2x

dt2C b

dx

dtC cx D g.t/; (9.7.1)

where a ¤ 0 and g is any function other than g.t/ � 0. If g.t/ � 0, then (9.7.1) is nolonger a nonhomogeneous equation but rather the homogeneous equation

ad2x

dt2C b

dx

dtC cx D 0: (9.7.2)

Now it is time to investigate how to solve (9.7.1).An example of a second-order nonhomogeneous linear equation is

x00 C 2x0 � 3x D �12:

This is (9.7.1) with a D 1, b D 2, c D �3, and g.t/ D �12. It is clear, with no needfor pencil-and-paper work, that a solution of this equation is x.t/ � 4. A functionthat is a solution of a nonhomogeneous differential equation and which contains noarbitrary constants is known as a particular solution. 10 Thus, x.t/ � 4 is a particularsolution. But then so is e �3t C 4 and the functions

2e�3t C 4; e�3t C et C 4; et C 4; 5et C 4;

just to list a few. In fact, this equation has infinitely many solutions of the form

x.t/ D c1e�3t C c2et C 4;

where c1 and c2 denote constants. A point to be made here is that we obtained theconstant solution x.t/ � 4 by inspection; that is, it is evident from a cursory glancethat this function satisfies the equation on .�1;1/.

9The term inhomogeneous is also commonly used.10It is also known as a particular integral of the nonhomogeneous equation.

9.7. NONHOMOGENEOUS LINEAR EQUATIONS 281

A second example of a nonhomogeneous linear differential equation is

4x00 C x D t

This equation results from setting a D 4, b D 0, c D 1, and g.t/ D t in (9.7.1). Aswith the first example, this equation has more than one particular solution. One thatcomes to mind with barely any thought is x.t/ D t . Clearly, this function satisfies theequation on .�1;1/ because its second derivative is always zero.

The reason for our concern with finding a particular solution of (9.7.1) is that ageneral solution of (9.7.1), as it turns out and as we will prove shortly, can be expressedas the sum of this particular solution and a general solution of equation (9.7.2). We call(9.7.2) the homogeneous differential equation associated with (9.7.1). 11

Solutions of Second-Order Nonhomogeneous Equations

Theorem 9.9. A general solution of

ax00 C bx0 C cx D g.t/ (9.7.3)

on an interval J isx.t/ D xh.t/C xp.t/; (9.7.4)

where xh.t/ is a general solution of

ax00 C bx0 C cx D 0 (9.7.5)

and xp.t/ is any particular solution of (9.7.3) on J .

Proof. Recall from Section 9.3.3 that the result of applying the linear differential op-erator

L D aD2 C bD C c

to a twice-differentiable function x is

LŒx� D .aD2 C bD C c/x D aD2x C bDx C cx D ax00 C bx0 C cx:

This allows us to write (9.7.3) as LŒx�D g.t/. This abbreviated form spares us a lot ofwriting both now and in the future.

Let xh.t/ denote a solution of the homogeneous equation (9.7.5). Let xp.t/ denotea solution of the nonhomogeneous equation (9.7.3) that we have somehow managed tofind (We will worry about how to do this later). Observe that for any function of theform xh.t/ C xp.t/, we have

LŒxh.t/ C xp.t/� D LŒxh.t/� C LŒxp.t/� D 0 C g.t/ D g.t/:

11The general solution of the associated homogeneous equation is also known as the complementaryfunction of the nonhomogeneous equation.


This says that xh.t/ C xp.t/ is a solution of LŒx� D g.t/, namely of the nonhomoge-neous equation (9.7.3).

Now let x represent any solution of (9.7.3) on an interval J . This means of coursethat LŒx.t/� D g.t/ for all t in J . Since the operator L is linear,

LŒx.t/ � xp.t/� D LŒx.t/� � LŒxp.t/� D g.t/ � g.t/ D 0:

This shows that the difference x.t/ � xp.t/ is a solution of LŒx� D 0, that is, of theassociated homogeneous equation

ax00 C bx0 C cx D 0:

Hence, x.t/ � xp.t/ D xh.t/ for some xh.t/. And so

x.t/ D xh.t/ C xp.t/:

This argument shows that any solution x.t/ of (9.7.3) can be expressed as the sum of ageneral solution xh.t/ of (9.7.5) and any particular solution xp.t/ of (9.7.3).

Example 9.22. Find a general solution of the nonhomogeneous equation

x00 C 2x0 � 3x D �12: (9.7.6)

Solution. Recall that xp.t/ � 4 is a particular solution on the interval .�1;1/. Theassociated homogeneous differential equation is

x00 C 2x0 � 3x D 0; (9.7.7)

which has the characteristic equation �2 C 2�� 3 D 0. Since its factorization is

.� C 3/.� � 1/ D 0;

the characteristic roots are � D �3 and � D 1. Therefore, a general solution of (9.7.7)is

xh.t/ D c1e�3t C c2et :

By Theorem 9.9, a general solution of (9.7.6) on .�1;1/ is

x.t/ D xp.t/ C xh.t/ D c1e�3t C c2et C 4:


4x00 C x D t: (9.7.8)

Solution. As we found earlier, xp.t/ D t is a particular solution of (9.7.8) on theinterval .�1;1/. The associated homogeneous differential equation is

4x00 C x D 0:

9.8. METHOD OF UNDETERMINED COEFFICIENTS 283

Since the characteristic equation is 4�2 C1 D 0, the characteristic roots are � D ˙i=2.Thus, the general solution of the associated homogeneous equation is

xh.t/ D c1 cost

2C c2 sin

t

2:

By Theorem 9.9, a general solution of (9.7.8) on .�1;1/ is

x.t/ D xh.t/ C xp.t/ D c1 cost

2C c2 sin

t

2C t:


x00 C 4x D 4t�1 C 2t�3 (9.7.9)


Solution. It is easy to mentally verify that xp.t/ D t�1 is a solution of (9.7.9) fort > 0. Since the characteristic roots are ˙2i , a general solution xh of the associatedhomogeneous equation

x00 C 4x D 0

isxh.t/ D c1 cos 2t C c2 sin 2t:

Consequently, a general solution of (9.7.9) on .0;1/ is

x.t/ D c1 cos 2t C c2 sin 2t C t�1:

9.8 Method of Undetermined Coefficients

The examples in Section 9.7 may give the false impression that the only way of findinga particular solution is by making an educated guess and keeping one’s fingers crossedthat the guess will actually turn out to be a solution. Unfortunately, particular solu-tions usually do not just somehow magically pop up in one’s head. In this section, weconsider some differential equations with particular solutions that are not as obvious asthey have been up to now. This necessitates developing a method whereby particularsolutions can be found without completely relying on guesswork. There may still besome guesswork involved as to the form of a particular solution—however, unlike theprevious equations, we will not usually be able to come up with a particular solutionwithout first doing some pencil-and-paper work. Before we say anymore, let us look atexamples of some more nonhomogeneous equations.

