second order 19 differential equations...19 second order differential equations scientists study the...

19SECOND ORDER

DIFFERENTIAL

EQUATIONS

Scientists study the oscillations of solar coronalloops (gigantic loops of plasma related todisturbances in the sun’s magnetic field).

I n this chapter we study second order linear differential equations.These equations have numerous applications throughout science and

engineering. Among the most important of these are the descriptionof mechanical vibrations, oscillating currents in electrical circuits, andthe phenomena of resonance discussed in Section 3.

19.1 Second Order Linear Differential Equations

A second order linear differential equations is an equation of the form

a(t)y′′ + b(t)y′ + c(t)y = f (t)

where a(t) is nonzero. The functions a(t), b(t), c(t) are the coefficientsof the equation. The function f (t) on the right is often called theforcing function. Note that in a linear equation, y and its derivativesoccur at most to the first power (there are no terms such as y2, (y′)2,yy′′, etc).

Since the independent variable is time in many applications, weshall use t instead of x as the independent variable throughout this

chapter. Thus y′ denotes the derivative y′ = dy/dt . We also assume that y′′ exists and is

In this chapter, we use t for the independentvariable and y′ denotes the derivative

y′ = dy

dt

continuous.The following result assures us that solutions of second order linear differential equa-

tions exist. In fact, infinitely many solutions exist and, as we saw previously for first orderdifferential equations, we can pick out one particular solution by specifying initial val-ues. For a second order equation, this is done by specifying the values y(t0) and y′(t0) ata point t = t0.

THEOREM 1 Theorem Existence and Uniqueness for the Initial Value ProblemAssume that a(t), b(t), c(t), and f (t) are continuous on an open interval (a, b) andthat a(t) �= 0 for all t ∈ (a, b). Let t0 ∈ (a, b). Then the initial value problem

a(t)y′′ + b(t)y′ + c(t)y = f (t), y(t0) = y0, y′(t0) = y1

has precisely one solution y(t) defined for all t ∈ (a, b).

In the first three sections of this chapter, we consider second order linear differentialequations with constant coefficients. Thus, we shall study

ay′′ + by′ + cy = f (t) 1

where a, b, c are constants with a �= 0 and f (t) is a continuous function.

1

2 C H A P T E R 19 SECOND ORDER DIFFERENTIAL EQUATIONS

Homogeneous Equations

Eq. (1) is called homogeneous if the forcing function f (t) is zero. Otherwise, the equationis called inhomogeneous. Inhomogeneous equations are discussed in the next section. Therest of this section is devoted to homogeneous equations (a, b, c are constants):

ay′′ + by′ + cy = 0 2

There is a straightforward procedure for finding all solutions (2), based on the followingthree facts.

(A) If y1 and y2 are solutions, then the linear combination y = c1y1 + c2y2 is also asolution for all constants c1 and c2.

A linear combination of two functions y1(t)

and y2(t) is a function of the form

y(t) = c1y1(t) + c2y2(t)(B) If y1 and y2 are independent solutions (defined below), then every solution is of theform y = c1y1 + c2y2.

(C) With one exception (the case of double roots described below), there exist independentsolutions either of the form eλt or of the form eαt cos ωt and eαt cos ωt .

To explain (A)-(C) greater detail, we introduce a useful way of rewriting Eq. (1) interms of the “differential operator"

D = d

dt

By definition, Dy denotes the derivatived

dty = y′. Similarly, D2y denotes y′′, etc. We

may then form the operator

P(D) = aD2 + bD + c

By definition, P(D) applied to y is

P(D)y = (ad2

dt+ b

d

dt+ c)y = ay′′ + by′ + cy

So in this notation,

ay′′ + by′ + cy = 0 is written P(D)y = 0

For example, 2y′′ + 3y′ − y = 0 is written (2D2 + 3D − 1)y = 0.Now we can verify (A), that is, a linear combination of solutions is a solution. This is

based on the fact that P(D) is a linear operator, by which we mean that for any functionsy1, y2 and constants c1, c2, the following property holds (see Exercise 26):

P(D)(c1y1 + c2y2) = c1P(D)y1 + c2P(D)y2 3

Now if y1 and y2 are solutions, that is, if P(D)y1 = 0 and P(D)y2 = 0, then

P(D)(c1y1 + c2y2) = c1P(D)y1 + c2P(D)y2 = 0

This shows that y = c1y1 + c2y2, is also a solution, verifying (A).Next, we state (B) in a precise manner. Two functions y1(t) and y2(t) are called

independent (or linearly independent) if neither is a constant multiple of the other, thatis: both are non-zero and y2(t) �= cy1(t) for all constants c. Fact (B) has the followingSee Exercise 27 for a proof of Theorem 2 based

on Theorem 1.precise statement.

S E C T I O N 19.1 Second Order Linear Differential Equations 3

THEOREM 2 Theorem If y1(t) and y2(t) are independent solutions of (2), then everysolution y(t) can be expressed (in just one way) as a linear combination

y(t) = c1y1(t) + c2y2(t)

We refer to the expression y(t) = c1y1(t) + c2y2(t) as a general solution. Each choiceof constants c1 and c2 gives us a particular solution.

To find solutions of the form described in (C), observe that for any λ,

Deλt = d

dteλt = λeλt

D2eλt = d2

dt2eλt = λ2eλt

and therefore

P(D)eλt = (ad

dt2+ b

d

dt+ c)eλt = (aλ2 + bλ + c)︸︷︷︸

P(λ)

eλt

In other words,Eq. (4) remains true when λ = a + bi is acomplex number. The complex exponential isinterpreted using Euler’s formula

e(a+bi)t = ea(cos ωbt + i sin ωbt)

as discussed in the Appendix on ComplexNumbers.

P(D)eλt = P(λ)eλt 4

This gives us a way of finding exponential solutions. Consider P(D) = D2 + 3D + 2.Then

P(D)eλt = (λ2 + 3λ + 2)︸︷︷︸P(λ)

eλt = (λ − 1)(λ + 2)eλt

For λ = −2, P(−2) = 0 and thus

P(D)e−2t = P(−2)e−2t = 0

Therefore, y = e−2t is a solution of y′′ + 3y′ + 2y = 0.To carry this out in general, we define the characteristic equation

P(λ) = aλ2 + bλ + c = 0 5

It follows from (4) that

P(D)eλt = 0 if and only if P(λ) = 0.

In other words, y = eλt is a solution precisely when λ is a root of the characteristicequation.

Eq. (5) has two roots, given by the quadratic formula:

λ1, λ2 = −b ± √b2 − 4ac

2a

We consider the three possible cases separately: real roots, double root, and complex roots.

Case 1: Distinct Real Roots. The roots λ1, λ2 are real and distinct if b2 − 4ac > 0. Inthis case, the exponential functions

y1 = eλ1t , y2 = eλ2t


are independent solutions of P(D)y = 0 (independent because eλ1t and y2 = eλ2t arenot constant multiples of each other if λ1 �= λ2). By Theorem 2, every solution is a linearcombination of the two exponential solutions, so the general solution is

General Solution (distinct real roots): y = c1eλ1t + c2e

λ2t

EXAMPLE 1 Initial Value Problem - Real Roots Solve the initial value problem

2y′′ − 3y′ − 2y = 0, y(0) = 3

2, y′(0) = −1

8

Solution

Step 1. Solve the Characteristic Equation.The characteristic equation

2λ2 − 3λ − 2 = (2λ + 1)(λ − 2) = 0

has roots λ = − 12 , 2.

Step 2. Write Down General Solution.The functions y = e− 1

2 t and y = e2t are independent solutions. The general solutionis

y = c1e− 1

2 t + c2e2t

2

4

2t

yy2 = e2t

y1 = e t /2

y = e + e2t12

−2

− t54

14

FIGURE 1 Solutions of2y′′ − 3y′ − 2y = 0.

Step 3. Use Initial Conditions.Observe that

y(0) = c1e12 ·0 + c2e

2·0 = c1 + c2

y′(0) = − 12c1e

− 12 ·0 + 2 · c2e

2·0 = − 12c1 + 2c2

Therefore, the initial conditions give us two equations:

c1 + c2 = 3

26

− 12c1 + 2c2 = −1

87

By Eq. (6), c2 = 32 − c1. Substituting in Eq. (7), we obtain

− 12c1 + 2c2 = − 1

2c1 + 2(3

2− c1) = 3 − 5

2c1 = −1

8

This gives c1 = 25 (3 + 1

8 ) = 54 and c2 = 3

2 − c1 = 14 . The desired particular solution

is (see Figure 1)

y = 5

4e− 1

2 t + 1

4e2t

Case 2: Double Root.The characteristic equation has a double root if b2 − 4ac = 0. Thedouble root is

λ = − b

2a


In this case there is only one exponential solution, namely y = eλt . However, it canbe checked that y = teλt is also a solution (see Exercise 28). Since eλt and teλt areindependent, the general solution is

General Solution (double root): y = c1eλt + c2te

λt = (c1 + c2t)eλt

EXAMPLE 2 Initial Value Problem - Double Root Solve the initial value problem

4y′′ − 4y′ + y = 0, y(1) = 0, y′(1) = 1.

y = e t /2

y = (t − 1)e(t − 1)/2

y = te t /2

−4−8 4t

4

y

FIGURE 2 Solutions of4y′′ − 4y′ + y = 0.

Solution

Step 1. Solve the Characteristic Equation.The characteristic equation has a double root λ = 1

2 since

4t2 − 4t2 + 1 = 4(t − 1

2)2 = 0

Step 2. Write Down General Solution.

y = (c1 + c2t)e12 t

Step 3. Use Initial Conditions.The first initial condition gives

y(1) = (c1 + c2)e1/2 = 0

and thus c2 = −c1. Next,

y′(t) = 1

2(c1 + c2t)e

12 t + c2e

12 t = (

1

2c1 + c2 + 1

2c2t)e

12 t

y′(1) =(

12c1 + 3

2c2

)e

12 = 1

Substituting c2 = −c1, we obtain

( 12c1 − 3

2c1)e12 = −c1e

12 = 1 ⇒ c1 = −e− 1

2 , c2 = e− 12

Our particular solution is (see Figure 2)

y = (−e− 12 + e− 1

2 t)e12 t = (t − 1)e

12 (t−1)

Case 3: Complex Roots. If b2 − 4ac < 0, then the roots are complex:

λ1, λ2 = −b ± √b2 − 4ac

2a= −b

2a±

√4ac − b2

2ai

We set

α = −b

2a, ω =

√4ac − b2

2a

and write the roots as follows:

λ1 = α + iω λ2 = α − iω


As in the case of real roots, the exponential functions y = eλ1t and y = eλ2t are solutionsof P(D)y = 0. However, these are complex exponentials, so we interpret them usingEuler’s formula:

eλ1t = e(α+ωi)t = eαt eiωt = eαt (cos ωt + i sin ωt)

eλ2t = e(α−ωi)t = eαt e−iωt = eαt (cos ωt − i sin ωt)

Since we want real solutions, we form the linear combinations:

eαt cos ωt = 1

2(eλ1t + eλ2t ) eαt sin ωt = 1

2i(eλ1t − eλ2t )

These are independent real solutions, and thus the general solution is

General Solution (complex roots): y = eαt (c1 sin ωt + c2 cos ωt)

We can avoid using Euler’s formula by verifyingdirectly that eαt cos ωt and eαt sin ωt satisfy thedifferential equation (see Exercise 29). However,the approach using Euler’s formula revealsclearly that the cases of real and complex rootsare based on the same underlying principle.

EXAMPLE 3 Initial Value Problem - Complex Roots Solve the initial value problem

4y′′ + 4y′ + 13y = 0, y(0) = 3, y′(0) = 3

2.

1

−1

3

2 4 6x

y

FIGURE 3 Graph of the solutiony(t) = e−t/2(3 cos

√3t + √

3 sin√

3t).

Solution

Step 1. Solve the characteristic equation.The roots of 4t2 + 4t + 13 = 0 are

−4 ± √(−4)2 − 4(4)(13)

8= −1

2± √

3 i

Step 2. Write down a general solution.A general solution is

y(t) = e− 12 t (c1 cos

√3t + c2 sin

√3t)

Step 3. Use initial conditions.We have

y(0) = e0(c1 cos 0 + c2 sin 0) = c1

so the first initial condition y(0) = 3 gives us c1 = 3. Therefore

y(t) = e− 12 t (3 cos

√3t + c2 sin

√3t)

y′(t) = −1

2e− 1

2 t (3 cos√

3t + c2 sin√

3t)

+ e− 12 t (−3

√3 sin

√3t + √

3c2 cos√

3t)

y′(0) = −3

2+ √

3c2 = 3

2

We obtain c2 = √3. The desired particular solution is (see Figure 3)

y(t) = e− 12 t

(3 cos

√3t + √

3 sin√

3t)

8


There is a second way of writing the general solutions in the case of complex rootsusing the identity (see Exercise 31)

c1 cos ωt + c2 sin ωt = C cos(ωt − θ) 9

where C =√

c21 + c2

2, cos θ = c1/C, sin θ = c2/C. The value of θ itself if given by

θ =

⎧⎪⎪⎨⎪⎪⎩

tan−1 c2c1

if c1 > 0tan−1 c2

c1+ π if c1 < 0

π2 if c1 = 0 and c2 > 0−π

2 if c1 = 0 and c2 < 0

10

Thus the general solution may be writtenEq. (11) shows that y(t) has the shape of acosine wave whose amplitude grows (or decaysif α < 0) exponentially like eαt . y(t) = Ceαt cos(ωt − θ) (C, θ any constants) 11

EXAMPLE 4 Write the solution in Example 3 in the form (11).

