Purcell Edisi 9 Bab 15 Persaman Diferensial

775

Differential EquationsCHAPTER 1515.1 Linear

HomogeneousEquations

15.2 NonhomogeneousEquations

15.3 Applications ofSecond-OrderEquations

15.1Linear Homogeneous EquationsWe call an equation involving one or more derivatives of an unknown function adifferential equation. In particular, an equation of the form

in which denotes the kth derivative of y with respect to x, is called an ordinarydifferential equation of order n. Examples of differential equations of orders 1, 2,and 3 are

If, when is substituted for y in the differential equation, the resultingequation is an identity for all x in some interval, then is called a solution of thedifferential equation. Thus, is a solution to since

for all x.We call the general solution of the given equation, since it canbe shown that every solution can be written in this form. In contrast,is called a particular solution of the equation.

Differential equations appeared earlier in this book, principally in three sec-tions. In Section 3.9, we introduced the technique called separation of variables andused it to solve a wide variety of first-order equations. In Section 6.5, we solved thedifferential equation of exponential growth and decay, and in Section 6.6we studied first-order linear differential equations and some applications.

In this chapter, we consider only nth-order linear differential equations, that is,equations of the form

where . (Note that y and all its derivatives occur to the first power.) This iscalled a linear equation because, if it is written in operator notation,

then the operator in brackets is a linear operator. Thus, if L denotes this operatorand if f and g are functions and c is constant, then

That L has these properties follows readily from the corresponding properties forthe derivative operators .

Of course, not all differential equations are linear. Many important differentialequations, such as

are nonlinear. The presence of the exponent 2 on y is enough to spoil the linearity,as you may check. The theory of nonlinear differential equations is both compli-cated and fascinating, but best left for more advanced courses.

dy

dx+ y2 = 0

D, D2, Á , Dn

L(cf) = cL(f)

L(f + g) = L(f) + L(g)

CDnx + a1(x)Dn-1x + Á + an-1(x)Dx + an(x) Dy = k(x)

n Ú 2

y(n) + a1(x)y(n-1) + Á + an-1(x)y¿ + an(x)y = k(x)

y¿ = ky

2 cos x + 102 cos x + C

f¿(x) + 2 sin x = -2 sin x + 2 sin x = 0

y¿ + 2 sin x = 0f(x) = 2 cos x + 10f(x)

f(x)

d3y

dx3 + adydxb2

- ex = 0

d2y

dx2 + 3xdy

dx- 2y = 0

y¿ + 2 sin x = 0

y(k)

F (x, y, y(1), y(2), Á , y(n)) = 0

776 Chapter 15 Differential Equations

Second-Order Linear Equations A second-order linear differential equa-tion has the form

In this section, we make two simplifying assumptions: (1) and are con-stants, and (2) is identically zero. Thus, our initial task is to solve

A differential equation for which is said to be homogeneous.To solve a first-order equation required one integration and led to a general

solution with one arbitrary constant. By analogy, we might expect that solving asecond-order equation to involve two integrations and thus that the general solu-tion would have two arbitrary constants. Our expectations are correct. In fact, asecond-order homogeneous linear differential equation always has two fundamen-tal solutions and , which are independent of each other (i.e., neitherfunction is a constant multiple of the other). By the linearity of the operator

is also a solution. Moreover, it can be shown that every solution has this form.

The Auxiliary Equation Because it seems likely that willbe a solution to our differential equation for an appropriate choice of r.To test thispossibility, we first write the equation in the operator form

(1)

Now

The latter expression is zero, provided

(2)

Equation (2) is called the auxiliary equation for (1) (note the similarity inform). It is an ordinary quadratic equation and can be solved by factoring or, ifnecessary, by the Quadratic Formula. There are three cases to consider, correspon-ding to whether the auxiliary equation has two distinct real roots, a single repeatedroot, or two complex conjugate roots.

r2 + a1r + a2 = 0

= erx (r2 + a1r + a2)

= r2erx + a1rerx + a2e

rx

(D2 + a1D + a2)erx = D2(erx) + a1D(erx) + a2 e

rx

(D2 + a1D + a2 )y = 0

erxDx(erx) = rerx,

C1u1(x) + C2u2(x)

D2 + a1 D + a2 ,

u2(x)u1(x)

k(x) = 0

y– + a1 y¿ + a2 y = 0

k(x)a2(x)a1(x)

y– + a1(x)y¿ + a2(x)y = k(x)

■ EXAMPLE 1 Find the general solution to

SOLUTION The auxiliary equation

has the two roots and Since and are independent solutions, thegeneral solution to the differential equation is

■y = C1e-3x + C2e

-4x

e-4xe-3x-4.-3

r2 + 7r + 12 = (r + 3)(r + 4) = 0

y– + 7y¿ + 12y = 0.