Example 9.25. Find a particular and then a general solution of

x00 C 4x D �20e6t : (9.8.1)


Solution. The form of a particular solution of a nonhomogeneous equation obviouslydepends on the function on its right-hand side as we will explore here and subsequentexamples. For this equation that function is an exponential function. The fact thatderivatives of exponential functions are also exponential functions suggests that a par-ticular solution has the form

xp.t/ D Ae6t : (9.8.2)

Let us consider this possibility and call it a trial particular solution. Now substitute itand its second derivative into (9.8.1) to see if the coefficient A can be assigned a valueto really make it a particular solution. These substitutions give

x00p.t/ C 4xp.t/ D 36Ae6t C 4Ae6t D 40Ae6t :

Clearly the right-hand side is equal to the right-hand side of (9.8.1) for A D �1=2.Therefore,

xp.t/ D �1

2e6t

is a particular solution of (9.8.1). Since xh.t/ D c1 cos 2tCc2 sin 2t is the general solu-tion of the associated homogeneous differential equation, it follows from Theorem 9.9that a general solution of (9.8.1) is

x.t/ D xh.t/C xp.t/ D c1 cos 2t C c2 sin 2t � 1

2e6t :

Before continuing with more examples, let us reflect on how we obtained the par-ticular solution in Example 9.25. First we managed to come up with the so-called trialparticular solution (9.8.2) by asking what type of function would generate the right-hand side of the differential equation when the differential operator L D D 2 C 4 isapplied to it. At that point, the value of the coefficient A was unknown. But then itsvalue was found by substituting the trial particular solution (9.8.2) into the differentialequation (9.8.1) to ascertain what the value of A would have to be in order for it totruly be a solution. We explore this approach further by considering more examples.


x00 � 2x0 � 3x D 2 � 9t2: (9.8.3)

Solution. The left-hand side of (9.8.3) is LŒx� where L is the differential operatorL D D2 � 2D � 3. The right-hand side of (9.8.3) is a quadratic function. Applying L

to a quadratic function yields another quadratic function. This suggests that a particularsolution of (9.8.3) has the form

xp.t/ D At2 C Bt C C: (9.8.4)

Even though the t-term is missing from the right-hand side of (9.8.3), it would beassuming too much not to expect such a term in xp.t/. Applying L to xp.t/, we have

LŒxp.t/� D x00p.t/ � 2x0

p.t/ � 3xp.t/

D 2A � 2.2At C B/ � 3.At2 C Bt C C /

D �3At2 C .�4A � 3B/t C .2A � 2B � 3C /:


Comparing this with the right-hand side of (9.8.3) shows that x p.t/ is a solution pro-vided

�3At2 C .�4A � 3B/t C .2A � 2B � 3C / D �9t2 C 2:

Equating coefficients of like terms, we obtain the following three simultaneous equa-tions in terms of the three undetermined coefficients A, B, and C :

�3A D �9

�4A � 3B D 0

2A � 2B � 3C D 2:

The solution of these simultaneous equations is: A D 3, B D �4, C D 4. Therefore,a particular solution is

xp.t/ D 3t2 � 4t C 4: (9.8.5)

Since the associated homogeneous equation is

x00 � 2x0 � 3x D 0;

the characteristic equation is �2 � 2�� 3 D 0. Its roots are � D �1 and � D 3. Hence,a general solution of the associated homogeneous equation is

xh.t/ D c1e�t C c2e3t :


x.t/ D c1e�t C c2e3t C 3t2 � 4t C 4:

To recapitulate, we found the particular solution (9.8.5) of the nonhomogeneousequation (9.8.3) by first coming up with (9.8.4) as the form of a trial (potential) partic-ular solution, which was based on the form of the right-hand side of (9.8.3). At thatpoint, the coefficients A, B, C had not yet been determined. But then their values weredetermined by substituting the trial particular solution (9.8.4) into (9.8.3) to ascertainwhat values would truly make it a particular solution.

The technique that we used to find particular solutions of both of the nonhomoge-neous equations (9.8.1) and (9.8.3) is known as the Method of Undetermined Coeffi-cients. The essence of the method is:

1. Determine a trial particular solution based on the form of the right-hand side ofthe nonhomogeneous equation.

2. Substitute the trial particular solution into the equation to determine the valuesof its coefficients.

The choice of a trial particular solution is not always so straightforward as it was in theprevious examples. Slight complications may arise as we shall see shortly in the nextset of examples.


Slight Complications

The previous examples suggest that the technique will work whenever the functiong on the right-hand side of (9.7.3) is either an exponential function or a polynomial.Consider

ax00 C bx0 C cx D ke˛t ; (9.8.6)

where a, k , and ˛ are nonzero constants. The obvious choice for a trial particularsolution is

xp.t/ D Ae˛t (9.8.7)

since the result of L D aD2 C bD C c operating on it is

LŒAe˛t � D aD2.Ae˛t /C bD.Ae˛t /C cAe˛t D .a˛2 C b˛ C c/Ae˛t :

Since the right-hand side is an exponential function, (9.8.7) is a particular solution of(9.8.6) if

A D k

a˛2 C b˛ C c

and a˛2 C b˛C c ¤ 0. On the other hand, (9.8.7) is not a particular solution of (9.8.6)if a˛2 C b˛ C c D 0; for then it would be a solution of the associated homogeneousequation (9.7.5). This means that when � D ˛ is a root of the characteristic equation

a�2 C b�C c D 0

some sort of adjustment to the trial particular solution xp.t/ D Ae˛t will have to bemade as the next example demonstrates.


x00 � 2x0 D e2t : (9.8.8)

Solution. The function e 2t on the right-hand side suggests that a particular solution of(9.8.8) will be

xp.t/ D Ae2t (9.8.9)

for some value of A. Unfortunately,

x00p.t/ � 2x0

p.t/ D 4Ae2t � 2.2Ae2t / D 0:

That is, this trial particular solution fails in that we obtain 0 instead of a multiple ofe2t . In fact, it is a solution of the corresponding homogeneous equation, namely,

x00 � 2x0 D 0: (9.8.10)

This could have been ascertained earlier had we first found a general solutionof (9.8.10),which because the characteristic roots are 0 and 2, is

xh.t/ D c1 C c2e2t :


Notice that (9.8.9) is a member of this family of solutions: this is seen by letting c 1 D 0

and c2 D A.So what is a suitable form of a trial particular solution? The answer lies with the

derivatived

dt

hte2t

iD 2te2t C e2t :

That is to say, we can still generate the term e2t by differentiating te 2t . This suggestsreplacing (9.8.9) with the trial particular solution

xp.t/ D Ate2t : (9.8.11)

Differentiating, we get

x0p.t/ D 2Ate2t C Ae2t

x00p.t/ D 4Ate2t C 2Ae2t C 2Ae2t D 4Ate2t C 4Ae2t :

Thus,

x00p.t/ � 2x0

p.t/ D 4Ate2t C 4Ae2t � 2.2Ate2t C Ae2t / D 2Ae2t :

Comparing with the right-hand side of (9.8.8), we see that (9.8.11) is a particular solu-tion if A D 1=2. Hence, by Theorem 9.9, a general solution of (9.8.8) is

x.t/ D c1 C c2e2t C 1

2te2t :

Now suppose g is a polynomial of degree n. If the constant c in the operatorL D aD2 C bD C c is not 0, then a particular solution is also a polynomial of degreen, since L maps polynomials of degree n to other polynomials of degree n. However,L D aD2 C bD maps a polynomial of degree n to one of lesser degree. Consequently,a particular solution will not be a polynomial of degree n when c D 0. Note that 0 isa characteristic root only when c D 0. Put another way, a particular solution of (9.7.3)is a polynomial of degree n unless 0 is one of the characteristic roots. Since the casec D 0 involves additional complications, it will be considered in later examples.

Nonhomogeneous terms that typically appear in physical applications but whichwe have not yet considered are linear combinations of sine and cosine functions. Thenonhomogeneous term in the differential equation in the next example involves thecosine function.

Example 9.28. Find a particular solution of

x00 C 4x0 C 4x D �3 cos t: (9.8.12)

Then find its general solution.