Solution The solution in Example 3 is y = e− 12 t

(3 cos

√3t + √

3 sin√

3t)

. We have

C =√

c21 + c2

2 =√

32 + (√

3)2 = √12 = 2

√3

θ = tan−1 c2

c1= tan−1

√3

3= tan−1 1√

3= π

6

Therefore,

y(t) = 2√

3e− 12 t cos(

√3t − π

6)

19.1 SUMMARY

• A second order linear differential equation has the form

a(t)y′′ + b(t)y′ + c(t)y = f (t)

The equation is called homogeneous if f (t) = 0 and inhomogeneous otherwise. Thisequation may be rewritten

P(D)y = f (t)

where D = d

dtand P(D) = aD2 + bD + c.

• The general second order homogeneous linear differential equation with constant coef-ficients has the form

ay′′ + by′ + cy = 0 12

where a, b, c are constants and a �= 0.• The general solution (12) depends on the nature of the roots of the characteristic equationP(λ) = aλ2 + bλ + c = 0.


Roots of Characteristic Equation General Solution

distinct real roots λ1, λ2 c1eλ1t + c2e

λ2t

double root λ (c1 + c2t)eλt

complex roots λ = α ± ωi eαt (c1 cos ωt + c2 sin ωt)

• In case of complex roots, the general solution may be written in the form

e−αt (c1 cos ωt + c2 sin ωt) = Ceαt cos(ωt − θ)

where C =√

c21 + c2

2 and tan θ = c2/c1. The value of θ itself is given in (10).

19.1 EXERCISES

Preliminary Questions1. Which of the following differential equations are linear?

(a) ty′′ − y2y′ + 4y = et (b) ty′′ − t2y′ + 4y = et

(c) ty′′ − t2y′ + 4y = ey (d) y′y′′ − t2y′ + 4y = et

2. Which of the following are linear differential equations with con-stant coefficients?

(a) y′′ + 2ty = t2 (b) 4y′′ − 2y′ + 7y = t2

(c) y′′ − 17y2 = 0 (d) 2y′′ = 7y′ + y

3. What is the characteristic equation of y′′ + 6y′ + 5y = 0?

4. It is claimed that y = t3 is a solution of t2y′′ − 6y = 0. How canyou verify this claim?

5. Find two independent solutions of y′′ − 5y′ + 6y = 0.

6. Every solution of y′′ − 9y = 0 may be written as a linear combi-nation of e3t and e−3t . Describe the general solution of y′′ + 9y = 0.

Exercises1. Show that y(t) = t4 − 6t2 + 3 satisfies y′′ − ty′ + 4y = 0.

2. Verify that y(t) = t3 − 9t2 + 18t − 6 is a solution of the initialvalue problem

ty′′ + (1 − t)y′ + 3y = 0 y(0) = −6, y′(0) = 18

3. Verify that y(t) = (1 − 2t2)e−t2/2 is a solution to the initial valueproblem

y′′ = (t2 − 5)y y(0) = 1, y′(0) = 0

4. Show that y = c1t + c2t ln t is a general solution of

t2y′′ − ty′ + y = 0

Find the particular solution satisfying y(1) = 4, y′(1) = 3.

5. Consider y′′ − 5y′ + 6y = 0.

(a) What is the characteristic equation and what are its roots?

(b) Verify directly that y = e2t and y = e3t are solutions.

(c) Let y be a solution satisfying y(0) = 1, y′(0) = 0. Show thaty = c1e2t + c2e3t where c1 + c2 = 1 and 2c1 + 3c2 = 0. Then solvefor c1 and c2.

6. Consider y′′ − 4y + 13y = 0.

(a) What is the characteristic equation and what are its roots?

(b) Verify directly that y = e2t sin 3t and y = e2t cos 3t are solutions.

(c) Let y(t) be the solution satisfying y(0) = 1, y′(0) = 0. Show thaty = e2t (c1 cos 3t + c2 sin 3t) where c1 = 1 and 2c1 + 3c2 = 0. Findy(t).

7. Write each of the following expressions in the form C cos(4t − θ)

for suitable constants C and θ .

(a) 3√

3 cos 4t + 9 sin 4t

(b) −3√

3 cos 4t + 9 sin 4t

(c) 3√

3 cos 4t − 9 sin 4t

(d) −3√

3 cos 4t − 9 sin 4t

8. Solve the initial value problem

y′′ − 2y′ + 2y = 0, y(0) = 1, y′(0) = 2

and write your answer in the form y(t) = Cet cos(t − θ).

Exercises 9–16: Find the general solution of the differential equation.

9. y′′ + 6y′ + 8y = 0 10. y′′ + 25y = 0

11. y′′ − 25y = 0 12. y′′ − 2y′ − 3y = 0

13. y′′ + 2y′ + 2y = 0 14. y′′ − 4y′ + 4y = 0

15. 2y′′ − 2y′ + 5y = 0. 16. 4y′′ + 12y′ + 9y = 0

Exercises 17–24: Find the particular solution of the differential equa-tion satisfying the initial conditions.

17. y′′ − 4y = 0; y(0) = 1, y′(0) = 0

18. y′′ − 4y = 0; y(1) = 0, y′(1) = 1

S E C T I O N 19.2 Inhomogeneous Equations 9

19. 2y′′ + y′ − 3y = 0; y(0) = 1, y′(0) = −1

20. y′′ − 6y′ + 13y = 0; y(0) = 2, y′(0) = 1

21. 4y′′ + 20y′ + 25y = 0; y(0) = 2, y′(0) = 0

22. 2y′′ + 6y′ = 0; y(1) = 3, y′(1) = 9

23. 3y′′ − 5y′ − 2y = 0; y(1) = 2, y′(1) = −2

24. 9y′′ + 18y′ + 10y = 0; y(π) = 3, y′(π) = −3

25. Show that for k �= 0, cosh(kt) and sinh(kt) are independent so-lutions of y′′ − k2y = 0. Conclude that a general solution is givenby

y = c1 cosh(kt) + c2 sinh(kt)

This is an alternate to the general solution y = c1ekt + c2e−kt

26. Verify (3), which states that for all constants c1, c2,

P(D)(c1y1 + c2y2) = c1P(D)y1 + c2P(D)y2

Further Insights and Challenges27. In this exercise, we prove Theorem 2: if y1 and y2 are independentsolutions of

ay′′ + by′ + cy = 0

then every solution y may be expressed as a linear combination of y1and y2.

(a) Use the uniqueness part of Theorem 1 to prove that the vectors

u = ⟨y1(0), y′

1(0)⟩, v = ⟨

y2(0), y′2(0)

⟩are not multiples of each other. It follows that u and v span R2, that is,every vector w in R2 is of the form w = c1u + c2v.(b) Set w = ⟨

y(0), y′(0)⟩. Let c1, c2 be constants such that w =

c1u + c2v and set g(t) = c1y1(t) + c2y2(t). Use Theorem 1 to provethat y(t) = g(t).

28. Double Roots Suppose that the characteristic equation of ay′′ +by′ + cy = 0 has a double root λ. Verify directly that y = teλt is asolution of ay′′ + by′ + cy = 0. Hint: in the case of a double root,λ = −b/(2a).

29. Complex Roots Suppose that the characteristic equation of ay′′ +by′ + cy = 0 has complex roots λ = α ± ωi where α = −b/(2a)

and ω = 12a

√4ac − b2. Verify directly that y = eαt cos ωt and

y = eαt sin ωt are solutions. Suggestion: a nice approach is to showfirst that the function u(t) = y(t)e−αt satisfies

u′′ + ω2u = 0

30. Let y(t) be a solution of ay′′ + cy′ = 0.

(a) Suppose that a = 2, c = 3. Calculate limt→∞ y(t) where y(t) satis-

fies y(0) = 10, y′(0) = 6.

(b) Prove that if a and c are positive, then limt→∞ y(t) = K where

K = y(0) + ac y′(0).

31. Use the addition law for the cosine function to verify the identity(9) with the given values of C and θ . Recall that by definition, tan−1 x

takes values in (−π2 , π

2 ).

32. Uniqueness via Energy This exercise proves the uniqueness ofsolutions to the initial value problem (Theorem 1) for

ay′′ + by′ + cy = 0

in the case where a, b, c ≥ 0. Define the energy E(t) of a solution by

E(t) = ay′(t)2 + cy(t)2

(a) Prove that E′(t) = −2b(y′)2. Hint: Use the Chain Rule to calculateE′(t).(b) Conclude that E(t) is a nonincreasing function of t such thatE(t) ≥ 0 for all t .

(c) Use (b) to show that if y(t0) = y′(t0) = 0, then y(t) = 0 for all t .

(d) Prove that if y1 and y2 are solutions satisfying the same initialconditions:

y1(t0) = y2(t0), y′1(t0) = y′

2(t0)

then y1 = y2. Hint: Apply (c) to y = y1 − y2.

19.2 Inhomogeneous Equations

We now consider an inhomogeneous linear differential equation where the forcing functionf (t) is nonzero:

ay′′ + by′ + cy = f (t) 1

As before, we set P(D) = aD2 + bD + c and write (1) as P(D)y = f .There is an important relationship between solutions of (1) and solutions of the asso-

ciated homogeneous equation

ay′′ + by′ + cy = 0


In fact, the next theorem tells us that if we find a single particular solution yp(t) of (1), thenwe obtain all solutions by adding on solutions of the associated homogeneous equation.

THEOREM 1 Let yp(t) be a particular solution of (1). If yh(t) is any solution of theassociated homogeneous equation, then

y(t) = yp(t) + yh(t)

is also solution of (1). Furthermore, every solution is obtained as a sum y(t) = yp(t) +yh(t) in this way.

Proof If P(D)yp = f and P(D)yh = 0, then

P(D)(yp + yh) = P(D)yp + P(D)yh = f + 0 = f

This proves that y = yp + yh is a solution of the inhomogeneous equation. On the otherhand, if y(t) is any solution of (1), set g(t) = yp(t) − yh(t). Then

P(D)g = P(D)y − P(D)yp = f (t) − f (t) = 0

Thus g(t) is a solution of the associated homogeneous equation and y(t) = yp(t) + g(t)

as required.

By Theorem 1, if y1 and y2 are independent solutions of the associated homogenousequation, then (1) has general solution

y(t) = yp(t) + c1y1(t) + c2y2(t) (c1, c2 any constants)

This leaves us with the problem of finding a single particular solution yp. Solutionsare guaranteed to exist (by Theorem 1 in Section 1), but in many cases it is impossibleto write them down in terms of elementary functions. However, this is possible for cer-tain special classes of forcing functions. In the remainder of this section we discuss twoimportant techniques for finding such explicit solutions.

Method of Undetermined Coefficients

This method can be used whenever the forcing function is an exponential function, poly-nomial, sine or cosine (or when it is sum of products of such functions). We shall focuson the following cases:

(i) f (t) = Aeαt for some constants A, α

(ii) f (t) = A cos βt + B cos βt for some constants A, B, and β

(iii) f (t) = eαt (A cos βt + B cos βt) for some constants A, B, α, β

(iv) Any polynomial f (t) = Amtm + Am−1tm−1 + · · · + A1t + A0

In each case, we look for a particular “trial" solution yp(t) consisting of terms of the sameform as the forcing function and its derivatives (see Table 1 in the section summary).This strategy works because, for these special classes of functions, P(D)yp has the same


general form as yp itself. For example, if yp is an exponential function, then P(D)yp isan exponential of the same type. If yp is a sum of sines and cosines, then so is P(D)yp,etc.

