Diff EQ Chapter-4

294

CHAPTER 4 Second-Order LinearDifferential Equations

4.1 The Harmonic Oscillator

The Undamped Oscillator

1. 0x x+ = , ( )0 1x = , ( )0 0x =

The general solution of the harmonic oscillator equation 0x x+ = is given by

( )( )

1 2

1 2

cos sin

sin cos .

x t c t c t

x t c t c t

= +

= − +

Substituting the initial conditions ( )0 1x = , ( )0 0x = , gives

( )( )

1

2

0 1

0 0

x c

x c

= =

= =

so 1 1c = , 2 0c = . Hence, the IVP has the solution ( ) cosx t t= .

2. 0x x+ = , ( )0 1x = , ( )0 1x =

The general solution of the harmonic oscillator equation 0x x+ = is given by

( )( )

1 2

1 2

cos sin

sin cos .

x t c t c t

x t c t c t

= +

= − +


( )( )

1

2

0 1

0 1

x c

x c

= =

= =

or 1 2 1c c= = . Hence, the IVP has the solution

( ) cos sinx t t t= + .

Higher-Order Linear Differential Equations

SECTION 4.1 The Harmonic Oscillator 295

In polar form, this would be

( ) 2 cos4

x t t π⎛ ⎞= −⎜ ⎟⎝ ⎠

.

3. 9 0x x+ = , ( )0 1x = , ( )0 1x =

The general solution of the harmonic oscillator equation 9 0x x+ = is given by

( )( )

1 2

1 2

cos3 sin3

3 sin 3 3 cos3 .

x t c t c t

x t c t c t

= +

= − +


( )( )

1

2

0 1

0 3 1

x c

x c

= =

= =

so 1 1c = , 213

c = . Hence, the IVP has the solution

( ) 1cos3 sin33

x t t t= + .


( ) ( )10 cos 33

x t t δ= −

where 1 1tan3

δ −= . This would be in the first quadrant.

4. 4 0x x+ = , ( )0 1x = , ( )0 2x = −


( )( )

1 2

1 2

cos2 sin 2

2 sin 2 2 cos2 .

x t c t c t

x t c t c t

= +

= − +

Substituting the initial conditions ( )0 1x = , ( )0 2x = − , gives

( )( )

1

2

0 1

0 2 2

x c

x c

= =

= = −

so 1 1c = , 2 1c = − . Hence, the IVP has the solution

( ) cos 2 sin 2x t t t= − .


( ) 2 cos 24

x t t π⎛ ⎞= +⎜ ⎟⎝ ⎠

.

296 CHAPTER 4 Higher-Order Linear Differential Equations

5. 16 0x x+ = , ( )0 1x = − , ( )0 0x =


( )( )

1 2

1 2

cos4 sin 4

4 sin 4 4 cos4 .

x t c t c t

x t c t c t

= +

= − +

Substituting the initial conditions ( )0 1x = − , ( )0 0x = , gives

( )( )

1

2

0 1

0 4 0

x c

x c

= = −

= =

so 1 1c = − , 2 0c = . Hence, the IVP has the solution

( ) cos4x t t= − .

6. 16 0x x+ = , ( )0 0x = , ( )0 4x =


( )( )

1 2

1 2

cos4 sin 4

4 sin 4 4 cos4 .

x t c t c t

x t c t c t

= +

= − +

Substituting the initial conditions ( )0 0x = , ( )0 4x = , we get

( )( )

1

2

0 0

0 4 4

x c

x c

= =

= =

so 1 0c = , 2 1c = . The IVP has the solution

( ) sin 4x t t= .

7. 216 0x xπ+ = , x (0) = 0, ( )0x π=

2

016

1πω = = 4π

x = c1 cos 4π t + c2 sin 4π t

x = −4π c1 sin 4π t + 4π c2 cos 4π t

x(0) = 0 = c1

(0)x = π = 4π c2 c2 = 14

x = 1 sin 44

tπ


8. 24 0x xπ+ = , x (0) = 1, ( )0x π=

2

0 4 2π πω = =

x = c1 cos 4π t + c2 sin 4π t

x = 1 2sin cos2 2 2 2

c t c tπ π π π+

x(0) = 1 = c1

(0)x = π = 2 2c π , c2 = 2

x = cos 2sin2 2

t tπ π+

Graphing by Calculator

9. cos siny t t= +

The equation tells us 2T π= and because

0

2T πω

= , 0 1ω = . We then measure the delay

0

0.8δω

≈ which we can compute as the phase

angle ( )0.8 1 0.8δ ≈ = . The amplitude A can be

measured directly giving 1.4A ≈ . Hence,

( )cos sin 1.4cos 0.8t t t+ ≈ − .

Compare with the algebraic form in Problem 15.