Theorem A Distinct Real Roots

If and are distinct real roots of the auxiliary equation, then the generalsolution of is

y = C1er1x + C2e

r2x

y– + a1y¿ + a2y = 0r2r1

Section 15.1 Linear Homogeneous Equations 777

■ EXAMPLE 2 Find the solution of that satisfies

and

SOLUTION The auxiliary equation is best solved by theQuadratic Formula.

The general solution to the differential equation is, therefore,

The condition implies that Then

and so

We conclude that and

■

This is all fine if the auxiliary equation has distinct real roots. But what if it hasthe form

Then our method produces the single fundamental solution and we must findanother solution independent of this one. Such a solution is as we nowdemonstrate.

= 0

= (xr21 er1x + 2r1 e

r1x) - 2r1(xr1 er1x + er1x) + r21 xer1x

(D2 - 2r1 D + r21)xer1x = D2(xer1x) - 2r1 D(xer1x) + r21 xer1x

xer1x,er1x

r2 - 2r1 r + r21 = (r - r1)2 = 0

y = 12 e

(1+22)x - 12 e

(1-22)x

C1 = 12

22 = y¿(0) = C1(1 + 22) - C1(1 - 22) = 2C122

y¿ = C1(1 + 22)e(1+22)x - C1(1 - 22)e(1-22)x

C2 = -C1. y(0) = 0

y = C1e(1+22)x + C2e

(1-22)x

r =-b ; 2b2 - 4ac

2a=

2 ; 24 + 42

= 1 ; 22

r2 - 2r - 1 = 0

y¿(0) = 22.

y(0) = 0y– - 2y¿ - y = 0

■ EXAMPLE 3 Solve

SOLUTION The auxiliary equation has 3 as a repeated root. Thus,

■

Finally, we consider the case where the auxiliary equation has complex conju-gate roots. The simple equation

with auxiliary equation and offers a hint. Its fundamentalsolutions are easily seen to be sin and cos You can check by direct differen-tiation that the general situation is as follows.

bx.bxroots ; bir2 + b2 = 0

(D2 + b2)y = 0

y = C1e3x + C2xe

3x

y– - 6y¿ + 9y = 0.Consider the second-order differen-tial equation

with auxiliary equation

The latter equation may have tworeal roots and one real root or two complex roots

Solution toRoots Differential Equation

+ C2 eax sin bx

y = C1 eax cos bxa ; bi

y = C1 er1x + C2 xe

r1xr1 = r2

y = C1 er1x + C2 e

r2xr1 Z r2

a ; bi.r1,r2,r1

r2 + a1r + a2 = 0

y– + a1y¿ + a2 = 0

Summary

Theorem B A Single Repeated Root

If the auxiliary equation has the single repeated root then the general solu-tion of is

y = C1er1x + C2xe

r1x

y– + a1 y¿ + a2 y = 0r1,

Theorem C Complex Conjugate Roots

If the auxiliary equation has complex conjugate roots then the generalsolution of is

y = C1eax cos bx + C2e

ax sin bx

y– + a1 y¿ + a2 y = 0a ; bi,


■ EXAMPLE 4 Solve

SOLUTION The roots of the auxiliary equation are Hence, the general solution is

■

Higher-Order Equations All of what we have done extends to higher-orderlinear homogeneous equations with constant coefficients. To solve

find the roots of the auxiliary equation

and make the obvious generalizations of the second-order case. For example, if theauxiliary equation is

then the general solution to the differential equation is

■ EXAMPLE 5 Solve

SOLUTION The auxiliary equation is

with roots and a double root of 0. Hence, the general solution is

■y = C1 + C2x + C3e5x + C4e

-4x

-4, 5,

r4 - r3 - 20r2 = r2(r - 5)(r + 4) = 0

d4y

dx4 -d3y

dx3 - 20d2y

dx2 = 0.

y = C1er1x + (C2 + C3x + C4x

2)er2x + eax[C5 cos bx + C6 sin bx]

(r - r1) (r - r2)3 [r - (a + bi)][r - (a - bi)] = 0

rn + a1 rn-1 + Á + an-1 r + an = 0

y(n) + a1 y(n-1) + Á + an-1 y¿ + an y = 0

y = C1 e2x cos 3x + C2 e

2x sin 3x

2 ; 3i.r2 - 4r + 13 = 0

y– - 4y¿ + 13y = 0.

Concepts Review1. The auxiliary equation corresponding to the differential

equation is ________ . This equationmay have two real roots, a single repeated root, or ________ .

2. The general solution to is ________ .y =(D2 - 1)y = 0

(D2 + a1 D + a2) y = 03. The general solution to is

________ .

4. The general solution to is ________ .y =(D2 + 1) y = 0

y =(D2 - 2D + 1) y = 0

Problem Set 15.1In Problems 1–16, solve each differential equation.