Solution. The presence of a cosine function on the right-hand side of (9.8.12) suggeststhe trial particular solution

xp.t/ D A cos t: (9.8.13)


Let us see if this really is a particular solution for an appropriate value of A by substi-tuting A cos t for the dependent variable x:

x00p C 4x0

p C 4xp D d2

dt2.A cos t/C 4

d

dt.A cos t/ C 4.A cos t/

D �A cos t � 4A sin t C 4A cos t D 3A cos t � 4A sin t:

Equating right-hand sides, we have

3A cos t � 4A sin t D �3 cos t:

Clearly, A D 0 doesn’t solve this equation. And if it had a nonzero solution, that wouldmean

3A C 3 ¤ 0 and � 4A ¤ 0;

while.3A C 3/ cos t � 4A sin t � 0;

which implies that cos t is a multiple of sin t . Certainly that is not true. Consequently,this trial particular solution is not really a solution at all.

The problem lies with our choice (9.8.13) as the trial particular solution since

L D D2 C 4D C 4

operating on it generates sin t as well as cos t . It turns out that we were too restrictivein our choice. We should have allowed ourselves a little more latitude by including asine term to complement the cosine term as follows:

xp.t/ D A cos t C B sin t: (9.8.14)

So do values of A and B exist so that xp.t/ is a particular solution of (9.8.12)? Toanswer this, differentiate xp.t/ twice:

x0p.t/ D �A sin t C B cos t

x00p.t/ D �A cos t � B sin t:

Now multiply xp.t/ and its derivatives by the appropriate coefficients of the operatorL D D2 C 4D C 4 and add the results:

x00p.t/ D �A cos t � B sin t

4x0p.t/ D 4B cos t � 4A sin t

C 4xp.t/ D 4A cos t C 4B sin t

LŒxp.t/� D .3A C 4B/ cos t C .3B � 4A/ sin t

This layout using a column format to align like terms rather than stringing them outin a single row, as was previously done, safeguards against accidentally omitting anyof them. The function xp.t/ is a particular solution if LŒxp.t/� D �3 cos t ; this meansthat A and B must be given values so that


3A C 4B D �3

�4A C 3B D 0:

It is left as an exercise to verify that the solution of this algebraic system is

A D � 9

25; B D �12

25

Therefore, a particular solution of (9.8.12) is

xp.t/ D � 9

25cos t � 12

25sin t:

Since the characteristic equation �2 C 4� C 4 D 0 has the repeated root � D �2, ageneral solution of (9.8.12) is

x.t/ D c1e�2t C c2te�2t � 9

25cos t � 12

25sin t:

The lesson to be taken from Example 9.28 is

When a cosine function is part of a trial particular solution, it must alwaysbe paired up with the sine function that has the same argument as thecosine function and vice versa.

Example 9.29. Find a particular and then general solution of

x00 C x D sin t: (9.8.15)

Solution. Due to the presence of the sine function in (9.8.15), let’s heed the lesson ofExample 9.28 that sine and cosine functions must appear together in any trial particularsolution. So, the form of a trial particular solution ought to be

xp.t/ D A sin t C B cos t: (9.8.16)

Substituting it into the left-hand side of (9.8.15), we obtain

x00p.t/ C xp.t/ D d2

dt2.A sin t C B cos t/C .A sin t C B cos t/

D �A sin t � B cos t C A sin t C B cos t D 0:

The right-hand side is 0: the A’s and B’s have cancelled out. Nothing of this sorthas occurred in the previous examples. Have we made a mistake? No matter whatvalues are assigned to A and B, our trial particular solution will never be a solution of(9.8.15). In fact, what we have unwittingly shown here is that (9.8) is a solution of theassociated homogeneous equation

x00 C x D 0; (9.8.17)


irrespective of the values of A and B.There is a lesson to be learned here. Since the characteristic equation is �2 C1 D 0,

the characteristic roots are � D ˙i . Hence, the general solution of (9.8.17) is

xh.t/ D c1 cos t C c2 sin t:

This is precisely the form of our trial solution (9.8.16), aside from the letters usedfor the coefficients. Had we started with finding a general solution of (9.8.17) first,we could have avoided mistakenly thinking that is the form of a particular solution of(9.8.15). As a result, a good rule of thumb is

Before looking for a particular solution of a nonhomogeneous equation,first find a general solution of the associated homogeneous equation.

So, if (9.8.16) doesn’t work, what does? We need to search for another trial par-ticular solution xp.t/ so that when L D D2 C 1 operates on it, a term involving somemultiples of sin t results. In view of the product rule, that is the case for the functiont cos t . Again as sine and cosine functions should be paired together, let’s try

xp.t/ D At sin t C Bt cos t:

By the product rule,

x0p.t/ D At cos t C A sin t � Bt sin t C B cos t

x00p.t/ D �At sin t C A cos t C A cos t � Bt cos t � B sin t � B sin t:

Thus,

x00p.t/ D �At sin t � Bt cos t C 2A cos t � 2B sin t

C xp.t/ D At sin t C Bt cos t

LŒxp.t/� D C 2A cos t � 2B sin t

From this, we see that xp.t/ is a particular solution if A and B have values so thatLŒxp.t/� D sin t . This is the case for A D 0 and B D �1=2. Thus,

xp.t/ D �1

2t cos t

is a particular solution of (9.8.15). So a general solution is

x.t/ D c1 cos t C c2 sin t � 1

2t cos t:

Let us summarize what we have learned so far from the previous examples. Asidefrom the exceptions noted above, some of the more common trial particular solutionsxp.t/ of nonhomogeneous equations are listed in Table 9.1.


x00 � 2x0 D 8te2t : (9.8.18)


Table 9.1: Particular Solutions of LŒx� D g.t/

g.t/ xp.t/

aekt Aekt

pn.t/| Antn C An�1tn�1 C � � � C A1t C A0

a cos!t C b sin!tA cos!t C B sin!t

(always include both terms even if a D 0 or b D 0)

ekt pn.t/ ekt�Antn C An�1tn�1 C � � � C A1t C A0

�ekt .a cos!t C b sin!t / ekt

�A cos!t C B sin!t

�pn.t/.a cos!t C b sin!t/

�Antn C An�1tn�1 C � � � C A1t C A0

�cos!t

C �Bntn C Bn�1tn�1 C � � � C B1t C B0

�sin!t

†pn.t/ denotes a polynomial of degree n: pn.t/ D antn C an�1tn�1 C � � � C a1t C a0.

Solution. According to a rule of thumb noted earlier, we should first find a generalsolution of the associated homogeneous differential equation

.D2 � 2D/x D 0: (9.8.19)

Since the characteristic equation is �.� � 2/ D 0, the characteristic roots are �1 D 0

and �2 D 2. And so a general solution of (9.8.19) is

xh.t/ D c1 C c2e2t :

The right-hand side of (9.8.18) is 8te 2t , which in Table 9.1 is of the form

g.t/ D ekt pn.t/;

where k D 2 and n D 1. According to the entry to its right, a trial particular solutionof (9.8.18) has the form

xp.t/ D ektAntn C An�1tn�1 C � � � C A1t C A0

D e2t .A1t C A0/ :

However, let us write xp.t/ D .At C B/e2t in order to avoid dealing with subscripts.But, as it turns out, this expression will not work. The reader should try it in orderto gain experience in recognizing potential errors of this type. The problem is thatthe term Be2t is a solution of the associated homogeneous equation (9.8.19). We canrectify this by replacing xp.t/ with t � xp.t/. In other words, a new trial particularsolution is

xp.t/ D .At2 C Bt/e2t : (9.8.20)


Note that neither of its terms is a solution of (9.8.19). 12 We leave it as an exercise tocheck that

x00p � 2x0

p D .4At C 2A C 2B/e2t :

Consequently, by matching coefficients we obtain the simple system:

4A D 8;

2A C 2B D 0:

Solving, we find that a particular solution of (9.8.18) is

xp.t/ D .At2 C Bt/e2t D .2t2 � 2t/e2t :

Therefore, general solution of (9.8.18) is

x.t/ D c1 C c2e2t C 2.t2 � t/e2t :

Theorem 9.10 shows how to build a trial particular solution from Table 9.1 when theright-hand side of the nonhomogeneous equation is a mixture of some of the differenttypes of functions listed in the left-hand column of the table.