To apply this method when the forcing function is an exponential function, as in theequation

ay′′ + by′ + cy = Ceαt

we take a trial function of the form

yp = Aeαt (A an unknown constant)

Then

P(D)yp = (aD2 + bD + cD)(Aeαt ) = P(α)Aeαt

So yP satisfies P(D)yp = Ceαt if P(α)A = C. If P(α) in nonzero, then A = C/P (α)

and we obtain the particular solution

yp = C

P(α)eαt

EXAMPLE 1 Exponential Forcing Function Solve the initial value problem

y′′ − 4y′ + 3y = 5e−4t , y(0) = 4, y′(0) = −1 2

Solution

Step 1. Find the coefficients of a “trial" solution.We look first for a solution of the form: yp(t) = Ae−4t . Let P(D) = D2 − 4D + 3.Then P(−4) = 35 and

P(D)yp = P(D)(Ae−4t ) = P(−4)Ae−4t = 35Ae−4t

So yp is a particular solution if P(D)yp = 35Ae−4t = 5e−4t . Thus A = 5/35 = 1/7and we obtain

yp(t) = 1

7e−4t

Step 2. Solve the homogeneous equation.The roots of the characteristic equation

P(λ) = λ2 − 4λ + 3 = (λ − 1)(λ − 3) = 0

are λ = 1, 3. The homogeneous equation has general solution

yh(t) = c1et + c2e

3t

Step 3. Write down the general solution.The general solution of Eq. (2) is

y(t) = yp(t) + yh(t) = 1

7e−4t + c1e

t + c2e3t


Step 4. Find the particular solution.We have

y(t) = 1

7e−4t + c1e

t + c2e3t

y′(t) = −4

7e−4t + c1e

t + 3c2e3t

The initial conditions yield the equations

y(0) = 1

7+ c1 + c2 = 4

y′(0) = −4

7+ c1 + 3c2 = −1

Subtract the second equation from the first to obtain

5

7− 2c2 = 5 ⇒ c2 = −15

7

The first equation then yields

1

7+ c1 − 15

7= 4 ⇒ c1 = 6

Thus, the solution to the initial value problem is (see Figure 1)

2

4

−1 1t

y

FIGURE 1 The particular solutiony(t) = 1

7 e−4t + 325 et − 17

7 e3t

ofy′′ − 4y′ + 3y = 5e−4t .

y(t) = 1

7e−4t + 6et − 15

7e3t

EXAMPLE 2 Polynomial Forcing Function Find the general solution of the differen-tial equation

y′′ − 4y′ + 3y = 3t2 − 5.

Solution

Step 1. Find the coefficients of a “trial" solution.The forcing function f (t) = 3t2 − 5 is a polynomial of degree two, so we consider atrial solution of the same type

yp(t) = At2 + Bt + C

We have

y′p(t) = 2At + B

y′′p(t) = 2A

y′′p − 4y′

p + 3yp = (2A) − 4(2At + B) + 3(At2 + Bt + C)

= 3At2 + (−8A + 3B)t + (2A − 4B + 3C)︸︷︷︸this must equal f (t)

Therefore, yp is a particular solution if

3At2 + (−8A + 3)t + (2A − 4B + 3C) = 3t2 − 5


Equate coefficients of powers of t on both sides:

3A = 3 − 8A + 3B = 0 2A − 4B + 3C = −5

The first equation yields A = 1 and the second equation then yields B = 83 . By the

third equation,

C = 1

3(−5 − 2A + 4B) = 1

3(−5 − 2 + 32

3) = 11

9

This yields the particular solution

yp(t) = t2 + 8

3t + 11

9

Step 2. Solve the homogeneous equation.As in Example 1, the associated homogeneous equation y′′ − 4y′ + 3y = 0 has generalsolution yh(t) = c1e

t + c2e3t .

Step 3. Write down the general solution.Our inhomogeneous equation y′′ + 2y′ − 8y = 3t2 − 5 has general solution

y(t) = yp(t) + yh(t) = t2 + 8

3t + 11

9+ c1e

t + c2e3t

Solutions of inhomogeneous equations are additive in the following sense: if y1(t)

and y2(t) satisfy

ay′′1 + by′

1 + cy1 = f (t)

ay′′2 + by′

2 + cy2 = g(t)

then for all constants A, B, y(t) = Ay1(t) + By2(t) is a solution ofThe additivity of solutions described here isoften referred to as the Principle ofSuperposition. It holds for all linear differentialequations and plays a fundamental role inphysical applications.

ay′′ + by′ + cy = Af (t) + Bg(t)

EXAMPLE 3 Additivity Find a particular solution of

y′′ − 4y′ + 3y = 3t2 − 5t + 45e−4t 3

Solution According to Examples 1 and 2,

y1(t) = t2 + 8

3t + 11

9is a solution of y′′ − 4y′ + 3y = 3t2 − 5t

y2(t) = 1

7e−4t is a solution of y′′ − 4y′ + 3y = 5e−4t

Thus 9y2 satisfies y′′ − 4y′ + 3y = 45e−4t and the sum

yp(t) = y1(t) + 9y2(t) = t2 + 8

3t + 11

9+ 9

7e−4t y2(t)

is a particular solution of Eq. (3).


EXAMPLE 4 Exponential-trigonometric forcing function Find any particular solutionof

y′′ + 6y′ + 8y = 10e5t cos 7t.

Solution Although the forcing function has no sine term, our trial solution must includeboth cosine and sine:

yp = e5t (A cos 7t + B sin 7t)

After simplification, we obtain

y′p = e5t

((5A + 7B) cos 7t + (−7A + 5B) sin 7t

)y′′p = e5t

((−24A + 70B) cos 7t − (70A + 24B) sin 7t

)y′′p + 6y′

p + 8yp = 14e5t((A + 8B) cos 7t + (−8A + B) sin 7t

)We choose A and B so that

14e5t((A + 8B) cos 7t + (−8A + B) sin 7t

)= 10e5t cos 7t

The coefficient of sin 7t on the left must equal zero. Thus −8A + B = 0 or B = 8A.Matching the coefficients of cos 7t , we obtain

14(A + 8B) = 14(A + 64A) = 910A = 10 ⇒ A = 1

91, B = 8

91

This yields the particular solution

yp(t) = e5t (1

91cos 7t + 8

91sin 7t)

The method we have used so far must be modified when the forcing function f (t)

contains terms that satisfy the associated homogeneous equation. To see why, considerthe equation

y′′ − y = et

If we were to proceed as before, we would try to find a solution of the form y = Aet . Butthis is not a solution for any value of A because

y′′ − y = Aet − Aet = 0 �= et

Here is a general rule: if f (t) satisfies the associated homogeneous equation, multiplyFurther modification is required if the forcingfunction is tk times a solution of the associatedhomogeneous equation for some k ≥ 1. We omitdiscussion of these cases.

each term of the previous trial solution by t (or t2 if the characteristic equation has adouble root), as in the following examples.

EquationSolutions ofhomogeneousequation

Trial solution

y′′ − 3y′ + 2y = et y = et , e2t yp(t) = Atet instead of et

y′′ − 2y′ + y = et y = et , tet yp(t) = At2et instead of et

y′′ − 2y′ + y = tet y = et , tet yp(t) = At3et instead of tet

y′′ + y = cos t y = sin t, cos t yp(t) = At cos t + Bt sin t

y′′ = 1 y = 1, t yp(t) = At2


EXAMPLE 5 Exponential forcing function - an exceptional case Find the general so-lution of y′′ − 4y′ + 3y = e3t .

Solution f (t) = e3t is a solution of the associated homogeneous equation since 3 isa root of the characteristic equation λ2 − 4λ + 3λ = (λ − 3)(λ − 1) = 0. Therefore, weuse the trial function yp(t) = Ate3t :

y′p(t) = A(1 + 3t)e3t

y′′p(t) = A(6 + 9t)e3t

y′′p − 4y′

p + 3yp = A(6 + 9t)e3t − 4A(1 + 3t)e3t + 3Ate3t = 2Ae3t

So yp(t) is a solution if 2Ae3t = e3t . Thus A = 12 and the general solution is

y(t) = 1

2te3t + c1e

t + c2e3t

EXAMPLE 6 Polynomial forcing function - an exceptional case Find the general so-lution of y′′ + 7y′ = 1.

Solution Since y = 1 satisfies y′′ + 7y′ = 1, we use the trial solution yp(t) = At .Substitute yp in the differential equation:

y′′p + 7y′

p = 0 + 7A = 1

So A = 17 and yp(t) = 1

7 t . The characteristic equation λ2 + 7λ = 0 has roots λ = 0, −7,and the general solution is

y(t) = 1

7t + c1 + c2e

−7t

In the next example, we use the theory we have developed to analyze the followingphysical situation∗. Consider a tube rotating counterclockwise about its midpoint in aplane at a constant angular velocity of ω radians per second (Figure 2 (A)). If we assumethat the tube is horizontal at time t = 0, then it makes an angle θ = ωt with the horizontalat time t . Our goal is to determine the motion of a small ball of mass m inside the tubethat is free to slide back and forth without friction.

The position of the ball at time t is determined by its distance r(t) from the midpointof the tube, which we place at the origin. Gravity exerts a force −mg on the ball. The onlyforce acting on the ball in the radial direction is the radial component of gravity:

F = −mg sin θ = −mg sin ωt

Newton’s Law of Motion F = ma, can be used to show that r(t) satisfies the differentialequation

mr ′′ − mω2r = −mg sin ωt 4

The term mω2r , referred to as the centrifugal force, is a so-called “fictitious force" thatappears because system is rotating.

∗This example is adapted from p. 380-381 in Ordinary Differential Equations by M. Tenenbaum and H. Pollard,Dover Publications, NY 1963


r(t)

−mg

(A) (B)

FIGURE 2

EXAMPLE 7 Ball is a Rotating Tube Describe the motion of the ball in the rotatingtube. Is it possible for the ball to remain inside the tube indefinitely?

Solution By the Method of Undetermined Coefficients, (4) has a particular solution ofthe form rp(t) = A sin ωt :

r ′′p − ω2rp = −Aω2 sin ωt − ω2(A sin ωt) = −g sin ωt

Thus A = 12g/ω2 and

rp(t) = g

2ω2sin ωt

The roots of the characteristic equation λ2 − ω2 = 0 are λ = ±ω, so the general solutionis

r(t) = c1eωt + c2e

−ωt + g

2ω2sin ωt

Since ω > 0, the term eωt tends to infinity as t → ∞. So if c1 �= 0, then r(t) increasesuntil the ball flies out of the tube. The ball remains inside the tube indefinitely if and onlyif c1 = 0. In this case,

r(t) = c2e−ωt + g

2ω2sin ωt

Note that since the first term tends to zero as t → ∞, r(t) approaches the particularsolution rp(t).

The particular solution rp(t) is the unique periodic solution of (4). It is interesting todetermine the ball’s path in this case. Since θ = ωt , the path satisfies the polar equation

r = 1

2(g/ω2) sin θ . This is the equation of a circle of radius 1

4g/ω2 shown in Figure 2 (B).

Variation of Parameters

A second method for finding a particular solution to

ay′′ + by′ + cy = f (t) 5

consists of seeking a particular solution of the form

yp(t) = u1(t)y1(t) + u2(t)y2(t) 6


where y1(t) and y2(t) are independent solutions of the associated homogeneous equation.This method is called variation of parameters because we replace the constants in thegeneral solution y = c1y1 + c2y2 of the homogeneous equation with functions u1 and u2.

To describe the method, we define the Wronskian determinant W for any two func-tions y1 and y2:

W =∣∣∣∣ y1 y2

y′1 y′

2

∣∣∣∣ = y1y′2 − y′

1y2

In Exercise 54, we verify that if y1 and y2 are solutions of the homogeneous equation,then function

yp(t) = u1(t)y1(t) + u2(t)y2(t)

satisfies (5) if u1 and u2 satisfy the differential equations

u′1 = −y2f

aWu′

2 = y1f

aW7

Thus, we obtain a particular solution if we can solve these equations for u1 and u2. As youmay expect, this can be done explicitly only for certain choices of forcing function f (t).

When it applies, the Method of IndeterminateCoefficients is easier to carry out than Variationof Parameters. However, Variation of Parametersapplies to more general types of forcingfunctions and also to linear differential equationswhose coefficients are not necessarily constant.But keep in mind that in many cases, it is notpossible to express the solutions in terms ofelementary functions.

EXAMPLE 8 Find the general solution of y′′ + y = sec t .

Solution

Step 1. Find independent solutions of the associated homogeneous equation.Two independent solutions of solutions of

y′′ + y = 0

are y1(t) = cos t and y2(t) = sin t .

Step 2. Calculate the Wronskian.