1. 2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14. y(4) - y = 0

y(4) + 3y–¿ - 4y– = 0

y– + y¿ + y = 0

y– + 2y¿ + 2y = 0

y– + 9y = 0; y = 3, y¿ = 3 at x = p>3y– + 4y = 0; y = 2 at x = 0, y = 3 at x = p>4y– + 6y¿ - 2y = 0

y– - 4y¿ + y = 0

y– + 10y¿ + 25y = 0

y– - 4y¿ + 4y = 0

y– - 3y¿ - 10y = 0; y = 1, y¿ = 10 at x = 0

y– + 6y¿ - 7y = 0; y = 0, y¿ = 4 at x = 0

y– + 5y¿ - 6y = 0y– - 5y¿ + 6y = 0

15.

16.

17. Solve and express your answer in terms ofthe hyperbolic functions cosh and sinh.

18. Show that the solution of

can be written as

19. Solve Hint: First showthat the auxiliary equation is

20. Solve and express your answer in the

form Hint: Let and

where c = 2C21 + C2

2.

cos g = C2>c,sin g = C1>cceax sin (bx + g).

y– - 2y¿ + 2y = 0

(r2 + r + 1)2 = 0.y(4) + 2y(3) + 3y– + 2y¿ + y = 0.

y = ebx AD1 cosh 2b2 + c2 x + D2 sinh 2b2 + c2 x B

d2y

dx2 - 2bdy

dx- c2y = 0

y– - 4y = 0

3(D2 + 1)(D2 - D - 6)4 y = 0

(D4 + 3D2 - 4)y = 0

Section 15.2 Nonhomogeneous Equations 779

21. Solve by first making the substi-tution

22. Show that the substitution transforms the Eulerequation to a homogeneous linearequation with constant coefficients.

23. Show that if and are distinct real roots of the auxil-iary equation, then is a solution of

24. Show that if are complex conjugate roots of theauxiliary equation, then is a so-lution of

25. Recall that complex numbers have the form where a and b are real. These numbers behave much like the real numbers, with the proviso that Show each of thefollowing:(a) Hint: Use the Maclaurin series for

cos u, and sin u.(b)

(c) Dx e(a+bi)x = (a + bi)e(a+bi)x

ea+bi = ea(cos b + i sin b)

eu,ebi = cos b + i sin b

i2 = -1.

a + bi,y– + a1y¿ + a2y = 0.


ax sin bxa ; bi

y– + a1 y¿ + a2 y = 0.y = C1e

r1x + C2er2x

r2r1

ax2 y– + bxy¿ + cy = 0x = ez

x = ez.x2 y– + 5xy¿ + 4y = 0 26. Let the roots of the auxiliary equation

be From Problem 25c, it follows, just as in the real case,that satisfies Show that this solution can be rewritten in the form

giving another approach to Theorem C.

Use a CAS to solve each of the following equations:

27.

28.

29.

30.

Answers to Concepts Review: 1. com-plex conjugate roots 2. 3.4. C1 cos x + C2 sin x

(C1 + C2x)exC1e

-x + C2ex

r2 + a1 r + a2 = 0;

3y– - 2y¿ + y = 0; y(0) = 2.5, y¿(0) = -1.5

2y– + y¿ + 2y = 0; y(0) = 0, y¿(0) = 1.25

y– + 5y¿ + 6.25y = 0; y(0) = 2, y¿(0) = -1.5

y– - 4y¿ - 6y = 0; y(0) = 1, y¿(0) = 2

CAS


ax sin bx

(D2 + a1D + a2)y = 0.y = c1e(a;bi)x + c2e(a–bi)xa ; bi.

r2 + a1r + a2 = 0

Proof The linearity of the operator L is the key element in the proof. Let andbe as described. Then

and so is a solution to (2).Conversely, let y be any solution to (2). Then

L(y - yp) = L(y) - L(yp) = k(x) - k(x) = 0

y = yp + yh

L(yp + yh) = L(yp) + L(yh) = k(x) + 0

yh

yp

Consider the general nonhomogeneous linear equation with constant coefficients

(1)

Solving this equation can be reduced to three steps:

1. Find the general solution

to the corresponding homogeneous equation (i.e., equation (1) with being identically zero), as described in Section 15.1.

2. Find a particular solution to the nonhomogeneous equation.3. Add the solutions from Steps 1 and 2.

We state the result as a formal theorem.

yp

k(x)

yh = C1u1(x) + C2u2(x) + Á + Cnun(x)

y(n) + a1y(n–1) + Á + an–1y¿ + any = k(x)

15.2Nonhomogeneous

Equations

Theorem A

If is any particular solution to the nonhomogeneous equation

(2)

and if is the general solution to the corresponding homogeneous equation,then

is the general solution of (2).

y = yp + yh

yh

L(y) = (Dn + a1Dn–1 + Á + an–1D + an) y = k(x)

yp


and so is a solution to the homogeneous equation. Consequently,can be written as plus a solution to the homogeneous equa-

tion, as we wished to show. ■

Now we apply this result to second-order equations.