π

Š1.5

1.5

t

T = 2π

A ≈ 1.4δω

≈ 0.8

y

0

10. 2cos siny t t= +

The equation tells us 2T π= and because

0

2T πω

= , 0 1ω = . We then measure the delay

0

0.5δω

≈ , which we can compute as the phase

angle ( )0.5 1 0.5δ ≈ = . The amplitude A can be

measured directly giving 2.2A ≈ . Hence,

( )2cos sin 2.2cos 0.5t t t+ ≈ − .

8Š4

Š2.5

2.5

4

T = 2πA ≈ 2.2

y

t

δω

≈ 0.50


11. 5cos3 sin 3y t t= +

The equation tells us that period is 23

T π= and

because 0

2T πω

= , 0 3ω = . We then measure the

delay 0

0.05δω

≈ , which we can compute as the

phase angle ( )3 0.05 0.15δ ≈ = . The amplitude A

can be measured directly giving 5.1A ≈ . Hence,

( )5cos3 sin3 5.1cos 3 0.15t t t+ ≈ − .

–5

5

tπ

T = 2π A ≈ 5.1/3

y δω

≈ 0.050

12. cos3 5sin3y t t= +

The equation tells us the period is 23

T π= and

because 0

2T πω


delay 0

0.5δω

≈ , which we can compute as the

phase angle ( )0.5 3 1.5δ ≈ = . The amplitude A


( )cos3 5sin3 5.1cos 3 1.5t t t+ ≈ − .

3

Š5

5 A ≈ 5.1

y

T = 2π /3

t

δω

≈ 0.50

13. cos5 2sin5y t t= − +

equation tells us that period is 25

T π= and

because 0

2T πω


delay 0

or 0.48

δ πω

≈ , which we can compute as

the phase angle ( )5 0.4 2δ ≈ = . The amplitude A


( )cos5 2sin 2.2cos 5 2t t t− + ≈ − .

–2

2

t21

y

A ≈ 2.2T = 2π /5

δω

≈ 0.40


Alternate Forms for Sinusoidal Oscillations

14. We have

( ) ( ) ( ) ( )0 0 0 0 0

1 0 2 0

cos cos cos cos cos cos cos sin sincos sin

A t A t t A t A tc t c t

ω δ ω δ ω δ δ ω δ ω

ω ω

− = + = +

= +

where 1 cosc A δ= , 2 sinc A δ= .

Single-Wave Forms of Simple Harmonic Motion

15. cos sint t+

By Equation (4) 1 1c = , 2 1c = , and 0 1ω = . By Equation (5)

2A = , 4πδ =

yielding

cos sin 2 cos4

t t t π⎛ ⎞+ = −⎜ ⎟⎝ ⎠

.

(Compare with solution to Problem 9.)

16. cos sint t−

By Equation (4) 1 1c = , 2 1c = − , and 0 1ω = . By Equation (5)

2A = , 4πδ = −

yielding

cos sin 2 cos4

t t t π⎛ ⎞− = +⎜ ⎟⎝ ⎠

.

Because 1c is positive and 2c is negative the phase angle is in the 4th quadrant.

17. cos sint t− +

By Equation (4) 1 1c = − , 2 1c = , and 0 1ω = . By Equation (5)

2A = , 34πδ =

yielding

3cos sin 2 cos4

t t t π⎛ ⎞− + = −⎜ ⎟⎝ ⎠

.

Because 1c is negative and 2c is positive the phase angle is in the 2nd quadrant.


18. cos sint t− −

By Equation (5) 1 1c = − , 2 1c = − , and 0 1ω = . By Equation (6)

2A = , 54πδ =

yielding

5cos sin 2 cos4

t t t π⎛ ⎞− − = −⎜ ⎟⎝ ⎠

.

Because 1c and 2c are negative, the phase angle is in the 3rd quadrant.

Component Form of Harmonic Motion

Using ( )cos cos cos sin sinA B A B A B+ = − , we write:

19. ( ) ( ) ( ){ }2cos 2 2 cos2 cos sin 2 sin 2cos2t t t tπ π π− = − − − = −

20. 1 3cos cos cos sin sin cos sin3 3 3 2 2

t t t t tπ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

21. { }2 2 3 23cos 3 cos cos sin sin 3 cos sin cos sin4 4 4 2 2 2

t t t t t t tπ π π ⎧ ⎫⎧ ⎫ ⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − − = + = +⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎪ ⎪⎩ ⎭ ⎩ ⎭

22. 3 1cos 3 cos3 cos sin 3 sin cos3 sin 36 6 6 2 2

t t t t tπ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − − = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Interpreting Oscillator Solutions

23. 0x x+ = , ( )0 1x = , ( )0 0x =

Because 0 1ω = , we know the natural frequency is 12π

Hz and the period is 2π seconds. Using

the initial conditions, we find the solution (see Problem 1)

( ) cosx t t= ,

which tells us the amplitude is 1 and the phase angle 0δ = radians.