W = y1y′2 − y′

1y2 = cos t (cos t) − (− sin t)(sin t) = cos2 t + sin2 t = 1

Step 3. Solve the differential equations for u1 and u2.Since W = 1, the differential equation for u1 in (7) is

u′1 = −y2f

aW= −y2f = − sin t (sec t) = − tan t

The solution is u1 = − ∫tan t dt = ln | cos t | + C. We set C = 0 and take u1 =

ln | cos t | since we need just solution. Similarly, the differential equation for u2 is

u′2 = y1f

aW= y1f = cos t (sec t) = 1

The general solution is u2 = ∫dt = t + C. We take u2 = t . Thus

u1 = ln | cos t |, u2 = t


Step 4. Write down the particular and general solutions.

yp(t) = u1(t)y1(t) + u2(t)y2(t) = ln | cos t | cos t + t sin t

The general solution to the inhomogeneous differential equation is

y(t) =(

ln | cos t | cos t + t sin t)

+ c1 cos t + c2 sin t

19.2 SUMMARY

• We consider the inhomogeneous second order linear differential equation

ay′′ + by′ + cy = f (t) 8

where a, b, c are constants and f (t) is a non-zero “forcing" function. The associatedhomogeneous equation is the equation

ay′′ + by′ + cy = 0

• The general solution of Eq. (8) is

y(t) = yp(t) + c1y1(t) + c2y2(t)

where yp(t) is any particular solution and y1(t), y2(t) are independent solutions of theassociated homogeneous equation.• Method of Undetermined Coefficients: look for a particular solution having the sameform as the forcing function. The following trial solutions should be used if f (t) is not tk

times a solution of the associated homogeneous equation.

Forcing function f (t) Trial solution

eαt yp(t) = Aeαt

tmeαt yp(t) = A1eαt + A2teαt + · · · + Amtmeαt

cos βt or sin βt yp(t) = A cos βt + B sin βt

eαt cos βt or eαt sin βt yp(t) = eαt (A cos βt + B sin βt)

antn + an−1tn−1 + · · · a0 Antn + An−1tn−1 + · · · A0

TABLE 1

• If f (t) is a solution of the associated homogeneous equation, multiply the above trialfunction by t (or t2 if the characteristic equation has double roots).• The Wronskian of two functions y1 and y2 is the 2 × 2 determinant

W =∣∣∣∣ y1 y2

y′1 y′

2

∣∣∣∣ = y1y′2 − y′

1y2


• Variation of Parameters: Let y1 and y2 be independent solutions of the associatedhomogeneous equation and let u1 and u2 be solutions of the differential equations

u′1 = −y2f

aWu′

2 = y1f

aW9

Then yp = u1(t)y1(t) + u2(t)y2(t) is a particular solution of the inhomogeneous equation(8).

19.2 EXERCISES

Preliminary Questions1. What is the homogeneous equation associated to 4y′′ + 3y′ + 9y =

sin 2t?

Questions 2 - 4, refer to the inhomogeneous equation with non-zeroforcing function f (t):

ay′′ + by′ + cy = f (t) 10

2. Which of the following conclusions are correct? If y1 and y2 aresolutions of 10, then

(a) The sum y1 + y2 is also a solution of (10).(b) The sum y1 + y2 is a solution of the associated homogeneousequation.(c) The difference y1 − y2 is a solution of the associated homogeneousequation.

3. Suppose that y(t) is a solution of (10). Is 5y(t) also a solution of(10)? If not, which differential equation does 5y(t) satisfy?

4. Describe a general solution of (10) in terms of a particular solu-tion yp(t) and independent solutions y1(t) and y2(t) of the associatedhomogeneous equation.

5. Which trial function would be used in the method of undeterminedcoefficients applied to the equation y′′ + 7y′ + 10y = e9t .

6. Why is y = Ae2t not suitable as a trial function for y′′ − 7y +10y = e2t?

7. Is yp(t) = A sin 2t suitable as a trial function for y′′ − 7y′ +10y = sin 2t?

Exercises1. Find a solution of y′′ + 3y′ − 10y = e5t of the form y = Ae5t .

2. Find a solution of y′′ + 3y′ − 10y = 19 − 50t2 of the formy = At2 + Bt + C.

3. Show that y = A cos 2t + B sin 2t is a solution of y′′ + y′ − 6y =3 cos 2t − 2 sin 2t if

−10A + 2B = 3, 2A + 10B = 2

4. The general solution of y′′ + y′ − 2y = 0 is y = c1et + c2e−2t .Show that y′′ + y′ − 2y = et has a particular solution of the formy = Atet but does not have any solution of the form y = Aet .

In Exercises 5 - 24, use the method of undetermined coefficients to finda particular solution of the inhomogeneous differential equation.

5. 3y′′ + 5y = t3

6. y′′ − y′ + 3y = 9t2 + 6t

7. y′′ − y′ + y = t3

8. y′′ + 4y′ + y = e3t

9. y′′ − 3y′ + 8y = e−t

10. y′′ − 3y′ + 8y = 4 − 9e−t

11. y′′ − 3y′ + 8y = 4t − 9e−t

12. y′′ − 4y′ + 9y = et + 2e−t

13. y′′ + 3y′ + 10y = cos 2t .

14. y′′ + 3y′ = cos 5t − 13 sin 5t

15. 3y′′ − y′ = sin t

16. y′′ + y′ − 2y = cos 2t − 3 sin 2t

17. y′′ − 3y = et sin t

18. y′′ − 16y = 72t2e3t

19. y′′ + 16y = cos 4t − sin 4t

20. y′′ + y′ − 2y = et

21. y′′ + y = tet

22. y′′ − 2y′ = 130e−6t cos 4t

23. y′′ + 2y′ = 4

24. y′′ + 2y′ − 4y = tet

In Exercises 25 - 30, find the general solution of the inhomogeneousdifferential equation.

25. y′′ − 2y′ + 3y = sin 3t

26. y′′ + y′ = 12e−t


27. y′′ + 4y = 12 sin 2t

28. y′′ − 6y′ + 9y = t

29. y′′ − 2y′ + y = e−3t cos 2t

30. y′′ + 3y′ − 10y = 27t2e−t

In Exercises 31 - 38, solve the initial value problem.

31. y′′ − y = t , y(0) = 1, y′(0) = 1

32. y′′ + 2y′ + y = e2t , y(0) = −2, y′(0) = 3

33. y′′ − 5y′ − 6y = cos 2t , y(0) = 0, y′(0) = 1

34. y′′ − 6y′ + 9y = 36, y(0) = 2, y′(0) = 1

35. y′′ − 2y′ = e3t , y(0) = e, y′(0) = 0

36. y′′ − 2y′ − 8y = cos 2t , y(0) = −1, y′(0) = 12

37. y′′ − 5y′ + 4y = et , y(0) = 2, y′(0) = 3

38. y′′ − 2y′ − 3y = 120 cos 3t , y(0) = 4, y′(0) = 0

In Exercises 39 - 43, use variation of parameters to find a particularsolution of the inhomogeneous differential equation.

39. y′′ + y = e4t

40. y′′ + 4y = t

41. y′′ − 6y′ = 1

42. y′′ − 2y′ + y = t1/2et

43. y′′ − y′ = cos2 t

44. y′′ + 2y′ = t2et

45. y′′ + 4y′ + 4y = e−2t ln t

46. y′′ − 4y′ + 3y = (t sin t)e2t

47. A mass of m kg slides along a groove in a horizontal surface.Assume that the mass is subject to a friction force −rx′(t) wherex(t) is the mass’s position at time t . By Newton’s Second Law,mx′′(t) + rx′(t) = 0.

(a) Determine the general solution x(t) subject to the initial conditionx(0) = 0.(b) Prove that if the mass is given an initial velocity v0 = x′(0), then

limt→∞ x(t) = m

rv0

We interpret this as saying that the mass travels a distance mv0/r beforecoming to a halt. However, the actual solution x(t) describes a massthat moves more and more slowly, as it approaches its limit.

(c) Assume that m = 2 kg and r = 4 kg/s. How far does the masstravel if v0 = 1.5m/s?

48. In the situation of Exercise 47, suppose that the mass is subject toa constant force of F Newtons. In this case, Newton’s Law of Motiongives mx′′(t) + rx′(t) = F .

(a) Show that if x(0) = 0 and x′(0) = 0, then

x(t) = F

rt − Fm

(1 − e−rt/m

r2

)(b) Show that the mass has a terminal velocity of F/r , that is,lim→∞ x′(t) = F/r . Without friction, the velocity would increase indef-

initely since we would have mx′′(t) = F and x′(t) = (F/m)t .

Exercises 49 - 52 refer to example 7. Let r0 = r(0) denote the initialposition and r ′

0 = r ′(0) the initial radial velocity of the ball.

49. Suppose that r0 = 0.6 m and r ′0 = 0.

(a) Determine r(t) if ω = 6π rad/s.

(b) For which value of ω will the ball remain in the tube indefinitely?

50. Prove that the ball remains in the tube indefinitely if and only ifthe initial values satisfy

r0ω + r ′0 = g

2ω

51. Suppose that r0 = 0 and that ω = 3π rad/s. What initial radialvelocity r ′

0 must the ball have in order to executed periodic motion ina circle?

52. Suppose that the tube in Figure 2 (A) rotates on a horizontal tabletop without friction. In this case, r(t) satisfies (??) with the gravita-tional term −mg sin ωt replaced by zero. Show that only two types ofbehavior are possible: either the ball either flies out of the tube or itapproaches the origin as t → ∞.

53. Verify the following additivity property. if y1 and y2 are solutionsof

ay′′1 + by′

1 + cy1 = f (t)

ay′′2 + by′

2 + cy2 = g(t)

then y = y1 + y2 is a solution of

ay′′ + by′ + cy = f (t) + g(t)

Further Insights and Challenges54. Variation of Parameters. Show that if u1 and u2 satisfy (7), thenthe following two equations hold:

u′1y1 + u′

2y2 = 0 and u′1y′

1 + u′2y′

2 = a−1f

Then use these two equation to verify directly that yp = u1y1 + u2y2is a solution of ay′′ + by′ + cy = f if y1 and y2 are independent so-lutions of the homogeneous equation ay′′ + by′ + cy = 0.

S E C T I O N 19.3 Oscillations and Resonance 21

19.3 Oscillations and Resonance

Oscillations occur everywhere in the physical world, from the swinging of a pendulumand the tremors of an earthquake to the ubiquitous vibrations of atoms, molecules, andsubatomic particles. Many oscillating systems are modeled accurately by second order,constant coefficient differential equations, so we may use the techniques developed in thischapter to predict their behavior. Countless experiments have confirmed the validity ofthese mathematical predictions.

Our analysis of oscillations is divided into three cases:

1. simple harmonic motion

2. damped harmonic motion

3. forced oscillations

A typical example of simple harmonic motion is the motion of a mass m attachedto the end of the spring. Let x(t) be the position of the mass at time t , where x = 0 is

In this section, the unknown function is denotedx(t) instead of y(t).

the equilibrium position (Figure 1). Hooke’s Law, which is valid for small displacementsfrom equilibrium, states that the spring exerts a restoring force F(x) = −kx where k > 0is the spring constant (in units of force per unit distance). The velocity and accelerationof the mass are

v(t) = x′(t), a(t) = x′′(t)

Newton’s law, F = ma yields −kx(t) = mx′′(t) or

0Equilibrium

position

x

FIGURE 1 Mass m oscillating at the endof a spring

mx′′(t) + kx(t) = 0 1

This is the differential equation for simple harmonic motion. The characteristic equationmλ2 + k = 0 has two purely imaginary roots ±ω0i, where

ω0 =√

k

m

The general solution of (1) may be written in two ways:

x(t) = c1 cos ω0t + c2 sin ω0t or x(t) = C cos(ω0t − θ) 2

where C =√

c21 + c2

2 and tan θ = c2/c1.The maximum value of |x(t)|, called the amplitude, is equal to C. The constant ω0

(in radians per unit time) is called the natural angular frequency of the oscillator. Thefrequency of the oscillations (in cycles per second) is f = ω0/(2π) and the time requiredto complete one cycle, called the period, is equal to T = 2π/ω0.• Natural Angular frequency: ω0 =

√k

m• Frequency: f = ω0/(2π)

• Amplitude: C =√

c21 + c2

2

• Period: T = 2π

ω0

EXAMPLE 1 Simple Harmonic Motion A 1.5 kg mass is attached to a spring withspring constant k = 50 N/m. The mass is stretched 0.5 meters past equilibrium andreleased with initial velocity 1 m/s (directed away from equilibrium).

(a) Determine the motion of the spring.

(b) Write the motion in the form x(t) = A cos(ω0t − θ).