The Method of Undetermined Coefficients The results of the previoussection show us how to get the general solution to a homogeneous equation. Thework lies in finding a particular solution to the nonhomogeneous equation. Onesuch method of finding such a solution, the method of undetermined coefficients,involves making a conjecture, or educated guess, about the form of given theform of

It turns out that the functions most apt to occur in applications are poly-nomials, exponentials, sines, and cosines. For these functions, we offer a procedurefor finding based on trial solutions.

If Try

Modification. If a term of is a solution to the homogeneousequation, multiply the trial solution by x (or perhaps by a higherpower of x).

To illustrate the table, we suggest the appropriate trial solution in six cases.The first three are straightforward; the last three are modified because a term onthe right side of the differential equation is present in the solution to the homoge-neous equation.

1.

2.

3.

4.(2 is a solution to the homogeneous equation)

5.( is a solution to the homogeneous equation)

6.(sin 2x is a solution to the homogeneous equation)

Next we carry out the details in four specific examples.

■ EXAMPLE 1 Solve

SOLUTION The auxiliary equation has roots and 1, and so

To find a particular solution to the nonhomogeneous equation, we try

Substitution of this expression in the differential equation gives

2A + (2Ax + B) - 2(Ax2 + Bx + C) = 2x2 - 10x + 3

yp = Ax2 + Bx + C

yh = C1e-2x + C2e

x

-2r2 + r - 2 = 0

y– + y¿ - 2y = 2x2 - 10x + 3.

yp = Bx cos 2x + Cx sin 2xy– + 4y = sin 2xe4x

yp = Bxe4xy– - 3y¿ - 4y = e4x

yp = B2x3 + B1x

2 + B0xy– + 2y¿ = 3x2 + 2

yp = B cos x + C sin xy– + 4y = 2 sin x

yp = Be2xy– - 3y¿ - 4y = e2x

yp = B2x2 + B1x + B0y– - 3y¿ - 4y = 3x2 + 2

yp

k(x)

B cos bx + C sin bxb cos bx + c sin bx

BeaxbeaxBmx

m + Á + B1x + B0bmxm + Á + b1x + b0

yp =k(x) =

yp

k(x)k(x).

yp,

ypy = yp + (y - yp)y - yp

Section 15.2 Nonhomogeneous Equations 781

Equating coefficients of and 1, we find that

or Hence,

and

■

■ EXAMPLE 2 Solve

SOLUTION Since the auxiliary equation has roots and 3,we have

Note that is a solution to the homogeneous equation. Thus, we use themodified trial solution

Substituting in the differential equation gives

or

or, finally,

We conclude that and

■

If 3 had been a double root of the auxiliary equation in Example 2, we wouldhave used as our trial solution.

■ EXAMPLE 3 Solve

SOLUTION The homogeneous equation agrees with that of Example 2, and so

For the trial solution we use

Now

Hence, substitution of in the differential equation gives (after collecting terms)

Thus, and which imply that and We conclude that

■y = -765

cos 2x -465

sin 2x + C1e-x + C2e

3x

C = - 465.B = - 7

654B - 7C = 0,-7B - 4C = 1

(-7B - 4C) cos 2x + (4B - 7C) sin 2x = cos 2x

yp

D2yp = -4B cos 2x - 4C sin 2x

Dyp = -2B sin 2x + 2C cos 2x

yp = B cos 2x + C sin 2x

yp,

yh = C1e-x + C2e

3x

y– - 2y¿ - 3y = cos 2x.

Bx2e3x

y = 2xe3x + C1e-x + C2e

3x

B = 2

4Be3x = 8e3x

(9xBe3x + 6Be3x) - 2(3Bxe3x + Be3x) - 3Bxe3x = 8e3x

(Bxe3x)– - 2(Bxe3x)¿ - 3Bxe3x = 8e3x

yp

yp = Bxe3x

k(x) = 8e3x

yh = C1e-x + C2e

3x

-1r2 - 2r - 3 = 0

y– - 2y¿ - 3y = 8e3x.

y = -x2 + 4x -12

+ C1e-2x + C2e

x

yp = -x2 + 4x -12

A = -1, B = 4, and C = - 12.

-2A = 2, 2A - 2B = -10, 2A + B - 2C = 3

x2, x,


■ EXAMPLE 4 Solve

SOLUTION Combining the results of Examples 2 and 3 and using the linearityof the operator we obtain

■

The Method of Variation of Parameters A more general method thanthat of undetermined coefficients is the method of variation of parameters. If and are independent solutions to the homogeneous equation, then it can beshown (see Problem 23) that there is a particular solution to the nonhomogeneousequation of the form

where

We show how this method works in an example.