Superposition Principle for Nonhomogeneous Equations

Theorem 9.10. If xp.t/ is a solution of

ax00 C bx0 C cx D g.t/

and xq.t/ a solution ofax00 C bx0 C cx D h.t/;

then xp.t/C xq.t/ is a solution of

ax00 C bx0 C cx D g.t/ C h.t/:

Proof. By the hypotheses,

LŒxp.t/� D g.t/ and LŒxq.t/� D h.t/;

where LŒx� WD ax00 C bx0 C cx. Consequently,

LŒxp.t/C xq.t/� D LŒxp.t/� C LŒxq.t/� D g.t/ C h.t/:

This says that xp.t/ C xq.t/ is a solution of the differential equation

LŒx� D g.t/ C h.t/:

12Had this not been the case, we would have to multiply again by t .



x00 C x D 2t2 � 1 C sin t: (9.8.21)

Solution. We have already determined in Example 9.29 that the general solution of theassociated homogeneous differential equation is


According to Theorem 9.10, a particular solution of (9.8.21) is the sum of particularsolutions of the two nonhomogeneous equations

1. x00 C x D 2t2 � 1

2. x00 C x D sin t:

First let’s find a particular solution of (1). From Table 9.1, a trial particular solution isxp.t/ D At2 C Bt C C . Substituting this into (1), we get

x00p C xp D At2 C Bt C .2A C C /:

Setting the right-hand side equal to the right-hand side of (1), we obtain the equation

At2 C Bt C .2A C C / D 2t2 � 1:

Equating coefficients of like terms, we have

A D 2

B D 0

2A C C D �1;

which has the solution: A D 2, B D 0, C D �5. Thus, a particular solution of (1) is

xp.t/ D 2t2 � 5:

In Example 9.29, we already showed that a particular solution of (2) is

xq.t/ D �1

2t cos t:

Hence, it follows from the Superposition Principle that a particular solution of (9.8.21)is

xp.t/C xq.t/ D 2t2 � 5 � t

2cos t:

Therefore,

x.t/ D xh.t/C Œxp.t/ C xq.t/� D c1 cos t C c2 sin t C 2t2 � 5 � t

2cos t

is a general solution of (9.8.21).



5x00 C 3x0 � 2x D 14t3 � 12e�t C 6: (9.8.22)

Solution. The characteristic equation for the associated homogeneous equation

5x00 C 3x0 � 2x D 0 (9.8.23)

is5�2 C 3� � 2 D 0:

Since its roots are �1 D �1 and �2 D 25

, a general solution of (9.8.23) is

xh.t/ D c1e�t C c2e2t=5 (9.8.24)

Now we are ready to find a particular solution of (9.8.22). The first step is to inspectthe right-hand side of (9.8.22). We note that its terms can be rearranged as the sum ofthe cubic polynomial 14t 3 C 6 and the exponential function �12e �t . And so we havethe sum of two of the types of functions listed in Table 9.1 under the column headingg.t/.

First, let us find a particular solution of

5x00 C 3x0 � 2x D 14t3 C 6: (9.8.25)

Because of the cubic polynomial, common sense or Table 9.1 suggests the trial partic-ular solution

xp.t/ D At3 C Bt2 C C t C D: (9.8.26)

Comparing (9.8.26) and (9.8.24), we see that no term of (9.8.26) is a solution of theassociated homogeneous equation (9.8.23). Thus, (9.8.26) is the correct trial particularsolution. The result of substituting it and its first and second derivatives into (9.8.25) is

5x00p C 3x0

p � 2xp D� 2At3 C .9A � 2B/t2 C .30A C 6B � 2C t C .10B � 3C � 2D/: (9.8.27)

Equating the coefficients of like terms on the right-hand sides of (9.8.25) and (9.8.27),we obtain the system

�2A D 14

9A � 2B D 0

30A C 6B � 2C D 0

10B C 3C � 2D D 6:

The solution of the system is

A D �7; B D �63

2; C D �399

2; D D �1839

4:

9.9. METHOD OF VARIATION OF PARAMETERS 295

Hence, a particular solution of (9.8.25) is

xp.t/ D �7t3 � 63

2t2 � 399

2t � 1839

4: (9.8.28)

Now let us find a particular solution of

5x00 C 3x0 � 2x D �12e�t : (9.8.29)

Because of the exponential function on the right-hand side, x q.t/ D Ee�t seems likea potential trial particular solution. But when we compare x q.t/ with (9.8.24), we seethat it is a solution of (9.8.23) no matter the value of E. So it cannot be a solution of(9.8.29) as well. A way out of this predicament is to modify xq.t/ by multiplying it bythe independent variable t and then checking to see if

xq.t/ D Ete�t

is a solution of (9.8.23). Since (9.8.24) shows that it is not, let’s substitute it into theleft-hand side of (9.8.29). It is left as an exercise to verify that the result of this is

5x00q C 3x0

q � 2xq D �7Ee�t :

Thus, xq.t/ is a solution of (9.8.29) if �7E D �12. It follows that

xq.t/ D 12

7te�t (9.8.30)

is a particular solution of (9.8.29).By the Superposition Principle, a particular solution of (9.8.22) is the sum of the

particular solutions xp and xq, given by (9.8.28) and (9.8.30), respectively. Finally, byTheorem 9.9, a general solution of (9.8.22) is

x.t/ D xh.t/ C �xp.t/ C xq.t/

� Dc1e�t C c2e

2t5 � 7t3 � 63

2t2 � 399

2t � 1839

4C 12

7te�t :

9.9 Method of Variation of Parameters

In this section, we present another widely-used method for obtaining particular solu-tions of nonhomogeneous differential equations. It is called the Method of Variation ofParameters. This method is more involved than the Method of Undetermined Coeffi-cients. On the other hand, it can be used to solve a wider variety of nonhomogeneousequations. For instance, the Method of Undetermined Coefficients cannot handle theequation

x00 C x D sec t:


But the Method of Variation of Parameters can as we will demonstrate later in Exam-ple 9.38. Another case in point: the Method of Undetermined Coefficients is restrictedto equations with constant coefficients, the Method of Variation of Parameters is not.By and large, differential equations with variable coefficients do not have analytical so-lutions that can be expressed in terms of finitely many elementary functions. Even forthe equations that do, no general method exists for obtaining these solutions. However,special methods exist for some types of equations with variable coefficients. For ex-ample, in the previous chapter, we presented a method for finding analytical solutionsof homogeneous Cauchy-Euler equations. As a result, it turns out that the Method ofVariation of Parameters can be used to find particular solutions of nonhomogeneousCauchy-Euler equations. In fact, this method will enable us to find a particular solu-tion of a nonhomogeneous equation (with constant or with variable coefficients) if (i)a general analytical solution of its associated homogeneous equation exists and can befound and if (ii) the integrals resulting from applying the method can be evaluated.