Solution

1.12

0.5

−1.12

x(t)

1 2 3t

FIGURE 2 Simple Harmonic Motion ofthe Spring in Example 1:x(t) = 1.12 cos(5.77t − 1.11)

(a) The position x of the mass satisfies Eq. (1) with m = 1.5 and k = 50:

1.5x′′ + 50x = 0

The angular frequency is

ω0 =√

k

m=

√50

1.5≈ 5.77

By Eq. (2), the general solution is

x(t) = c1 cos 5.77t + c2 sin 5.77t

We use the initial conditions to determine the constants c1 and c2:

x(0) = c1 cos 0 + c2 sin 0 = c1 = 0.5

x′(0) = −c1 sin 0 + c2 cos 0 = c2 = 1

Therefore

x(t) = 0.5 cos 5.77t + sin 5.77t

(b) As shown in Section 1, we can rewrite the solution as x(t) = C cos(ω0t − θ), where

C =√

c21 + c2

2 =√

.52 + 12 ≈ 1.12, tan θ = c2

c1= 1/0.5 = 2

Since c1 > 0, we choose θ in the interval [0, π2 ). Thus θ = tan−1 2 ≈ 1.11 and (Figure 2)

x(t) = 1.12 cos(5.77t − 1.11)

CONCEPTUAL INSIGHT We are able to solve for x(t) because the restoring force F(x) =−kx in Hooke’s Law is linear in x. However, Hooke’s Law is valid only if the springis not stretched too far. Some physical systems, such as a pendulum undergoing largeoscillations and certain molecular vibrations, are more accurately modeled by nonlinearrestoring forces such as F(x) = −kx + cx3. The resulting oscillatory motion, calledanharmonic, is more complicated than simple harmonic motion and cannot be describedexplicitly in terms of elementary functions.

The differential equation of simple harmonic motion does not take into account fric-tional forces, which diminish and eventually extinguish the oscillations of any physicalsystem (unless new energy is added to the system). In many situations, it is reasonableto assume that the frictional force is proportional to velocity x′(t) and thus is equal to−rx′(t) where r > 0 is the frictional constant (the minus sign appears because the frictionacts opposite to the direction of motion). The total force acting on the mass becomesF = −rx′ − kx and Newton’s law gives −rx′(t) − kx(t) = mx′′(t) or

mx′′(t) + rx′(t) + kx(t) = 0 3

This is the equation of a damped oscillator or damped harmonic motion. It is nothing morethan a homogeneous second order differential equation (of the type studied in Section 2)with the added restriction that the coefficients m, r , and k be positive.


CONCEPTUAL INSIGHT The motion of a damped oscillator depends on the size of thefrictional forces. If the frictional forces are small (that is, if r is small), we wouldexpect the oscillations to die out slowly over time (Figure 3). The oscillator is said tobe underdamped. But if r is large enough, the damping may be so great that the massis never able to oscillate (Figure 4). Instead, it simply slows down as it approaches theequilibrium position (Figure 4). In this case, the oscillator is said to be overdamped.

x(t)

t

FIGURE 3 Underdamped oscillations decayexponentially

A

x(t)

t

FIGURE 4 Overdamped motion: amplitudedecays exponentially

We now show that overdamping occurs if the roots of the characteristic equation arereal and underdamping occurs if they are complex. The roots of the characteristic equation

mλ2 + rλ + k = 0

are

λ1, λ2 = −r ± √r2 − 4mk

2m4

Case 1: Real roots (Overdamping) The roots are real if r2 − 4mk > 0. Since m, r , andk are positive, we have r >

√r2 − 4mk. It follows that both λ1 and λ2 are negative.

Therefore, the general solution of (3) is a sum of decaying exponentials:

x(t) = c1eλ1t + c2e

λ2t (λ1, λ2 both negative)

In other words, x(t) decreases to zero exponentially without oscillating as t → ∞.However, x(t) may pass through zero exactly once under suitable initial conditions (seeFigure 7 and Exercise 22).

Case 2: Complex roots (Underdamping) The roots are complex if r2 − 4mk < 0. Theroots then have imaginary part

ω =√

4mk − r2

2m=

√k

m− r2

4m25

and may be written

λ1 = − r

2m+ ωi, λ2 = − r

2m− ωi

The general solution of (3) is

x(t) = e− r2m

t(c1 cos ωt + c2 sin ωt

)or x(t) = Ce− r

2mt cos(ωt − θ)


The system oscillates with angular frequency ω but the amplitude e− r2m

tC decreasesexponentially as t → ∞.

FIGURE 5 A shock absorber is anoscillator that is critically damped.

−0.5

0.5

1

1 2 3 4

Underdamping

Return to equilibriumin critically damped case

Overdamping

Critical damping

t

x

FIGURE 6 Graphs of oscillatory motionwith fixed m and k.

Case 3: Double root (Critical Damping) This borderline case between underdampingand overdamping is called critical damping. It occurs if r2 − 4mk = 0. The generalsolution is

x(t) = (c1 + c2t)e− r

2mt

CONCEPTUAL INSIGHT Figure 6 shows why critical damping is important. An oscillator(with fixed m and k) approaches equilibrium most rapidly if r is chosen so that it iscritically damped (Exercise 26). For this reason, shock absorbers in motorcycles andautomobiles are designed to be critically damped (Figure 5). If the shock absorber wereunderdamped, the vehicle would bounce up and down repeatedly at each bump in theroad. But an overdamped shock absorber would take too long to return to equilibrium.

EXAMPLE 2 Underdamped vs. Overdamped Oscillation Let x(t) be the position ofa 2 kg mass oscillating at the end of a spring with constant k = 3 N/m. Assume thatx(0) = 2 m, x′(0) = −4 m/s. Compare the motion x(t) in the two cases:

(a) Frictional constant r = 5.

(b) Frictional constant r = 2.

Solution

(a) If r = 5, x(t) satisfies the differential equation (m = 2, k = 3),

2x′′ + 5x′ + 3x = 0

The characteristic equation 2λ2 + 5λ + 3 = (2λ + 3)(λ + 1) = 0 has roots λ = − 32 , −1.

The roots are real, so the oscillation is overdamped. The general solution is

x(t) = c1e− 3

2 t + c2e− t

The initial conditions yield

x(0) = c1e0 + c2e

0 = c1 + c2 = 2

x′(0) = −3

2c1e

0 − c2e0 = −3

2c1 − c2 = −4

The first equation yields c1 = 2 − c2 and the second equation then gives us

−3

2(2 − c2) − c2 = −4 ⇒ c1 = 4, c2 = −2

Therefore the particular solution (Figure 7) is

x(t) = 4e− 32 t − 2e−t

2

2 4 6t

x(t)

FIGURE 7 Overdamped Oscillation:

x(t) = 4e− 32 t − 2e−t

(b) If r = 2, then x(t) satisfies

2x′′ + 2x′ + 3x = 0


The characteristic equation 2λ2 + 2λ + 3 = 0 has roots −1

2± 1

2

√5i. Since the roots are

complex, the oscillation is overdamped. The general solution is

x(t) = e− 12 t (c1 cos

√5

2t + c2 sin

√5

2t)

The first initial condition yields

x(0) = c1 = 2

After simplification, and using c1 = 2, we have

x′(t) = 1

5e− 1

2 t ((−5 + √5B) cos

√5

2t − (5

√5 + B) sin

√5

2t)

x′(0) = −1 + 1√5B = −4 ⇒ B = −3

√5

2

2 4 6 8t

x(t)

FIGURE 8 Underdamped Oscillation:

x(t) = e− 12 t (2 cos

√5

2 t − 3√

5 sin√

52 t)

Therefore the particular solution (Figure 8) is

x(t) = e− 12 t (2 cos

√5

2t − 3

√5 sin

√5

2t)

Excursion: Analogy Between Circuits and SpringsA series RLC circuit is made up of three components: a charged capacitor (C), an inductor(L), and a resistor (R), connected as in Figure 9. A capacitor consists of two parallel

L C

R

FIGURE 9 An RLC Circuit

conducting plates separated by a small distance. An inductor (also called a solenoid ) is ahelical coil of tightly wound wire.

FIGURE 10 When a charged capacitor isconnected in series to an inductor andresistor, the current undergoes dampedoscillation. Image courtesy ofhttp://www.pha.jhu.edu/dept/lecdemo/EM-L2c.html

When a mass oscillates at the end of spring, there is a cyclical exchange betweenthe potential energy of the stretched or compressed spring and the kinetic energy of themoving mass. A similar cycle occurs in an RLC circuit where the exchange is betweenelectric potential energy and magnetic field energy (Figure 11). A flow of current causescharge to accumulate on the plates of the capacitor, thereby setting up an electric fieldbetween the plates. This corresponds to the stretching of a spring. When the charge onthe plates reaches a maximum, current begins to flows in the opposite direction. Thecapacitor’s electric potential energy decreases but the current flow through the inductorcreates a magnetic field, analogous to the kinetic energy of the mass. Similarly, when themagnetic field reaches maximum strength, it begins to collapse, inducing a current thatrecharges the capacitor but with opposite polarity.

Faraday’s Law of Induction is used to derive the following differential equation forthe quantity of charge Q = Q(t) on a plate of the capacitor at time t :

LQ′′ + RQ′ + 1

CQ = 0 6

Apart from notation, this is identical to the differential equation (3) for damped oscil-lation. We conclude that the charge Q(t) varies according to damped harmonic motion(Figure 10).

The current I = I (t) is equal to the rate at which charge flows through the circuit andthus I (t) = Q′(t). If we differentiate both sides of Eq. (6) with respect to t , we obtain

LI ′′ + RI ′ + 1

CI = 0 7


Thus, the current I (t) also varies according to damped harmonic motion.In SI units, charge is measured in coulombs (symbol: C), capacitance in farads (sym-

bol: F), resistance in ohms (symbol: capital omega �), inductance in henries (symbol: H),and current in amperes (symbol: A).

S

E L

C +Qmax

I = 0

x = 0

m

x = 0

m

= 0

x = 0

m

L

C

E L

C −Qmax

I = 0

L

C

I = ImaxI = Imax

Q = 0 Q = 0

− −− −

+ ++ +

+ ++ +

− −− −

max

m

x = 0x = 0

= 0max

FIGURE 11

One microfarad (μF) is 10−6 farads and onemillihenry (mH) is 10−3 henries.

EXAMPLE 3 Let Q(t) be the charge at time t is a series RLC circuit with inductanceL = 10 mH, capacitance C = 1.6 μF , and resistance R = 1.5 �. Determine:

(a) the differential equations satisfied by Q(t) and I (t).

(b) the frequency f of the current oscillation in the circuit.

(c) the general form of the current I (t) in the circuit.

Correspondence between the quantities R, L andC of an RLC circuit and r , m, k in a mass-springsystem:

RLC CircuitSpring-MassSystem

R rL m1/C k

ω0 = √k/m ω0 = 1/

√LC

Solution (a) In SI units, L = 10−2 H, R = 1.5, and C = 1.6 × 10−6. According to (6)and (7), the charge Q(t) and current I (t) satisfy the same equation:

LQ′′ + RQ′ + 1

CQ ⇒ 10−2Q′′ + 1.5Q′ + 1

1.6 × 10−6Q = 0

LI ′′ + RI ′ + 1

CI ⇒ 10−2I ′′ + 1.5Q′ + 1

1.6 × 10−6I = 0 8

(b) Compare (8) with the differential equation for damped harmonic motion:

LI ′′ + RI ′ + 1

CI = 0 ↔ mx′′ + rx′ + kx = 0

Formula (5) for the angular frequency of damped harmonic motion translates to a formulafor the angular frequency ω of the RLC system (see the table in the margin):

ω =√

k

m− r2

4m2︸︷︷︸damped harmonic motion

↔ ω =√

1

LC− R2

4L2︸︷︷︸RLC circuit


In our case,

ω =√

1

(10−2)(1.6 × 10−6)− 1.52

4(10−2)2≈ 7905 rad/s

The frequency of oscillation is f = ω/(2π) = 7905/(2π) or approximately 1258 cyclesper second.

(c) The roots of the characteristic equation of (8) are − R

2L± ωi ≈ −75 ± 7905i, so the

general solution of (8) is

I (t) = e−75t (c1 cos 7905t + c2 sin 7905t)

Forced Oscillations

Although friction causes oscillations to decay over time, the oscillations can be maintainedif an outside force is applied. Think of a child swinging back and forth on a swing. You cankeep the motion going indefinitely by giving the child a periodic push. If the magnitudeof the applied force at time t is f (t), then the oscillation x(t) satisfies the inhomogeneousdifferential equation for forced oscillations:

mx′′ + rx′ + kx = f (t) 9

A particularly important case is where the forcing function is sinusoidal, that is,f (t) = F0 cos ωt or f (t) = F0 sin ωt where F0 is a constant. Since the sine and cosinefunctions are horizontal shifts of each other, it suffices to consider the case where f (t) isa cosine:

mx′′ + rx′ + kx = F0 cos ωt 10

The angular velocity ω is called the driving frequency of the oscillator.We know, by the Method of Undetermined Coefficients, that (10) has a solution of

the form x = A cos(ωt − δ). To simplify the algebra involved in finding A and δ, we shallreplace the forcing function F0 cos ωt by the complex exponential F0e

iωt (interpretedusing Euler’s formula):

mx′′ + rx′ + kx = F0eiωt = F0(cos ωt + i sin ωt) 11

Now set P(D) = mD2 + rD + k and rewrite (11) as

P(D)x(t) = F0x(t) = F0eiωt 12

Let us look for a solution of (12) is the form

xc(t) = Aei(ωt−δ) (A, δ real numbers with A > 0)

Since P(D)xc(t) = P(iω)xc(t), (12) yieldsREMINDER The formula

P(D)eλt = P(λ)eλt

applies even if λ is complex.