■ EXAMPLE 5 Find the general solution of

SOLUTION The general solution to the homogeneous equation is

To find a particular solution to the nonhomogeneous equation, we set

and impose the conditions

When we solve this system of equations for and we obtain andThus,

(We can omit the arbitrary constants in the above integrations, since any solutionsand will do.) A particular solution is therefore

a result that is easy to check by direct substitution in the original differential equa-tion. We conclude that

■y = (ln ƒcos x ƒ) cos x + x sin x + C1 cos x + C2 sin x

yp = (ln ƒcos x ƒ) cos x + x sin x

v2v1

v2(x) = Ldx = x

v1(x) = L(-tan x) dx = ln ƒcos x ƒ

v¿2 = 1.v¿1 = -tan xv¿2,v¿1

-v¿1 sin x + v¿2 cos x = sec x

v¿1 cos x + v¿2 sin x = 0

yp = v1(x) cos x + v2(x) sin x

yh = C1 cos x + C2 sin x

y– + y = sec x.

v¿1u¿1 + v¿2u¿2 = k(x)

v¿1u1 + v¿2 u2 = 0

yp = v1(x)u1(x) + v2(x)u2(x)

u2(x)u1(x)

y = 2xe3x -765

cos 2x -465

sin 2x + C1e-x + C2e

3x

D2 - 2D - 3,

y– - 2y¿ - 3y = 8e3x + cos 2x.

Concepts Review1. The general solution to a nonhomogeneous equation has

the form where is a ________ and is the gener-al solution to the ________.

2. Thus, after noting that has the particu-lar solution we conclude that the general solution is

________.y =y = -1,

y– - y¿ - 6y = 6

yhypy = yp + yh,3. The method of undetermined coefficients suggests trying

a particular solution of the form __________________ for

4. The method of undetermined coefficients suggests tryinga particular solution of the form __________________ fory– - y¿ - 6y = e3x.

y =

y– - y¿ - 6y = x2.y =

Section 15.3 Applications of Second-Order Equations 783

Problem Set 15.2In Problems 1–16, use the method of undetermined coefficients tosolve each differential equation.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15.

16.

In Problems 17–22, solve each differential equation by variation ofparameters.

17. 18. y– - 4y = e2xy– - 3y¿ + 2y = 5x + 2

y– - 4y = 4 sin x; y = 4, y¿ = 0 when x = 0

y– - 5y¿ + 6y = 2ex; y = 1, y¿ = 0 when x = 0

y– + y¿ = ex + 3xy– + 9y = sin x + e2x

y– + 9y = sin 3xy– + 4y = 2 cos 2x

y– + 4y¿ = cos xy– - y¿ - 2y = 2 sin x

y– + 2y¿ + 2y = 3e-2xy– + 4y¿ + 3y = e-3x

y– + 6y¿ + 9y = 2e-xy– - 5y¿ + 6y = exy– + y¿ = 4xy– - 2y¿ + y = x2 + xy– + y¿ - 6y = 2x2y– - 9y = x

19. 20.

21. 22.

23. Let have solutions and and let Show that

Thus, if the conditions of the method of variation of parametershold,

24. Solve

Answers to Concepts Review: 1. particular solution to thenonhomogeneous equation; homogeneous equation2. 3.4. y = Bxe3x

y = Ax2 + Bx + C-1 + C1e-2x + C2e

3x

y– + 4y = sin3 x

L(yp) = (v1)(0) + (v2)(0) + (b)(0) + 0 + k(x) = k(x)

+ b(v¿1u1 + v¿2u2) + (v¿1u1 + v¿2u2)¿ + (v¿1u¿1 + v¿2u¿2)L(yp) = v1(u–1 + bu¿1 + cu1) + v2(u–2 + bu¿2 + cu2)

yp = v1u1 + v2 u2.u2,u1L(y) = y– + by¿ + cy = 0

y– - 5y¿ + 6y = 2exy– - 3y¿ + 2y =ex

ex + 1

y– + y = cot xy– + y = csc x cot x

Many problems in physics lead to second-order linear differential equations. Wefirst consider the problem of a vibrating spring under various assumptions. Thenwe return to and generalize an earlier application to electric circuits.

A Vibrating Spring (Simple Harmonic Motion) Consider a coiledspring weighted by an object A and hanging vertically from a support, as in Fig-ure 1a. We wish to consider the motion of the point P if the spring is pulled unitsbelow its equilibrium position (Figure 1b) and given an initial velocity of We as-sume friction to be negligible.

According to Hooke’s Law, the force F tending to restore P to its equilibriumposition at satisfies where k is a constant depending on the char-acteristics of the spring and y is the y-coordinate of P. But by Newton’s SecondLaw, where is the weight of the object A, a is the accelera-tion of P, and g is the constant acceleration due to gravity ( feet per secondper second). Thus,

is the differential equation of the motion. The solution y must satisfy the initialconditions and where and are the initial position andinitial velocity, respectively.