In Chapter 5 we derived the Variation of Parameters Formula for first-order nonho-mogeneous linear equations. Likewise we can derive a variation of parameters formulafor second-order nonhomogeneous linear equations. Let us review the method for first-order equations. Consider

dx

dtC 2tx D 5e�t2

: (9.9.1)

The Method of Variation of Parameters presupposes we have, or can find, a solution ofthe associated homogeneous equation, which in this case is

dx

dtC 2tx D 0: (9.9.2)

It is easy to show by separating variables or by using the integrating factor method thata general solution of (9.9.2) is

x.t/ D ke�t2

;

where k is an arbitrary constant. Now we replace k with a function u.t/ to look for aparticular solution of (9.9.1) of the form

xp.t/ D u.t/e�t2

: (9.9.3)

This may be viewed as varying the parameter (or varying the constant) k . Substitutingxp and its first derivative into (9.9.1) gives

x0p C 2txp D � � 2tue�t2 C u0e�t2�C 2tue�t2 D u0e�t2

:

Comparing the right-hand side with that of (9.9.1), we see (9.9.3) is a solution if

u0 D 5:

or u D 5t . Thus, a particular solution of (9.9.1) is

xp.t/ D u.t/e�t2 D 5te�t2

:


A general solution is

x.t/ D xh.t/ C xp.t/ D .5t C k/e�t2

:

Now let us use this idea to find out how to obtain a particular solution of the second-order linear nonhomogeneous equation

a2.t/d2x

dt2C a1.t/

dx

dtC a0.t/x D b.t/;

where a0, a1, a2, and b denote functions (or constants) that are continuous on anopen interval J . Moreover, we assume a2.t/ ¤ 0 for all t 2 J . As with first-orderequations, finding a particular solution depends on having a general solution of theassociated homogeneous equation

a2.t/d2x

dt2C a1.t/

dx

dtC a0.t/x D 0

for t 2 J . Under the hypothesis of continuity of the coefficients, solutions of thisequation exist. The result that allows us to make this claim is the following existenceand uniqueness theorem. No proof is given here since doing so would take us too farafield from the present discussion.

Existence and Uniqueness of Solutions

Theorem 9.11. Let a0, a1, a2, and b be continuous functions on an open intervalJ , where a2.t/ ¤ 0 for all t 2 J: For any t0 2 J and any real numbers � and �,a solution of the differential equation

a2.t/d2x

dt2C a1.t/

dx

dtC a0.t/x D b.t/ (9.9.4)

exists satisfying the initial conditions

x.t0/ D �; x0.t0/ D �: (9.9.5)

Furthermore, there is only one such solution and it is defined and satisfies (9.9.4)on all of J .

As we shall see shortly, we can express all solutions of the homogeneous equationassociated with (9.9.4) as a linear combination of two of its solutions. Moreover, anytwo will do as long as they are not proportional to one another. The mathematical termfor this is linear independence.


Linear Independent and Dependent Functions

Definition 9.4. Two functions f and g are said to be linearly independent on aninterval J if neither is equal to a constant times the other on the interval. Oth-erwise they are said to be linearly dependent on J ; that is, f and g are linearlydependent on J if there is a constant k such that either

f .t/ D kg.t/

org.t/ D kf .t/

for all t 2 J .

In the following examples, a denotes either a real number or �1 and b denotes eithera real number or C1.

Example 9.33. Consider the functions f .t/ D sin t and g.t/ D cos t on any interval.a; b/. Suppose a constant k exists such that sin t D k cos t for a < t < b. Fork ¤ 0 this is false; otherwise it would say that tan t is always equal to the constantk for a < t < b, except where cos t D 0. Obviously k cannot be equal to 0. Weconclude that such a k does not exist. Similarly there does not exist a constant k suchthat cos t D k sin t for a < t < b. Thus sin t and cos t are linearly independent on.a; b/.

Example 9.34. Let f .t/ D sin t and g.t/ � 0 on an interval .a; b/. Certainly noconstant k exists such that sin t D k � 0 on .a; b/. On the other hand, g.t/ D 0 � sin t

for all t 2 .a; b/. Therefore f and g are linearly dependent on .a; b/.

Example 9.35. The functions sin 2t and sin t cos t on linearly dependent on .a; b/.This follows immediately from the double-angle formula: sin 2t D 2 sin t cos t .

There is a convenient way to test if two functions are linear independent. It in-volves a function that is known as the Wronskian, named after the Polish-born Frenchmathematician Josef Hoene de Wronski (1778–1853).

Wronskian

Definition 9.5. The function

W .t/ WD f .t/g0.t/ � f 0.t/g.t/; (9.9.6)

where f and g are any two differentiable functions, is called the Wronskian of fand g.

When we deal with more than two functions, say with f , g, h, then we indicate theWronskian of two of them, say f and g, with the more explicit notation W Œf; g�.t/. A


convenient way to write the Wronskian is as a determinant. A 2 � 2 determinant—thedeterminant of a matrix with two rows and two columns—is computed by subtractingthe product of the entries on the secondary diagonal from the product of the entriesalong the main diagonal. For example, the Wronskian of f and g is

W Œf; g�.t/ Dˇˇf .t/ g.t/

f 0.t/ g0.t/

ˇˇ D f .t/g0.t/ � f 0.t/g.t/: (9.9.7)

Criterion for Linearly Independent Functions

Theorem 9.12. Let a pair of functions f and g be differentiable on an intervalJ . If there is at least one point in J where the Wronskian

W .t/ Dˇˇf .t/ g.t/

f 0.t/ g0.t/

ˇˇ (9.9.8)

is not zero, then f and g are linearly independent on J .

Proof. Suppose there is a point t0 2 J where the Wronskian is not zero:

W .t0/ Dˇˇf .t0/ g.t0/

f 0.t0/ g0.t0/

ˇˇ D f .t0/g

0.t0/ � f 0.t0/g.t0/ ¤ 0: (9.9.9)

To show this implies f and g are linearly independent on J , suppose the contrary,namely, that there is a constant k such that either

f .t/ D kg.t/ or g.t/ D kf .t/

for all t 2 J . Without loss of generality, suppose that f .t/ D kg.t/ (if otherwise,merely interchange the names of the functions). Differentiating, we get

f 0.t/ D kg0.t/

for all t 2 J . In particular, at t D t0, we have the pair of equations

f .t0/� kg.t0/ D 0

f 0.t0/� kg0.t0/ D 0:

Multiplying the first equation by g 0.t0/ and the second one by �g.t0/ and adding theresults, we get

f .t0/g0.t0/ � f 0.t0/g.t0/ D 0:

But this contradicts the supposition. Therefore f and g must be linearly independenton J .


We show next that under the hypotheses of continuity of the coefficients that theassociated homogeneous equation of (9.9.4) always has a pair of linearly independentsolutions.

Linearly Independent Solutions of Homogeneous Equations

Theorem 9.13. Let a0, a1, and a2 be continuous functions on an open intervalJ , where a2.t/ ¤ 0 for all t 2 J . Then two linearly independent solutions of thehomogeneous linear equation

a2.t/d2x

dt2C a1.t/

dx

dtC a0.t/x D 0 (9.9.10)

exist on the interval J .

Proof. Select any number t0 in J . By Theorem 9.11, there is a unique solution x1.t/

of (9.9.10) satisfying the initial conditions: x 1.t0/ D 0, x01.t0/ D 1. Likewise there is

a unique solution x2.t/ with x2.t0/ D 1 and x02.t0/ D 0. Suppose x1.t/ and x2.t/ are

linearly dependent on J . That is, suppose there is a constant k such that

x2.t/ D kx1.t/ for all t 2 J :

Setting t D t0, we obtain the false statement that 1 D k �0. Consequently, the solutionsx1.t/ and x2.t/ must be linearly independent on J .

It can be proven that a general solution of a homogeneous equation can be expressedas a linear combination of any two of its linearly independent solutions. This is thecontent of the next theorem.

General Solutions of Homogeneous Equations

Theorem 9.14. Let a0, a1, and a2 be continuous functions on an open intervalJ , where a2.t/ ¤ 0 for all t 2 J . Let x1.t/ and x2.t/ be linearly independentsolutions of (9.9.10) on J . If x.t/ is a solution of (9.9.10) on J , then constantsc1 and c2 exist such that

x.t/ D c1x1.t/ C c2x2.t/

for all t 2 J .