P(iω)Aeω−iδ = F0eiωt

This gives P(iω)Ae−iδ = F0 or

Ae−iδ = F0

P(iω)= F0

m(iω)2 + riω + k= F0

(k − mω2) + rωi


We must assume that r > 0 or that ω �= ω0. Otherwise the denominator is zero. Aftersome algebraic manipulations (Exercise 27), we find that A and δ may be written in termsof the following quantities

ω0 =√

k

mγ = r

mA(ω) = 1√

(ω20 − ω2)2 + γ 2ω2

as follows

A = A(ω)

(F0

m

), tan δ = γω

ω20 − ω2

13

We call A(ω) the amplitude factor and δ the phase shift. A precise formula for δ is givenin Theorem 1 below (see Figure 12). By Euler’s Theorem,

xc(t) = Aei(ωt−δ) = A(

cos(ωt − δ) + i sin(ωt − δ))

The real and imaginary parts of xc(t) satisfy the real and imaginary parts of (11). However,the real part of (11) is (10) and hence, the real part xp(t) = A cos(ωt − δ) of xc(t) is thedesired particular solution of (10). It is a sinusoidal wave whose amplitude A is constantin time. For this reason, xp(t) is called the steady-state solution. Here is a summary ofour results.

0

π2

π

FIGURE 12 Phase shift as a function of ω

THEOREM 1 Theorem Periodic Forcing Function - Steady State Solution Assumethat r > 0 or that ω �= ω0. Then the differential equation

mx′′ + rx′ + kx = F0 cos ωt 14

has steady-state solution

xp(t) = A(ω)

(F0

m

)cos(ωt − δ) 15

where

A(ω) = 1√(ω2

0 − ω2)2 + γ 2ω2, (ω0 =

√k

m, γ = r

m)

The phase shift δ is determined as follows:

δ =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

tan−1(

γω

ω20−ω2

)if ω < ω0

tan−1(

γω

ω20−ω2

)+ π if ω > ω0

π2 if ω = ω0

The general solution of (14) has the form

x(t) = xp(t) + xh(t)

where xh(t) is a solution of the associated homogeneous equation

mx′′ + rx′ + kx = 0

If r > 0, then xh(t) decays to zero exponentially. For this reason, xh(t) is called thetransient term. It dies out as t gets large and plays no role in the long term behavior ofthe forced oscillator. In other words, x(t) behaves like xp(t) for t large.


EXAMPLE 4 Steady-state and Transients Compare the steady-state solution withthe solution of the initial value problem

9x′′ + 6x′ + 37x = 18 cos 3t, x(0) = 0, x′(0) = 0

Solution The driving frequency is ω = 3.

Step 1. Compute the steady-state solution.We have ω2

0 = k/m = 37/9 and γ = r/m = 6/9 = 2/3. For ω = 3 we obtain

A(3) = 1√(ω2

0 − ω2)2 + γ 2ω2= 1√

((37/9) − 32)2 + (2/3)2(32)≈ 0.189

tan δ = γω

ω20 − ω2

= 2

(37/9) − 32≈ −0.409

Since ω > ω0, δ = tan−1(−0.409) + π = 2.753. Now apply (15) with F0/m =18/9 = 2:

xp(t) = (0.189)

(18

9

)cos(3t − 2.753) = 0.378 cos(3t − 2.753)

Step 2. Find the general solution.The characteristic equation 9λ2 + 6λ + 37 = 0 has roots − 1

3 ± 2i by the quadraticformula. Hence the solution of the associated homogeneous equation is

xh(t) = e− 13 t (c1 cos 2t + c2 sin 2t)

and the general solution is

x(t) = 0.378 cos(3t − 2.753) + e− 13 t (c1 cos 2t + c2 sin 2t)

0.5

5 10

x(t)x

t

xp(t)

FIGURE 13 Comparison of x(t) with thesteady state solution xp(t).

Step 3. Use initial conditions.We have

x′(t) = −1.134 sin(3t − 2.753) + e− 13 t (−2c1 sin 2t + 2c2 cos 2t)

− 1

3e− 1

3 t (c1 cos 2t + c2 sin 2t)

The initial conditions give us

x(0) = 0.378 cos(−2.753) + c1 = 0

x′(0) = −1.134 sin(−2.753) + 2c2 − 1

3c1 = 0

Substitute c1 = −0.378 cos(2.753) ≈ 0.350 in the second equation:

−1.134 sin(−2.753) + 2c2 − 1

3(0.350) = 0 ⇒ c2 ≈ −0.156

The solution to our initial value problem is

x(t) = 0.378 cos(3t − 2.753) + e− 13 t (0.350 cos 2t − 0.156 sin 2t)

Figure 13 shows that for t large, the transient terms dies out and x(t) is indistinguishablefrom xp(t).


CONCEPTUAL INSIGHT The graph of the amplitude factor A(ω) is called a resonancecurve. It is of great importance because it describes how the response of an oscillatorvaries with the angular frequency ω of the driving force. Figures 14 and 15 show typicalresonance curves. the precise shape depends m and k, but as long the system is not tooheavily damped (more precisely, if r ≤ √

2mk), there is a unique frequency ωr , calledthe resonant frequency, at which A(ω) takes on a maximum value.

Resonance is a fundamental concept that appears throughout science and engineer-ing. For example, it is the basis of all wireless communication. The atmosphere is filledwith countless electromagnetic signals of all different frequencies. How does a receivertune in to a signal of one given frequency ωr while ignoring all other signals? By em-ploying an oscillator (an RLC circuit) whose resonance curve is sharply peaked at ωr

(Figure 15). The oscillator responds to the desired signal because A(ωr) is large, but itis insensitive to signals whose frequencies ω are not too close to ωr because A(ω) issmall.

r

A( )

FIGURE 14 A resonance curve is a graph of theresponse A(ω) as a function of the drivingfrequency.

A( )

r

FIGURE 15 A more sharply-peaked resonancecurve occurs when the ratio

√km/r is large.

The next theorem gives a formula for the resonant frequency (verification is left asExercise 28).

THEOREM 2 Resonant Frequency If r ≤ √mk, then A(ω) takes on its maximum

value at the resonant frequency

ωr =√

ω20 − 1

2γ 2

The maximum value is

A(ωr) = 1

γ

√ω2 + 1

4

EXAMPLE 5 Determine, for the driven oscillator

2x′′ + 1

4x′ + 8x = 12 cos ωt

(a) the amplitude factor A(ω).

(b) the resonant frequency if ω = 1.8.


(c) the steady state solution xp(t) for ω = 1.8.

11.26

2

3

4A( )

A(1.8)

543

Resonant frequency

1.9981.81

FIGURE 16 Resonance curve forExample 5.

Solution (a) We have m = 2, r = 14 , and k = 8, and thus

ω0 =√

k

m=

√8

2= 2, γ = r

m= 1/4

2= 1

8

A(ω) = 1√(ω2

0 − ω2) + γ 2ω2= 1√

(22 − ω2)2 + ω2/64

(b) For ω = 1.8, the resonant frequency is (Figure 16)

ωr =√

ω20 − 1

2γ 2 =

√22 − 1

2(1

8)2 ≈ 1.998

(c) For ω = 1.8,

A(1.8) = 1√(22 − 1.82)2 + (1.8)2/64

≈ 1.26

tan δ = γω

ω20 − ω2

= ( 18 )(1.8)

22 − 1.82≈ 0.30

Thus δ = tan−1(0.30) ≈ 0.29 and since F0/m = 12/2 = 6, the steady state solution forω = 1.8 is

x(t) = (1.26)(6) cos(1.8t − 0.3) = 7.56 cos(1.8t − 0.3)

Excursion: Micro Diving BoardsA diving board is an example of an oscillator. When you jump off a diving board, the boardflaps up and down for a few seconds at its natural frequency fr (approximately 2–4 cyclesper second). You can “tune into” the diving board by jumping up and down at the end of

FIGURE 17 Microcantilever (circled)with tiny magnetic particles on the free endvibrates at its resonant frequency (Photocredit: Ying-Ju Wang, NIST ).

the board at its resonant frequency (which is very close to its natural frequency). If youdo this, you will achieve an oscillatory state of large amplitude. On the other hand, if youjump at a frequency far from fr , your oscillations will remain small.

Scientists in the field of nanotechnology use microscopic diving boards of lengthson the order of 10 to 100 microns (10−6 m) called microcantilevers to probe atomic andmolecular structures. These microcantilevers resonate at frequencies on the order of 105

cycles per second or more. In one experiment in 2006, scientists at NIST (National Instituteof Standards and Technology) attached tiny magnetic particles to the tip of a cantilever andinduced the cantilever to vibrate at its resonant frequency∗. As the particles vibrated upand down on the cantilever, they created an oscillating magnetic field which the scientistsused to make the atoms in a rubidium gas spin like wobbly tops. Techniques of this typemay eventually be used in high-performance magnetic sensors, power-efficient chip-scaleatomic devices such as clocks, or perhaps serve as components in quantum computers ofthe future.

∗Y-J. Wang, M. Eardley, S. Knappe, J. Moreland, L. Hollberg, and J. Kitching. Magnetic resonance in an atomicvapor excited by a mechanical resonator. Physical Review Letters, December 2006.


19.3 SUMMARY

• An undriven linear oscillator is modeled by the differential equation

mx′′ + rx′ + kx = 0 16

where m > 0, k > 0, and r ≥ 0. Let λ1, λ2 be the roots of the characteristic equationmλ2 + rλ + k = 0. Set

ω0 = √k/m

• Simple harmonic motion: mx′′ + ky = 0: λ1, λ2 = ±ω0i.General solution:

x(t) = c1 cos ω0t + c2 sin ω0t or x(t) = C cos(ω0t − θ)

• Damped harmonic motion (r > 0).

1. Underdamping (r2 − 4mk < 0): λ1, λ2 = − r

2m± ωi where ω =

√k

m− r2

4m2.

General solution:

x(t) = e− r2m

t (c1 cos ωt + c2 sin(ωt)) or x(t) = Ce− r2m

t cos(ωt − θ)

2. Overdamping (r2 − 4mk > 0): λ1, λ2 are real, negative, and distinct. General solution:

x(t) = c1eλ1t + c2e

λ2t

3. Critical damping (r2 = 4mk): the characteristic equation has a double root λ = − r2m

.General solution:

x(t) = (c1 + c2t)e−r2m

t

• Periodic forcing function:

mx′′ + rx′ + kx = F0 cos ωt

Steady-state solution:

xp(t) = A(ω)

(F0

m

)cos(ωt − δ)

γ = r

m, A(ω) = 1√

(ω20 − ω2)2 + γ 2ω2

, tan δ = γω

ω20 − ω2

– The general solution is a sum x(t) = xp(t) + xh(t) with transient term xh(t) (asolution of the associated homogeneous equation).

– As t → ∞, the transient term xh(t) decays exponentially. So for large t , x(t) ≈ xp(t)

and the system oscillates at the driving frequency ω.– A(ω) measures the response of the oscillator to the driving force. If r ≤ √

mk, then

maximum response occurs at the resonant frequency ωr =√

ω20 − 1

2γ 2.


19.3 EXERCISES

Preliminary Questions1. In simple (undamped) harmonic motion, the roots of the charac-

teristic equation are

(a) real and distinct(b) purely imaginary(c) complex numbers with positive real part

2. Determine if the oscillating system is underdamped, overdamped,or critically damped:

(a) x′′ + 4x′ + 3x = 0

(b) x′′ + 4x′ + 4x = 0

(c) x′′ + 4x′ + 5x = 0

3. Let ω0 be the natural angular frequency of a an oscillator with zerodamping and let ω be the angular frequency when the damping constantr is positive. Which of the following statements is correct?

(a) ω is larger than ω0.

(b) ω is smaller than ω0.

ExercisesIn Exercises 1-8, a mass of m kg oscillates at the end of a spring withspring constant k and frictional constant r . Let x(t) denote the positionof the mass at time t .

1. Assume that m = 800 g, k = 25 N/m, and r = 0. The mass isstretched 20 cm past equilibrium and released with zero initial veloc-ity.

(a) Set up the differential equation for x(t).

(b) Determine x(t). Note: the unit N/m is equal to one kg/sec2.

(c) Find the time t at which the mass first arrives at the equilibriumposition x = 0.