If we let then this equation takes the form

and has the general solution

The conditions and at determine the constants and Ifthe object is released with an initial velocity of 0, then Thus,

We say that the spring is executing simple harmonic motion with amplitude andperiod (Figure 2).2p>B y0

y = y0 cos Bt

C1 = y0 and C2 = 0.C2.C1t = 0y¿ = v0y = y0

y = C1 cos Bt + C2 sin Bt

d2y

dt2+ B2y = 0

B2 = kg>w = k>m,

v0y0y¿(0) = v0,y(0) = y0

wg

d2y

dt2= -ky. k 7 0

g = 32wF = ma = (w>g)a,

F = -ky,y = 0

v0.y0

15.3Applications of

Second-Order Equations

y y

P

P

O

A

A

y0

O

(b)(a)

Figure 1

Period =

Simple harmonic motion

y0

y

t

2π

B

Figure 2


■ EXAMPLE 1 When an object weighing 5 pounds is attached to the lowestpoint P of a spring that hangs vertically, the spring is extended 6 inches. The 5-pound weight is replaced by a 20-pound weight, and the system is allowed to cometo equilibrium. If the 20-pound weight is now pulled downward another 2 feet andthen released, describe the motion of the lowest point P of the spring.

SOLUTION The first sentence of the example allows us to determine the springconstant. By Hooke’s Law, where s is the amount in feet that the spring isstretched, and so or Now put the origin at the equilibrium pointafter the 20-pound weight has been attached. From the derivation just before theexample, we know that In the present case, and

We conclude that

The motion of P is simple harmonic motion, with period and amplitude 2 feet. That is, P oscillates up and down from 2 feet below 0 to 2 feet above 0 andthen back to 2 feet below 0 every seconds. ■

Damped Vibrations So far we have assumed a simplified situation, in whichthere is no friction either within the spring or resulting from the resistance of theair. We can take friction into account by assuming a retarding force that is propor-tional to the velocity The differential equation describing the motion thentakes the form

By letting and this equation can be written as

an equation for which the methods of Section 15.1 apply. The auxiliary equationfor this second-order linear differential equation is so the rootsare

We must consider the cases where is negative, zero, and positive.

Case 1: In this case, the roots are complex:

Notice that and will both be positive. The general solution of the differentialequation is thus

which can be written in the form (see Problem 15)

The factor called the damping factor, causes the amplitude of the motionto approach zero as (Figure 3a). ■t: q

e-at,

y = Ae-at sin (bt + g)

y = e-at (C1 cos bt + C2 sin bt)

ba

r = -E

2;i

224B2 - E2 = -a ; bi

E2 - 4B2 6 0

E2 - 4B2

-E ; 2E2 - 4B2

2

r2 + Er + B2 = 0,

d2y

dt2+ E

dy

dt+ B2y = 0

B2 = kg>w,E = qg>w

wg

d2y

dt2= -ky - q

dy

dt. k 7 0, q 7 0

dy>dt.

12 p L 1.57

12 p

y = 2 cos 4t

B2 = kg>w = (10)(32)>20 = 16.y0 = 2y = y0 cos Bt.

k = 10.5 = k(12),

ƒF ƒ = ks,


y

x

(a) (b) (c)

x x

y y

Figure 3

Case 2: In this case, the auxiliary equation has the double rootwhere and the general solution of the differential equation is

The motion described by this equation is said to be critically damped. ■

Case 3: The auxiliary equation has roots and and thegeneral solution of the differential equation is

It describes a motion that is said to be overdamped.The graphs in the critically damped and overdamped cases cross the t-axis at

most once and may look something like Figure 3b or 3c. ■

■ EXAMPLE 2 If a damping force with is imposed on the system ofExample 1, find the equation of motion.

SOLUTION and sowe must solve

The auxiliary equation has roots and thus

When we impose the conditions and at we find thatand Consequently,

■

Electric Circuits Consider a circuit (Figure 4) with a resistor (R ohms), an in-ductor (L henrys), and a capacitor (C farads) in series with a source of electromo-tive force supplying E(t) volts. The new feature in comparison to the circuits ofSection 6.6 is the presence of a capacitor. Kirchhoff’s Law in this situation says thatthe charge Q on the capacitor, measured in coulombs, satisfies

(1)

The current measured in amperes, satisfies the equation obtained bydifferentiating equation (1) with respect to t ; that is,

(2)

These equations can be solved by the methods of Sections 15.1 and 15.2 for manyfunctions E(t).