Note the importance of Theorem 9.14. It states that all solutions of (9.9.10) becomeknown once we have found two of its linearly independent solutions. In other words,if x1 and x2 are linearly independent solutions of (9.9.10), then every solution x.t/

of (9.9.10) can be expressed as a linear combination of these two solutions. For thisreason, we say that

xh.t/ D c1x1.t/C c2x2.t/: (9.9.11)

is a general solution of (9.9.10).



x00 � x D 0: (9.9.12)

Solution. It is easy to mentally verify that the functions x 1.t/ D et and x2.t/ D e�t

are solutions of (9.9.12) for t 2 .�1;1/. Since neither is a multiple of the other, theyare linearly independent on .�1;1/. Thus, by Theorem 9.14, a general solution of(9.9.12) on .�1;1/ is

x.t/ D c1et C c2e�t : (9.9.13)

Of course, this is the same result we would have gotten had Theorem 9.1 been used.

Engineers and physicists find it practical to express solutions of equations like(9.9.12) in terms of the so-called hyperbolic functions. The three most common onesare the hyperbolic cosine, hyperbolic sine, and hyperbolic tangent functions, whichare defined respectively by

cosh t WD et C e�t

2; sinh t WD et � e�t

2; tanh t WD et � e�t

et C e�t:

It is left as an exercise to show that

d

dtcosh t D sinh t

andd

dtsinh t D cosh t

and then to verify that x1.t/ D cosh t and x2.t/ D sinh t are solutions of (9.9.12)for t 2 .�1;1/. It is also left as an exercise to check that neither is a multipleof the other. Consequently, cosh t and sinh t are linearly independent on .�1;1/.Therefore, by Theorem 9.14,

x.t/ D c1 cosh t C c2 sinh t: (9.9.14)

is also a general solution of (9.9.12) on .�1;1/.To sum up, we have shown that both (9.9.13) and (9.9.14) are general solutions of

(9.9.12). This means that any solution of (9.9.12) can be expressed either as a linearcombination of the exponential functions e t and e�t or of the hyperbolic functionscosh t and sinh t . So we see that there is no unique general solution of a differentialequation. That is why we use the phrase “a general solution” rather than “the generalsolution.”

To find a particular solution of (9.9.4) on an interval J , we start with (9.9.11). Aswe have seen with the first-order equations in Chapter 5, the idea behind the variationof parameters method13 is to view c1 and c2 in (9.9.11) as parameters: constants thatwe vary to find a particular solution of the related nonhomogeneous equation (9.9.4).In other words, we replace c1 and c2 in (9.9.11) with unknown functions, say u 1.t/ andu2.t/, thereby changing (9.9.11) to

13or variation of constants method


xp.t/ D u1.t/x1.t/ C u2.t/x2.t/: (9.9.15)

We use the subscript “p” to denote a particular solution of (9.9.4), just as we use thesubscript “h” to denote a general solution of the associated homogeneous equation(9.9.10). Of course, any two functions will not do: u 1 and u2 will have to be selectedso that (9.9.11) solves (9.9.4). Now the issue is this: Do such functions exist—and ifthey do, how do we go about finding them?

To answer these questions, it helps to employ the differential operator

L D a2D2 C a1D C a0:

With this operator, (9.9.10) and (9.9.4) have the abbreviated forms

LŒx.t/� D 0 and LŒx.t/� D b.t/;

respectively.To see if an expression of the form (9.9.11) satisfies (9.9.4), we substitute it and its

first and second derivatives into (9.9.4). The derivatives are

x0p D .u1x0

1 C u2x02/C .u0

1x1 C u02x2/

x00p D .u1x00

1 C u2x002/C .u0

1x01 C u0

2x02/C d

dt.u0

1x1 C u02x2/:

Even though we are omitting the argument t for the sake of brevity, keep in mind thatu1, u0

1, x1, x0

1, x0

p, : : : are functions of t . In order to simplify matters by eliminatingsome of the terms in both derivatives, we impose the condition

u01x1 C u0

2x2 D 0: (9.9.16)

At this juncture, we do not know if this will work, but let us forge on and see what willhappen. Multiply xp and its derivatives by their respective coefficients and then addthe results as follows:

a2.t/x00p D a2u1x00

1C a2u2x00

2C a2.u

01x0

1C u0

2x0

2/

a1.t/x0p D a1u1x0

1C a1u2x0

2

C a0.t/xp D a0u1x1 C a0u2x2

LŒxp� D u1LŒx1�C u2LŒx2�C a2.u01x0

1C u0

2x0

2/

Since x1.t/ and x2.t/ are solutions of (9.9.10),

LŒx1.t/� D 0 and LŒx2.t/� D 0:

And so the last sum simplifies to

LŒxp� D a2.u01x0

1 C u02x0

2/:

We conclude that (9.9.11) is a particular solution of (9.9.4) provided there are functionsu1.t/ and u2.t/ that satisfy the condition we imposed earlier, namely (9.9.16), and thefollowing condition

a2.u01x0

1 C u02x0

2/ D b.t/: (9.9.17)


Note that, with this second condition satisfied, we have LŒxp.t/� D b.t/, which is sim-ply an abbreviated way of stating that the function xp defined by (9.9.11) is a solutionof equation (9.9.4).

If we can show that there are functions u1 and u2 with derivatives satisfying con-ditions (9.9.16) and (9.9.17), namely, the system of equations

x1u01 C x2u0

2 D 0

x01u0

1 C x02u0

2 D b.t/

a2.t/;

then our approach will work. And so we attempt to solve this system. First, with theintention of solving for the unknown u 0

1, let’s eliminate u0

2by multiplying the first

equation by x 02

and the second one by �x2 and adding the results:

x1x02u0

1C x2x0

2u0

2D 0

C � x01x2u0

1 � x2x02u0

2 D �b.t/x2.t/

a2.t/

u01.x1x0

2� x0

1x2/ D �b.t/x2.t/

a2.t/.

Similarly, with u01

eliminated, the result is

u02.x1x0

2 � x01x2/ D b.t/x1.t/

a2.t/:

We conclude that the system does have a solution on the interval J provided the con-dition

x1.t/x02.t/ � x0

1.t/x2.t/ ¤ 0 (9.9.18)

holds for all t 2 J . Supposing for the moment that it does hold, we can solve thesystem and then integrate obtaining

u1.t/ D �Z

b.t/x2.t/

a2.t/W .t/dt ; u2.t/ D

Zb.t/x1.t/

a2.t/W .t/dt

where we denote the left-hand side of (9.9.18) by W .t/, in which case, a particularsolution of (9.9.4) on J is given by the formula

xp.t/ D �x1.t/

Zb.t/x2.t/

a2.t/W .t/dt C x2.t/

Zb.t/x1.t/

a2.t/W .t/dt : (9.9.19)

Because of condition (9.9.18), whether or not (9.9.19) is defined on the interval J

and is really a particular solution depends on W .t/ being nonzero on J .Using the existence-and-uniqueness theorem (Theorem 9.11), the following result

can be proven.

Wronskian and Linear Independence

Theorem 9.15. Two solutions x1.t/ and x2.t/ of (9.9.10) on an interval J arelinearly independent if and only if W Œx 1; x2�.t/ ¤ 0 for all t 2 J .


Consequently, W .t/ in (9.9.19) is never zero on the interval J ; so (9.9.19) is truly aparticular solution of (9.9.4) on J . However, due to the complexity of (9.9.19) and thelikelihood of forgetting or misplacing the minus sign, we suggest not committing it tomemory. Instead it is easier to memorize the system of equations (9.9.16) and (9.9.17)because of their similarity. This is the same as (9.9.23) below.

Summing up the discussion in this section, we have the following method for ob-taining particular solutions of nonhomogeneous linear second-order equations.