2. Repeat exercise 1, but assume a frictional force with r = 4 kg/s.Is the system underdamped or overdamped?

3. Repeat Exercises 1 (a) and (b) with r = 12 kg/s. Is the systemunderdamped or overdamped? Does the mass pass through equilibriumat any time t > 0?

4. Assume that m = 4 kg, r = 8 kg/s, and k = 13 N/m. Find themotion x(t) of the mass under initial conditions x(0) = 6 m, x′(0) = 2m/s.

5. Assume that m = 0.2 kg, k = 25 N/m. Determine the angularfrequency ω of oscillation for the cases (a) r = 0 and (b) r = 2 kg/s.

6. In the case of damped oscillations, x(t) = Ce−αt cos(ωt − θ). Theamplitude at time t is Ce−αt . How long does it take for the amplitudeto decrease by 90% if m = 5 g, r = 2 g/s, and k = 6 g/s2.

7. At time t = 0, a 4-kg mass attached to a spring with k = 5 N/mand r = 12 kg/s is pulled 3

4 m beyond equilibrium and released (withzero initial velocity).

(a) Solve for the position x(t) of the mass.

(b) Show that the mass never passes through equilibrium.

(c) Show that the mass passes through equilibrium exactly once if itis given an initial velocity of −3 m/s. When does this occur?

8. Assume that m = 4 g and k = 9 g/s2. Assume further thatx(0) = 10 cm and x′(0) = 0. Determine the type of damping for thevalues r = 4, 12, 20 and plot the solutions x(t) on the same set of axes.

9. Apendulum, consisting of a mass m swinging from a cord of length� has moment of inertia I = m�2 (assuming the cord has zero mass).Gravity exerts a torque τ = −�mg sin θ and the law τ = Iθ ′′ yieldsθ ′′ = −(g/�) sin θ . If the angle θ is small, we may use the approxima-tion sin θ ≈ θ to obtain the differential equation θ ′′ + (g/�)θ = 0.(a) Solve for θ(t).(b) Determine θ(t) and find the period of oscillation if � = 2 m andthe mass is released at an angle θ0 = .25 rad with zero initial velocity.(c) Rather than a mass and cord, assume that a solid rod of mass m

swings about a pivot at one end. In this case, I = 13m�2. Write down

the differential equation for θ(t) (both the exact one and the small angleapproximation) and determine the period of oscillation if � = 2.

−mg sin

−mg

FIGURE 18

10. Match graphs (A)-(C) in Figure 19 with the following solutions ofthe overdamped oscillator x′′ + 4x′ + 3x = 0.(a) x = 5e−t − 4e−3t

(b) x = −e−t + 2e−3t

(c) x = 1

2e−t + 1

2e−3t

y

t t

y y

t

(A) (C)(B)

1 2 3 4 1 2 3 4

1 2 3 41

1 1

−1

2

FIGURE 19


11. Find the steady-state solution of the driven oscillator

1

4x′′ + 2x′ + 3x = 40 cos ωt 17

for ω = 2 and ω = 6 (See Example 5).

12. Solve (17) for ω = 2 with initial conditions x(0) = 2 andx′(0) = 4.

13. Determine the resonant frequency of the driven oscillator

x′′ + 4x′ + 9x = 24 cos ωt 18

Then find the steady state solution with ω = 3.

14. Find the solution of (18) with ω = 3 such that x(0) = 3,x′(0) = 10.

15. Make a plot of the resonance curve of the oscillator

5x′′ + 2x′ + 10x = 50 cos ωt 17

16. Oscillating Mass with Gravity A mass m oscillates at the end ofa vertical spring with spring constant k. Since gravity exerts a force−mg on the mass, the vertical position y(t) of the mass satisfies thedifferential equation

my′′ + ry′ + ky = −mg 18

where r ≥ 0 is a damping factor. Choose the origin of the vertical axisso that y = 0 is the equilibrium position of the end of the unstretchedspring.

(a) Show that y = f (t) − mgk

is a solution of Eq. (18), where f (t) isany solution to my′′ + ry′ + ky = 0. Thus, in the presence of gravity,the mass oscillates about the equilibrium position y = −mg/k ratherthan y = 0.

(b) Show that if r > 0, then limt→∞ y(t) = −mg

k.

Exercises 17–20 deal with (undriven) RLC circuits as described in thissection before Example 3.

17. Let Q(t) be the charge (in amperes) in an undriven RLC circuitwith L = 0.02H, R = 3�, and C = 3 × 10−5F.

(a) Find the frequency ω/(2π) (in cycles/s) of the damped oscillation.

(b) Write the solution in the form C cos(ωt − δ).

18. Determine the charge Q(t) in an RLC circuit with L = .01H,R = 2�, and C = 1.5 · 10−6F. Assume that Q(0) = 0 andQ′(0) = 100 C/sec.

19. Find the current I (t) at time t in an RLC circuit with L = 1.2H,R = 430�, and C = .0001F, assuming that I (0) = 10 A and I ′(0) = 0A/s.

20. Suppose that L = .02H and C = 5 × 10−5F in an RLC-circuit.How large must R be so that the oscillation is overdamped?

21. Let

x(t) = 2

85

(4 cos 100t + sin 100t + e−50t cos(50

√19t)

)(a) Verify that x(t) is a solution of

1

10x′′ + 10x′ + 5000x = 400 cos 100t

(b) Which part of x(t) is the steady state solution xp(t)?

(c) Plot x(t) and xp(t) on the same set of axis. Estimate thetime required for the transient motion to effectively disappear.

Further Insights and Challenges22. Assume that a mass-spring system is overdamped. Write the gen-eral solution in the form

x(t) = c1e−λ1t + c2e−λ2t

where λ1 > λ2 (so the characteristic equation has roots −λ1 and −λ2).

(a) Show that if the initial condition satisfies x(0) > 0, then x(t0) = 0has a solution for t0 > 0 if and only if c1 > −c2 > 0.

(b) Use (a) to show that if x(0) > 0, then x(t0) = 0 has a solution fort0 > 0 if and only if

x′(0) < −λ1x(0)

This shows that in an overdamped system, the mass passes throughequilibrium (and does so precisely once) only if it is released witha sufficiently large negative intial velocity x′(0) directed towards theequilibrium position.

23. Decay Rate Let λ > 0. We say that a function f (t) decays like

e−λt (or has exponential decay rate λ) if limt→∞ f (t)/e−λt exists and is

nonzero.

(a) Show that 2e−3t + 25e−7t decays like e−3t .

(b) Suppose that λ1, λ2 are negative and λ1 > λ2. Show that the func-tion x = c1eλ1t + c2eλ2t decays like eλ1 if c1 �= 0 and like eλ2t ifc1 = 0.

24. Consider x′′ + 7x′ + 12x = 0.

(a) The roots of the characteristic equation are real and negative. Com-pute them and label them −λ1 and −λ2 where λ1 > λ2.

(b) Graph the two solutions x1(t) = e−λ1t and x2(t) =12 (e−λ1t + e−λ2t ) on the same set over axes over [0, 1]. Which so-lution decays more rapidly?

25. Decay Rate as a Function of Friction If r , m, k are posi-

tive and r2 > 4mk, then the roots of the characteristic equation ofmx′′ + rx′ + kx = 0 are real and negative. Label the roots −λ1, −λ2where λ1 > λ2.

(a) Show that a solution has exponential decay rate λ2 unless it has thespecial form ce−λ1t .

(b) Prove that λ1 is an increasing function of r , λ2 is a decreasingfunction of r (Figure 20), and that

limr→∞ λ1 = ∞, lim

r→∞ λ2 = 0

S E C T I O N 19.4 Power Series Solutions 35

2

1

(double root)r

r

Roots

4mk

r2m

FIGURE 20 Roots as a function of friction r

(c) Explain: As r tends to ∞, the special solutions y = ce−λ1t (withc �= 0) decay more and more rapidly but all other solutions decay moreand more slowly.

(d) Show that a solution has the form x = ce−λ1t if and only ifx′(0) = −λ1x(0). Conclude that if r is large, then the decay to equilib-rium is very slow unless the initial velocity has just the right magnitudex′(0) = −λ1x(0).

26. Critical Damping The goal is to damp a mass-spring systemin such a way that for generic initial conditions, the mass returns toequilibrium as rapidly as possible. Use Exercise 25 to show that this isachieved when the system is critically damped.

27. Verify the formulas in (13).

28. Verify Theorem 2. Hint: find the minimum value of the denomina-tor of A(ω).

19.4 Power Series Solutions

Many differential equation arising in applications cannot be solved explicitly in terms ofelementary function. One way of studying solutions of these equations is to represent themas power series. This technique was discussed in Section 11.6 [ET Section 10.6] (reviewif necessary). The key idea is that a differential equation leads to relations that allow usto solve for the coefficients of the power series.

Recall that if a power series has a positive radius of convergence, then we maycompute its derivatives using termwise differentiation:

y = f (x) =∞∑

n=0

anxn = a0 + a1x + a2x

2 + a3x3 + · · · 1

y′ = f ′(x) =∞∑

n=0

(n + 1)an+1xn = a1 + 2a2x + 3a3x

2 + 4a4x3 + · · · 2

y′′ = f ′′(x) =∞∑

n=0

(n + 1)(n + 2)an+2xn = 2a2 + (3 · 2)a3x + (4 · 3)a4x

2 · · · 3

Note that

y(0) = a0, y′(0) = a1 4

EXAMPLE 1 Power Series for Sine and Cosine Find the power series solution ofy′′ + y = 0.

Solution

Step 1. Find the recursion relation.Substitute the power series (3) and (1) for y′′ and y in the differential equation:


y′′ + y =(

2a2 + (3 · 2)a3x + (4 · 3)a4x2 + · · ·

)+

(a0 + a1x + a2x

2 + · · ·)

= (2a2 + a0) + (6a3 + a1)x + (12a4 + a2)x2 + · · ·

=∞∑

n=0

(an + (n + 2)(n + 1)an+2)xn

= 0

This equation is satisfied only if the coefficient of xn is zero for all n ≥ 0. In otherwords, an + (n + 2)(n + 2)an+2 = 0. We obtain the recursion relation

an+2 = − an

(n + 2)(n + 1)5

Step 2. Apply the recursion relation.Eq. (5) relates an to an+2, so a0 determines all of the even coefficients a2, a4, a6 · · · .Similarly, a1 determines all of the odd coefficients a3, a5, a7, · · · . More precisely,applying (5) successively, we obtain

a2 = − a0

2 · 1, a4 = − a2

4 · 3= a0

4 · 3 · 2 · 1, a6 = − a4

6 · 5= − a0

6 · 5 · 4 · 3 · 2 · 1, . . .

The general pattern in

a2n = (−1)na0

(2n)!Similar reasoning shows that

a2n+1 = (−1)na1

(2n + 1)!Step 3. Determine the solutions.

We find that y(x) is a sum of two power series consisting of the even and odd powersof x. Call these y1(x) and y2(x):

y(x) = a0

(1 − x2

2! + x4

4! − · · ·)

+ a1

(x − x3

3! + x5

5! − · · ·)

= a0

∞∑n=0

(−1)nx2n

(2n)!︸︷︷︸call this y1(x)

+ a1

∞∑n=0

(−1)nx2n+1

(2n + 1)!︸︷︷︸call this y2(x)

The ratio test shows that these power series converge for all x, and therefore y1 andy2 each represent solutions. They are clearly independent, so the general solutiony = a0y1 + a1y1 has two arbitrary constants a0 and a1, as we should expect.

Step 4. Identify the solutions.We recognize that y1 and y2 are the Taylor series of cos x and sin x. This is not unex-pected, because we know that cos x and sin x are independent solutions of y′′ + y = 0.

To apply the power series method to a general second order linear differential equationof the form

y′′ + b(x)y′ + c(x) = f (x) 6


we need to know that the power series solutions converge. The next theorem (stated withoutproof) guarantees convergence, provided that b(x), c(x), and f (x) are represented by theirTaylor series. We will apply this theorem only in the case of power series centered at a = 0.

THEOREM 1 Assume that b(x), c(x), and f (x) are represented by their Taylor seriescentered at a in an interval I = {x : |x − a| < R}. Then for all initial values y0, y′

0,(6) has a unique solution y(x) such that y(a) = y0 and y′(a) = y′

0. Furthermore, y(x)

is represented in I by a convergent power series centered at a.

Eq. 7 is a special case of the Airy equationy′′ ± k2xy = 0, named after George BiddellAiry, who was British Royal Astronomer in theyears 1835-1881. Among other achievements, hecomputed the density of the Earth by swinging apendulum at the top and bottom of a deep mine.However, Airy was not a popular personality andwas criticized for some bad scientific decisions.For example, he urged his government to stopfunding the development by Babbage of acalculating engine (a mechanical forerunner ofthe modern computer), believing it to be aworthless project.