Ld2I

dt2+ R

dI

dt+

1CI = E¿(t)

I = dQ>dt,Ld2Q

dt2+ R

dQ

dt+

1CQ = E(t)

y = e-0.16t (2 cos 4t + 0.08 sin 4t)

C2 = 0.08.C1 = 2t = 0,y¿ = 0y = 2

y = e-0.16t (C1 cos 4t + C2 sin 4t)

-0.16 ; 4i,r = -0.16 ; 215.9744i Lr2 + 0.32r + 16 = 0

d2y

dt2+ 0.32

dy

dt+ 16y = 0

B2 = (10)(32)>20 = 16,E = qg>w = (0.2)(32)>20 = 0.32

q = 0.2

y = C1e-a1t + C2e

-a2t

-a2,-a1E2 - 4B2 7 0

y = C1e-at + C2te

-at

a = E>2-aE2 - 4B2 = 0

L

RS

CE

Figure 4


■ EXAMPLE 3 Find the charge Q and the current I as functions of time t in anRCL circuit (Figure 4) if and Assumethat and at (when the switch is closed).

SOLUTION By Kirchhoff’s Law as expressed in equation (1),

The auxiliary equation has roots

so

By inspection, a particular solution is Therefore, the generalsolution is

When we impose the given initial conditions, we find that and We conclude that

and, by differentiation, that

■I =dQ

dt= 2e-400t sin 300t

Q = 10-3 C2.4 - e-400t(2.4 cos 300t + 3.2 sin 300t) DC2 = -3.2 * 10-3.

C1 = -2.4 * 10-3

Q = 2.4 * 10-3 + e-400t (C1 cos 300t + C2 sin 300t)

Qp = 2.4 * 10-3.

Qh = e-400t (C1 cos 300t + C2 sin 300t)

-800 ; 2640,000 - 1,000,0002

= -400 ; 300i

d2Q

dt2+ 800

dQ

dt+ 250,000Q = 600

t = 0I = 0Q = 0E = 12.R = 16, L = 0.02, C = 2 * 10-4,

Concepts Review1. A spring that vibrates without friction might obey a law

of motion such as We say it is executing simple har-monic motion with amplitude ________ and period ________.

2. A spring vibrating in the presence of friction might obeya law of motion such as called damped har-monic motion. The “period” is still ________ , but now the ampli-tude ________ as time increases.

y = 3e-0.1t cos 2t,

y = 3 cos 2t.3. If the friction is very great, the law of motion might take

the form the critically damped case, in whichy slowly fades to ________ as time increases.

4. Kirchhoff’s Law says that a(n) ________ satisfies asecond-order linear differential equation.

y = 3e-0.1t + te-0.1t,

Problem Set 15.31. A spring with a spring constant k of 250 newtons per

meter is loaded with a 10-kilogram mass and allowed to reachequilibrium. It is then raised 0.1 meter and released. Find theequation of motion and the period. Neglect friction.

2. A spring with a spring constant k of 100 pounds per footis loaded with a 1-pound weight and brought to equilibrium. It isthen stretched an additional 1 inch and released. Find the equa-tion of motion, the amplitude, and the period. Neglect friction.

3. In Problem 1, what is the absolute value of the velocity ofthe moving weight as it passes through the equilibrium position?

4. A 10-pound weight stretches a spring 4 inches. Thisweight is removed and replaced with a 20-pound weight, which isthen allowed to reach equilibrium. The weight is next raised

1 foot and released with an initial velocity of 2 feet per seconddownward. What is the equation of motion? Neglect friction.

5. A spring with a spring constant k of 20 pounds per foot isloaded with a 10-pound weight and allowed to reach equilibrium.It is then displaced 1 foot downward and released. If the weightexperiences a retarding force in pounds equal to one-tenth thevelocity, find the equation of motion.

6. Determine the motion in Problem 5 if the retarding forceequals four times the velocity at every point.

7. In Problem 5, how long will it take the oscillations to di-minish to one-tenth their original amplitude?

8. In Problem 5, what will be the equation of motion if theweight is given an upward velocity of 1 foot per second at the mo-ment of release?


R = 106 Ω

C = 10–6 F

S

E = 1 V

Figure 5

C # 2 "10–6 FE # 120sin 377t

S

Figure 6

9. Using Figure 5, find the charge Q on the capacitor as afunction of time if S is closed at Assume that the capacitoris initially uncharged.

t = 0.

12. Using Figure 7, find the current as a function of time if thecapacitor is initially uncharged and S is closed at Hint: Thecurrent at will equal 0, since the current through an induc-tance cannot change instantaneously.

t = 0t = 0.

13. Using Figure 8, find the steady-state current as a functionof time; that is, find a formula for I that is valid when t is verylarge ( ).t: q

14. Suppose that an undamped spring is subjected to an ex-ternal periodic force so that its differential equation has the form

d2 y

dt2+ B2 y = c sin At, c 7 0

(a) Show that the equation of motion for is

(b) Solve the differential equation when (the resonancecase).