Method of Variation of Parameters

Let x1.t/ and x2.t/ be linearly independent solutions of

a2.t/x00 C a1.t/x

0 C a0.t/x D 0 (9.9.20)

on an interval J that excludes points where a2.t/ D 0. Then a particular solutionof

a2.t/x00 C a1.t/x

0 C a0.t/x D b.t/ (9.9.21)

on J of the formxp.t/ D u1.t/x1.t/ C u2.t/x2.t/ (9.9.22)

is obtained by completing the following steps:

1. Solve the system of equations

u01x1 C u0

2x2 D 0 (9.9.23)

u01x0

1 C u02x0

2 D b.t/

a2.t/

for u01 and u0

2.

2. Determine u1 and u2 by integrating u 01 and u0

2 with respect to t .

3. Substitute u1 and u2 into (9.9.22) to obtain a particular solution of (9.9.21).

Example 9.37. Find a particular solution of the Cauchy-Euler equation

t2x00 C tx0 � 4x D 6t (9.9.24)


Solution. In the notation of (9.9.21),

a2.t/ D t2; a1.t/ D t; a0.t/ D �4; and b.t/ D 6t:

In Example 9.16, it is shown that a general solution of the associated homogeneousequation

t2x00 C tx0 � 4x D 0;


on .0;1/ is

xh.t/ D c1

t2C c2t2:

Then letting x1.t/ D t�2 and x2.t/ D t2, (9.9.23) becomes

t�2u01 C t2u0

2 D 0 (9.9.25)

�2t�3u01 C 2tu0

2 D 6t

t2:

Now solve for u01.t/ and u0

2.t/. With the aim of eliminating u 0

2, let’s multiply the first

equation by �2 and the second one by t , thereby obtaining

�2t�2u01 � 2t2u0

2 D 0

�2t�2u01 C 2t2u0

2 D 6:

Adding the two equations gives

�4t�2u01 D 6 ) u0

1 D �3

2t2:

Integrating, we find that

u1 D �3

2

Zt2 dt D �1

2t3:

Using this in the first equation of system (9.9.25) gives

t�2�� 3

2t2�C t2u0

2 D 0 ) u02 D 3

2t2:

Thus,

u2 D 3

2

Zt�2 dt D �3

2t�1:

Since

u1.t/x1.t/C u2.t/x2.t/ D �1

2t3 � t�2 � 3

2t�1 � t2 D �2t;

a particular solution of (9.9.24) is

xp.t/ D �2t:

In the derivation of the formulas, the use of u1 and u2 made it easy to rememberthat u1 is paired with x1 and u2 with x2. Nonetheless, it is awkward dealing withsubscripts when actually solving equations. For this reason, we will replace u1 and u2

with u and v, respectively, from now on.


x00 C x D sec t (9.9.26)

on the interval Œ0; �=2/.


Solution. The associated homogeneous equation is

.D2 C 1/x D 0: (9.9.27)

Since its characteristic equation is �2 C 1 D 0 with roots � D ˙i , a general solutionof (9.9.27) is


Now we look for a particular solution of the form

xp.t/ D u.t/ cos t C v.t/ sin t;

where the derivatives of u and v satisfy the system of equations

u0 cos t C v0 sin t D 0

�u0 sin t C v0 cos t D sec t:

Solving for u and v using Cramer’s rule,14 we obtain

u0 D

ˇˇ 0 sin t

sec t cos t

ˇˇˇ

ˇ cos t sin t

� sin t cos t

ˇˇ

D sin t � sec t

cos2 t C sin2 tD � sin t

cos t

and

v0 D

ˇˇ cos t 0

� sin t sec t

ˇˇˇ

ˇ cos t sin t

� sin t cos t

ˇˇ

D cos t � sec t D 1:

Integrating, we obtain

u D �Z

sin t

cos tdt D ln

ˇcos t

ˇ D ln�

cos t�

and v DZ

dt D t:

Note that we can drop the absolute bars because cos t > 0 for 0 � t < �=2. Hence, aparticular solution of (9.9.26) is

xp.t/ D ln.cos t/ � cos t C t sin t:


x.t/ D c1 cos t C c2 sin t C ln.cos t/ � cos t C t sin t:


x00 � 6x0 C 9x D t�1e3t (9.9.28)

on the interval .�1; 0/.

14See Problem 80 on page 310.


Solution. . The associated homogeneous equation is

.D2 � 6D C 9/x D 0: (9.9.29)

Since the characteristic equation is

.�2 � 6�C 9/ D .� � 3/2 D 0;

� D 3 is a repeated characteristic root. Consequently, (9.9.29) has the linearly inde-pendent solutions e3t and te3t , which means

xh.t/ D c1e3t C c2te3t :

is a general solution. Now we look for a particular solution of (9.9.28) of the form

xp.t/ D u.t/e3t C v.t/te3t ;

where the derivatives of u and v satisfy the system of equations

u0e3t C v0te3t D 0

u03e3t C v0.3te3t C v0e3t / D t�1e3t

1:

Now multiply each equation of the system by exp.�3t/ to simplify it:

u0 C tv0 D 0

3u0 C .3t C 1/v0 D t�1:

Using Cramer’s rule and integrating, we obtain

u0 D

ˇˇ 0 t

t�1 3t C 1

ˇˇˇ

ˇ 1 t

3 3t C 1

ˇˇ

D 0 � 1

.3t C 1/ � 3tD �1 ) u D �t

and

v0 D

ˇˇ 1 0

3 t�1

ˇˇˇ

ˇ 1 t

3 3t C 1

ˇˇ

D t�1 ) v DZ

t�1 dt D ln jt j D ln.�t/:

Therefore, a particular solution of (9.9.28) is

xp.t/ D Œ�t C t ln.�t/� e3t :

Consequently, a general solution of (9.9.28) is

x.t/ D Œc1 C c2t � t C t ln.�t/� e3t :

Combining the middle two terms and replacing c2 � 1 with c2, it simplifies to

x.t/ D Œc1 C c2t C t ln.�t/� e3t :


Problems

Something else an academic education will do for you. If you go along with itany considerable distance, it’ll begin to give you an idea what size mind you have.What it’ll fit and, maybe, what it won’t. After a while, you’ll have an idea whatkind of thoughts your particular size mind should be wearing. For one thing, itmay save you an extraordinary amount of time trying on ideas that don’t suit you,aren’t becoming to you. You’ll begin to know your true measurements and dressyour mind accordingly.

Mr. Antolini’s advice to Holden CaulfieldThe Catcher in the Rye by J. D. Salinger (ch. 24)

General Solutions

1. Verify any function of the form (9.3.16),where �1 and �2 are the characteristicroots (9.3.4), is a solution of (9.3.15) bysubstituting it into (9.3.15).

2. Verify any function of the form (9.3.19),where � D � b

2ais a repeated charac-

teristic root, is a solution of (9.3.18) bysubstituting it into (9.3.18).

Second-Order Homogeneous Equations

In Problems 3 through 26, find a general so-lution of the differential equation. Differentia-tion is with respect to the independent variablet unless otherwise indicated.

3.d2x

dt2� 9

dx

dtD 0

4. x00 C 3x0 � 10x D 0

5. 9d2p

ds2C 4p D 0

6. 15d2y

dx2C 44

dy

dx� 36y D 0

7. 2x00 D 3x � 5x0

8. 2y00 � 7y0 D 0

9.d2y

dt2� dy

dtC 7y D 0

10. 20y00 � 33y0 � 27y D 0

11. 4d2Q

dt2� 12

dQ

dtC 9Q D 0

12.d2x

dt2C 5x D 0

13. 12Rs C Ps � 35s D 0

14. 3d2x

dt2C 7

dx

dtD 4x

15. 2y00 D �2y0 � 3y

16. x00 C 4x0 C 4x D 0

17. 2x00 D 3x � x0

18. 3x00 � 4x0 C 12x D 0

19. x00 � 6x0 C 13x D 0

20. x00 � 4x D 0

21. 9 Rx C 30 Px C 25x D 0

22. 9d2y

dx2C 18

dy

dxC 10y D 0

23. 4d2y

dx2� 20

dy

dxC 25y D 0

24.d2p

ds2D 5

dp

ds

25. 2x00 D x0 � 3x

26. 3x00 C 7x0 � 20x D 0

Second-Order Initial Value Problems

In Problems 27 through 31, find the solution ofthe given initial value problem. Differentiationis with respect to the variable t unless other-wise indicated.