FIGURE 1 A caricature of Airy publishedin the magazine Vanity Fair in 1875.

EXAMPLE 2 Airy Functions Find the two independent power series solution of

y′′ + xy = 0 7

satisfying y(0) = 1, y′(0) = 0 and y(0) = 0, y′(0) = 1

Solution

Step 1. Find the recursion relation.Substitute the power series for y′′ and y in the differential equation:

y′′ + xy =∞∑

n=0

(n + 1)(n + 2)an+2xn + x

∞∑n=0

anxn

=(

2a2 + (3 · 2)a3x + (4 · 3)a4x2 + · · ·

)+

(a0x + a1x

2 + a2x3 + · · ·

)= 2a2 + ((3 · 2)a3 + a0)x + ((4 · 3)a4 + a1)x

2 + ((5 · 4)a5 + a2)x3 + · · ·

= 0

The coefficient of xn must be zero for all n ≥ 0, giving us a2 = 0 and for n ≥ 1,

(n + 2)(n + 1)an+2 + an−1 = 0 ⇒ an+2 = − an−1

(n + 2)(n + 1)

We rewrite this relation, replacing n by n + 1 to obtain the recursion formula

an+3 = − an

(n + 3)(n + 2)8

This condition relates an to an+3. So a0 determines the coefficients a3, a6, a9 · · · . Sim-ilarly, a1 determines a4, a7, a10, · · · and a2 determines a5, a8, a11, · · · . In particular,since a2 = 0, (8) implies

a2 = a5 = a8 = a11 = · · · = 0

In other words, a3n+2 = 0 for all n ≥ 0.Step 2. Use first initial condition.

The initial condition y(0) = a0 = 1 together with (8) gives us

a3 = − a0

3 · 2= − 1

3 · 2

a6 = − a3

6 · 5= 1

6 · 5 · 3 · 2

a9 = − a6

9 · 8= − 1

9 · 8 · 6 · 5 · 3 · 2


In general, the coefficient of x3n is the product of all the whole numbers from 3n downto 1 with every third number omitted. The condition y′(0) = a1 = 0, and (8) yield:

a1 = 1 ⇒ a4 = 0 ⇒ a7 = 0 ⇒ · · ·In other words, a3n+1 = 0 for all n, and since a3n+2 = 0 by Step 1, we obtain thesolution

y1(x) = 1 − x3

3 · 2+ x6

6 · 5 · 3 · 2− x9

9 · 8 · 6 · 5 · 3 · 2

Step 3. Use Second initial condition.The initial condition y(0) = a0 = 0 together with (8) gives us

a0 = 0 ⇒ a3 = 0 ⇒ a6 = 0 ⇒ · · ·In other words, a3n = 0 for all n. The initial condition y′(0) = a1 = 1 and (8) yield:

a4 = − 1

4 · 3

a7 = − a3

7 · 6= 1

7 · 6 · 4 · 3

a10 = − a7

10 · 9= − 1

10 · 9 · 7 · 6 · 4 · 3

Again, since a3n+2 = 0 by Step 1, we obtain the solution

y2(x) = x − x4

4 · 3+ x7

7 · 6 · 4 · 3− x10

10 · 9 · 7 · 6 · 4 · 3· · ·

The coefficient of x3n+1 is the product of all the whole numbers from 3n + 1 down to1 with every third number omitted.

By Theorem 1, the power series for y1 and y2 converge for all x, and thus we obtain ourtwo independent solutions.

The partial sums of a power series solution are polynomials that provide approximatesolutions to the differential equation. Figure 2 compares the graph of y2(x) with partialsum through degree 19:

T19(x) = x − x4

12+ x7

504− x10

45360+ x13

7076160− x16

1698278400+ x19

580811212800

We see that T19(x) provides a good approximation on the interval [0, 3] (in fact, the erroron this interval is at most 0.0002).

5 10 15

3

2

T19(x)

y

x

y1(x)1

−1FIGURE 2 Comparison of the Airyfunction y2(x) and T19(x).


The coefficients of a power series solution do not always satisfy a recursion relation,so it may be impossible to find a simple formula for the coefficients. However, if thecoefficients b(x) and c(x) in (6) are represented by their Taylor series, then we can calculatea finite number of coefficients, and thereby approximate the solutions to any desired degreeof accuracy on a given finite interval.

EXAMPLE 3 Find the terms of degree ≤ 3 in the power series solution of the initialvalue problem

y′′ − exy = 0, y(0) = 1, y′(0) = 1

Solution We make use of the Taylor series for ex :

ex = 1 + x + 1

2x3 + 1

6x3 + higher order terms

Since y(0) = a0 = 1 and y′(0) = a1 = 1, we have

y(x) = 1 + x + a2x2 + a3x

3 + higher order terms

y′′(x) = 2a2 + 6a3x + 12a4x2 + 20a5x

3 + higher order terms

We substitute these series in the differential equation, ignoring all terms of degree fourand higher. We use the symbol ≈ to indicate that these terms have been dropped. Thenthe equation y′′ = exy becomes

2a2 + 6a3x + 12a4x2 + 20a5x

3 ≈ (1 + x + 1

2x2 + 1

6x3)(1 + x + a2x

2 + a3x3)

Expand the right-hand side and drop powers of degree four and higher:

(1 + x + 1

2x2 + 1

6x3)(1 + x + a2x

2 + a3x3) ≈ 1 + 2x + (a2 + 3

2)x2 + (a3 + a2 + 2

3)x3

Thus we obtain

2a2 + 6a3x + 12a4x2 + 20a5x

3 = 1 + 2x + (a2 + 3

2)x2 + (a3 + a2 + 2

3)x3

Equate coefficients on both sides:

2a2 = 1, 6a3 = 2, 12a4 = a2 + 3

2, 20a5 = a3 + a2 + 2

3

Although we only need to find a2 and a3, these equations also gives us a4 and a5:

a2 = 1

2, a3 = 1

3, a4 = 1

12(a2 + 3

2) = 1

6, a5 = 1

20(a3 + a2 + 2

3) = 3

40

Therefore

y(x) = 1 + x + 1

2x2 + 1

3x3 + 1

6x4 + 3

40x5 + higher order terms


19.4 SUMMARY

• To find a power series solution of a second order differential equation, substitute thepower series for y, y′, and y′′ in the differential equation and solve for the coefficients.

y =∞∑

n=0

anxn = a0 + a1x + a2x

2 + a3x3 + · · · 9

y′ =∞∑

n=0

(n + 1)an+1xn = a1 + 2a2x + 3a3x

2 + 4a4x3 + · · · 10

y′′ =∞∑

n=0

(n + 1)(n + 2)an+2xn = 2a2 + (3 · 2)a3x + (4 · 3)a4x

2 + · · · 11

Note that y(0) = a0 and y′(0) = a1.• The resulting power series for y(x) need not have a nonzero radius of convergence.However, if the differential equation has the form

y′′ + a(x)y′ + b(x) = f (x)

and if b(x), c(x) and f (x) are represented by their Taylor series on an interval {x : |x| <

R}, then the power series for y(x) converges in the same interval.

19.4 EXERCISES

Preliminary Questions1. What is the coefficient of x5 in the power series for y′′(x) (in terms

of the coefficients an for y(x)).

2. What are a2 and a3 if y′′ + y′ + y = 0 and a0 = a1 = 4?

3. Rewrite an+2 = an−2

(n + 3)(n − 4), replacing n by n + 2.

4. What must be assumed about c(x) to guarantee that a power seriessolution y(x) of y′′ + y′ + c(x)y = 0 converges for |x| < 10?

Exercises1. Find the power series solution of the initial value problem

y′′ − y = 0, y(0) = 1, y′(0) = 0

Identify the resulting function y(x).

2. Let y(x) be a power series solution of the initial value problem

y′′ − xy = 0, y(0) = 1, y′(0) = 0

(a) Show that the coefficients of y(x) satisfy

an+3 = an

(n + 3)(n + 2)(n ≥ 0)

Hint: see Example 2.(b) Show that the only non-zero coefficients are a0, a3, a6, . . . . De-scribe a formula for these coefficients.(c) Write out the first four non-zero terms of the power series.

3. Let y(x) be a power series solution of

y′′ − 2xy′ + y = 0 12

(a) Find the recursion relation satisfied by the coefficients of y(x).

(b) Show that the following power series is a solution:

y1(x) = 1 − 1

2!x2 − 3

4!x4 − 7 · 3

6! x6 − 11 · 7 · 3

8! x8 − · · ·

The coefficient of x2k for k ≥ 2 is

a2k = − (4k − 5)(4k − 9) · · · 11 · 7 · 3

(2k)!(c) Find a power series solution y2(x) containing only odd powers ofx.

(d) Find the power series solution satisfying y(0) = 2, y′(0) = 4.

4. Let y(x) be a power series solution of

y′′ + x

1 − x2y′ − 1

1 − x2y = 0 13


(a) Show that the coefficients of y(x) satisfy

an+2 = (n − 1)2

(n + 2)(n + 1)an (n ≥ 1)

Hint: multiply (13) by (1 − x2).(b) Prove that a3 = a5 = a7 = · · · = 0.(c) Determine the solution with initial values y(0) = 0, y′(0) = 1.(d) Find a formula for the coefficients a2n in terms of a0.(e) Determine the solution with initial values y(0) = 1, y′(0) = 0.This power series converges only for |x| < 1.

5. Find the terms of degree ≤ 5 in the power series solution of

y′′ − xy = sin x, y(0) = 1, y′(0) = 1


y′′ + xy′ + (cos x)y = 1, y(0) = 3, y′(0) = −2


y′′ + xy′ = ex, y(0) = 1, y′(0) = 0

8. Find the terms of degree ≤ 4 in the power series representation ofthe general solution of

y′′ + (sin x)y′ + y = 0

9. Show that x2y′′ − xy′ + (1 − x)y = 0 has a unique power seriessolution satisfying y′(0) = 1.

10. Show that x2y′′ + (1 + x3)xy′ − y = 0 has a unique power seriessolution satisfying y′(0) = 1.

11. Let μ be a constant. Solutions of the differential equation

(1 − x2)y′′ − 2xy′ + μy = 0 14

are called Legendre functions.

(a) Find the recursion relation expressing an+2 in terms of an.

(b) Find the power series solution to (14) for μ = 12 satisfyingy(0) = 0 and y′(0) = −3.

(c) Show that (14) has a polynomial solution y if and only if μ =m(m + 1) for some whole number m. Show further that the degree ofy is m and that y is even (involving only even powers of x) if m is evenand odd polynomial if m is odd.

(d) The Legendre Polynomial Pm(x) is the polynomial solution forμ = m(m + 1) satisfying Pm(1) = 1. Calculate P4(x) and P5(x).Legendre polynomials are used to describe the states of the hydrogenatom and in many other physical applications.

Further Insights and ChallengesExercises 12 - 15 refer to the Bessel differential equation:

y′′ + 1

xy′ + (1 − n2

x2)y = 0 15

12. (a) Show that for each whole number n, (15) has a unique solutionwith power series expansion

yn(x) = xn + cn+2xn+2 + cn+4xn+4 + · · ·(b) The function Jn(x) = yn(x)/(2nn!) is called the Bessel functionof the first kind of order n. Show that

Jn(x) =∞∑

k=0

(−1)k

22k+nk!(n + k)!x2k+n

1 J0(x)

J1(x)

−0.4

1082 64 12

FIGURE 3 The first two Bessel functions of the first kind

13. Prove the relations

d

dxJ0(x) = −J1(x),

d

dxxJ1(x) = xJ0(x)

Hint: show that J ′0 satisfies (15) with n = 1 and that x−1(xJ1)′ satisfies

(15) with n = 0.

14. Figure 3 suggests that the zeroes of J0(x) and J1(x) alternate or“separate each other", that is, between any two consecutive zeros of J0there lies precisely one root of J1(x). In this exercise we prove this fact.Let α and β be consecutive zeroes of J0(x). Assume that α, β > 0.

(a) Use Exercise 13 and Rolle’s Theorem to prove that J1(x) has aroot x0 such that α ≤ x0 ≤ β.

(b) Prove that x0 �= α, β. Thus x0 lies between α and β. Hint: if α is aroot of both J0(x) and J1(x), then J0(α) = 0 and J ′

0(α) = 0. Derive acontradiction from the Uniqueness Theorem for initial value problems,which implies that J0(x) = 0 for all x > 0.

(c) Repeat this argument to show that between any two zeroes of J1(x)

there lies a zero of J0(x).

15. Prove the following generalization of the relations in Exercise 15:

d

dxx−nJn(x) = −x−nJn+1(x),

d

dxxn+1Jn+1(x) = xn+1Jn(x)

second order 19 differential equations...19 second order differential equations scientists study the...

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