(c) What happens to the amplitude of the motion in part (b)when

15. Show that can be written in theform Hint: Let and

16. Show that the motion of part (a) of Problem 14 is period-ic if is rational.

17. Refer to Figure 9, which shows a pendulum bob of mass msupported by a weightless wire of length L. Derive the equationof motion; that is, derive the differential equation satisfied by .Suggestion: Use the fact from Section 11.7 that the scalar tangen-tial component of the acceleration is where s measuresarc length in the counterclockwise direction.

d2s>dt2,u

B>Acos g = C2>A.

A = 2C21 + C2

2, sin g = C1>A,A sin (bt + g).C1 cos bt + C2 sin bt

t: q?

A = B

y = C1 cos Bt + C2 sin Bt +c

B2 - A2 sin At

A Z B

S

C = 10–7 F

L = 10–2 H

E = 20 V

Figure 7

L = 3.5 H

R = 1000 Ω

C # 2 " 10–6 FE # 120sin 377t

Figure 8

mg

L

θ

Figure 9

18. The equation derived in Problem 17 is nonlinear, but forsmall it is customary to approximate it by the equation

Here where G is a universal constant, M is the massof the earth, and R is the distance from the pendulum to the cen-ter of the earth. Two clocks, with pendulums of length and and located at distances and from the center of the earth,have periods and respectively.

(a) Show that

(b) Find the height of a mountain if a clock that kept perfecttime at sea level with inches hadto have its pendulum shortened to inches to keepperfect time at the top of the mountain.

Answers to Concepts Review: 1. 3; 2. decreases3. 0 4. electric circuit

p;p

L = 80.85L = 81(R = 3960 miles)

p1

p2=R1 2L1

R22L2

.

p2,p1

R2R1

L2L1

g = GM>R2,

d2u

dt2+g

Lu = 0

u

10. Find the current I as a function of time in Problem 9 if thecapacitor has an initial charge of 4 coulombs.

11. Use Figure 6.(a) Find Q as a function of time. Assume that the capacitor is

initially uncharged.(b) Find I as a function of time.


L = 1 H

R = 2 Ω

E = 1 V C # 12 F

Figure 1

15.4 Chapter Review

Concepts Test

Respond with true or false to each of the following assertions. Beprepared to justify your answer.

1. is a linear differential equation.

2. is a linear differential equation.

3. is a solution of

4. The general solution to should in-volve eight arbitrary constants.

5. is a linear operator.

6. If and are two solutions to then is also a solution.

7. The general solution to is

8. If and are solutions to the linear differentialequation then is a solution to

9. The equation has a particular solutionof the form

10. An expression of the form can al-ways be written in the form

Sample Test Problems

In Problems 1–13, solve each differential equation.

1. Suggestion: Let

2.

3. when

4. 5. y– - y = 14y– + 12y¿ + 9y = 0

x = 0y– - 3y¿ + 2y = 0, y = 0, y¿ = 3

y– - y = 0

u = dy>dx.d2y

dx2 + 3 dy

dx= ex

A sin (bt + g).C1 cos bt + C2 sin bt

yp = B sin 3x + C cos 3x.y– + 9y = 2 sin 3x

L(y) = 0.u1(x) - u2(x)L(y) = f(x),

u2(x)u1(x)

y = C1e-x + C2xe

-x + C3x2e-x.

y–¿ + 3y– + 3y¿ + y = 0

C1u1(x) + C2u2(x)f(x),y– + a1y¿ + a2y =u2(x)u1(x)

D2

CD2 + aD + b D3 y = 0

2y¿ - y2 = 1.y = tan x + sec x

y– + x2 y = 0

y– + y2 = 0

6. 7.

8. when

9. 10.

11. 12.

13.

14. Suppose that glucose is infused into the bloodstream of apatient at the rate of 3 grams per minute, but that the patient’sbody converts and removes glucose from its blood at a rate pro-portional to the amount present (with constant of proportionali-ty 0.02). If is the amount present at time t and (a) write the differential equation for Q;(b) solve this differential equation;(c) determine what happens to Q in the long run.

15. A spring with a spring constant k of 5 pounds per foot isloaded with a 10-pound weight and allowed to reach equilibrium.It is then raised 1 foot and released.What are the equation of mo-tion, the amplitude, and the period? Neglect friction.

16. In Problem 15, what is the absolute value of the velocityof the moving weight as it passes through the equilibrium posi-tion?

17. Suppose the switch of the circuit in Figure 1 is closed atFind I as a function of time if C is initially uncharged. (The

current at will equal zero, since current through an induc-tance cannot change instantaneously.)

t = 0t = 0.

Q(0) = 120,Q(t)

y(4) - 4y– + 4y = 0

y(4) - 3y– - 10y = 0y–¿ + 2y– - 8y¿ = 0

y– + y = sec x tan xy– + 6y¿ + 25y = 0

x = 0y– + 4y = 0; y = 0, y¿ = 2

y– + 4y¿ + 4y = e-2xy– + 4y¿ + 4y = 3ex

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