27. x00 C x0 D 0;

x.0/ D 2; x0.0/ D 1

28. x00 � 2x0 C x D 0;

x.0/ D 0; x0.0/ D �1

PROBLEMS 309

29. x00 � 2x0 C 2x D 0;

x.�/ D 1; x0.�/ D 0

30. y00 C 5y0 � 14y D 0;

y.0/ D 5; y0.0/ D 1

31. Rx C x D 0;

x.�/ D �5; x0.�/ D 2

Complex Exponential Functions

32. Use Euler’s formula to derive the double-angle formulas:

sin 2t D 2 sin t cos t

cos 2t D cos2 t � sin2 t:

33. Use (9.4.6) and (9.4.10) to derive thehalf-angle formulas:

cos2 t D 1 C cos 2t

2

sin2 t D 1 � cos 2t

2:

34. Verify (9.4.13) with eit and e�it de-fined by (9.4.11) and (9.4.12), respec-tively.

Cauchy-Euler Equations

In Problems 35 through 38, find a general so-lution of each equation on the interval .0;1/.

35. t2 d2x

dt2C t

dx

dtC x D 0

36. 4t2x00 � 3tx0 � 2x D 0

37. 5t2x00 � 3tx0 C 3x D 0

38. t2 d2x

dt2� 5t

dx

dtC 9x D 0

39. (a) Find a general solution of

t2x00 � tx0 C x D 0:

(b) Use the answer to part (a) to findthe solution satisfying the initialconditions:

x.1/ D 3; x0.1/ D 1:

40. Solve the initial value problem:

t2 d2x

dt2C 3t

dx

dtC 2x D 0I

x.1/ D 2; x0.1/ D 5:

41. Prove: If x.t/ is a solution of the Cauchy-Euler equation (9.5.2) for t > 0, thenthe function y.t/ WD x.�t/ is a solu-tion for t < 0.

Homogeneous Equations of Higher Order

In Problems 42 through 47, find a general so-lution of the equation. The prime notation in-dicates differentiation with respect to t unlessotherwise indicated.

42. x000 C x00 � 8x0 � 12x D 0

43.d4x

dt4�4

d3x

dt3C5

d2x

dt2�4

dx

dtC4x D 0

44. «x C Rx C 4 Px C 4x D 0

45. y000�y00C4y0�4y D 0

�y0 D dy

dx

�

46.d3x

dt3C 2

d2x

dt2� dx

dt� 2x D 0

47. 2x000 � 3x00 C 8x0 C 5x D 0

Higher-Order Initial Value Problems

In Problems 48 through 50, find a solution ofthe initial value problem. The prime notationindicates differentiation with respect to t un-less otherwise indicated.

48. 4x000 � 11x0 C 10x D 0;

x.0/ D �37; x0.0/ D 0; x00.0/ D 74

49. x000 � x0 D 0Ix.0/ D 6; x0.0/ D 1; x00.0/D 2

50. 2x.4/ � x000 � 9x00 C 4x0 C 4x D 0Ix.0/ D 10; x0.0/ D �12; x00.0/ D 0,

x000.0/ D 2


Nonhomogeneous Equationsand General Solutions

In Problems 51 through 53, find a generalsolution.

51. 3x00 � 5x0 D 6 � 10t ,

given that xp.t/ D t2 is a particular so-lution on .�1;1/

52.d2y

dx2C y D 5e2x ,

given that yp.x/ D e2x is a particularsolution on the interval .�1;1/

53. x00 � 4x0 C 4x D et

Method of Undetermined Coefficients

In Problems 54 through 66, find a particularsolution and a general solution using the Methodof Undetermined Coefficients.

54. x00 C 3x0 C 2x D sin t

55. x00 C 3x0 D 12 cos 2t

56. x00 C 9x D 6t

57. x00 � 3x0 C 2x D 5e2t � sin t

58. x00 � 4x0 D 3et=2 � 2e4t C 1

59. 15y00 � 6y0 � 48y D 10x2 � 13x

60. x00 C 9x D 4 sin 3t � cos 3t � 27

61. 4x00 C 4x0 C x D 9tet

62. 6x00 C 3x0 � 30x D 10et

63. x00 C 4x0 C 7x D 14 � 7t

64. x00 C x0 � 2x D �8et

65. x00 � 4x0 C 4x D e2t

66. 4x00 � 12x D t sin 2t

Method of Variation of Parameters

In Problems 67 through 72, find a particularsolution and a general solution using the Methodof Variation of Parameters.

67. x00 C x D csc t for 0 < t � �=2

68. x00 � 4x0 C 4x D e2t

t2for t > 0

69. t2 d2x

dt2C t

dx

dtC x D t for t > 0

70. t2 d2x

dt2C6t

dx

dtC4x D 18t2 C5t �12

for t > 0

71. x00 C 4x D 8 csc 2t

72. Rx C x D tan t for ��=2 < t < �=2

Deciding on an Appropriate Method

In Problems 73 through 78, find a general solu-tion of the differential equation. Use whatevermethod is most appropriate. However, there isno need to use any method if a particular so-lution immediately comes to mind from inspec-tion of the equation.

73. t2x00 � 3tx0 C 3x D t for t > 0

74. t2x00 � 5tx0 C 9x D 8t � 8

75. x00 C x D 1

sin t � cos2 ton .0; �=2/

76. x00 C x0 C x D cos2 t ,

Hint. Use one of the half-angle formu-las.

77. x00 C x D 4 � t2 C tan t

for ��=2 < t < �=2

Hint. Apply the Superposition Principleto divide up the right-hand side in or-der to use both the undetermined coeffi-cients and variation of parameters meth-ods.

78.d2x

dt2� x D 10 sin2 t

Initial Value Problem

79. Solve the initial value problem:

t2x00 � 5tx0 C 9x D �36Ix.1/D 2; x0.1/ D �5:

Cramer’s Rule

80. Consider the system of linear algebraicequations

ax C by D e

cx C dy D f;

PROBLEMS 311

where x and y are the unknowns. Doesthis system always have solutions? Ex-plain why it has a unique solution if andonly if ˇ

ˇa b

c d

ˇˇ D ad � bc ¤ 0:

Show that this solution expressed in termsof determinants is

x D

ˇˇe b

f d

ˇˇˇ

ˇa b

c d

ˇˇ; y D

ˇˇa e

c f

ˇˇˇ

ˇa b

c d

ˇˇ:

This method of using determinants tosolve systems is known as Cramer’s rule.

Wronskian

81. For the functions

x1.t/ D cos t; x2.t/ D sin t;

x3.t/ D e� t ; x4.t/ D e�t ; and

x5.t/ D te�t ;

compute the following Wronskians:

(a) W Œx1; x2�.t/

(b) W Œx3; x4�.t/

(c) W Œx4; x5�.t/.

Mass and Two Springs Problem

82. Two springs with spring constants k1

and k2 are attached to a mass m andtwo stationary supports as shown below.Assume that friction is negligible.

If m is pulled x0 units to the right ofits equilibrium position (x D 0) andthen released so that m has no initial ve-locity, determine the position x of m attime t and its frequency of oscillation.


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