24. 0x x+ = , ( )0 1x = , ( )0 1x =

Because 0 1ω = radians per second, we know the natural frequency is 12π

Hz (cycles per

second), and the period is 2π . Using the initial conditions, we find the solution (see Problem 2)

( ) 2 cos4

x t t π⎛ ⎞= −⎜ ⎟⎝ ⎠

,

which tells us the amplitude is 2 and the phase angle is 4πδ = radians.


25. 9 0x x+ = , ( )0 1x = , ( )0 1x =


Hz (cycles per

second), and the period is 23π . Using the initial conditions, we find the solution (see Problem 3)

( ) ( )10 cos 33

x t t δ= −

where 1 1tan3

δ −= , which tells us the amplitude is

103

and the phase angle is

1 1tan 0.32183

δ −= ≈ radians.

26. 4 0x x+ = , ( )0 1x = , ( )0 2x = −


Hz (cycles per second),

and the period is π. Using the initial conditions, we find the solution (see Problem 4)

( ) 2 cos 24

x t t π⎛ ⎞= +⎜ ⎟⎝ ⎠

,

which tells us the amplitude is 2 and the phase angle is 4πδ = − radians.

27. 16 0x x+ = , ( )0 1x = − , ( )0 0x =



and the period is 2π . Using the initial conditions, we find the solution (see Problem 5)

( ) ( )cos 4x t t π= − ,

which tells us the amplitude is 1 and the phase angle is δ π= radians.


28. 16 0x x+ = , ( )0 0x = , ( )0 4x =



and the period is 2π . Using the initial conditions, we find the solution (see Problem 6)

( ) cos 42

x t t π⎛ ⎞= −⎜ ⎟⎝ ⎠

,

which tells us the amplitude is 1 and the phase angle is 2πδ = radians.

29. 216 0x xπ+ = , x (0) = 0, ( )0x π=

From Problem 7, x = 1 sin 44

tπ

Amplitude = 14

, x = 1 cos 44 2

t ππ⎛ ⎞−⎜ ⎟⎝ ⎠

, phase angle δ = 2π , and period T =

0

2 2

4

ππω

= = 8.

30. 24 0x xπ+ = , x (0) = 1, ( )0x π=

4r2 + π2 = 0

r = 2

iπ± x = 1 2cos sin

2 2c t c tπ π

+ 1 = c1

1 2sin cos2 2 2 2

x c t c tπ π π π= − + π = 22

cπ

x = cos 2sin2 2

t tπ π+ c2 = 2

Amplitude: A = 21 2 5+ =

x = 5 cos 1.112

tπ⎛ ⎞−⎜ ⎟⎝ ⎠

Relating Graphs

31. (a) See graph, next page.

(b) 0.25 0x x+ = 0km

ω = = 0.5

From (4) x = c1 cos 0.5t + c2 sin 0.5t

x(0) = 0 ⇒ c1 = 0 so x(t) = c2 sin 2t⎛ ⎞

⎜ ⎟⎝ ⎠

and 2( ) cos2 2c tx t ⎛ ⎞= ⎜ ⎟

⎝ ⎠.


Alternatively, you could use (5)

x = A cos (0.5t − δ)

x(0) = 0 ⇒ δ = 2π so x(t) = A cos

2 2t π⎛ ⎞−⎜ ⎟

⎝ ⎠ = A sin

2t⎛ ⎞

⎜ ⎟⎝ ⎠

,

and ( ) cos2 2A tx t ⎛ ⎞= ⎜ ⎟

⎝ ⎠.

(c) See graph

Graph for b) and d) Graph for a)

(d) Amplitudes are approximately 2 5, , ,3 2 3 6A A A A , and A

Phase Portraits

For comparison of phase portraits, the main observation is that the elliptical shape depends on ω0, which is k in all of these problems because 0x kx+ = .

If ω0 = 1, trajectories are circular. As ω0 increases above 1, ellipses become taller and thinner. As ω0 decreases from 1 to 0, ellipses become shorter and wider.

The aspect ratio of max

max

xx

= 1ω

.

Other observations include: • All these phase portraits show closed elliptical trajectories that circulate clockwise. • The trajectory of Problem 33 has a greater radius than that of Problem 32 because the initial

condition is further from the origin. • The trajectories in Problems 36 and 37 are on the same ellipse with different starting points

that give different solution equations.


32. 0x x+ = 010⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

From Problem 1 x(t) = cos(t),

so ( )x t = −sin(t).

33. 0x x+ = 011⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

From Problem 2 x(t) = cos(t) + sin(t),

so ( )x t = −sin(t) + cos(t).

34. 9 0x x+ = 011⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

From Problem 3, x(t) = cos 3t + 1 sin 3 ,3

t

so ( )x t = −3 sin 3t + cos 3t.


35. 4 0x x+ = 012

⎡ ⎤= ⎢ ⎥−⎣ ⎦

x

From Problem 4, x(t) = cos 2t − sin 2t,

so ( )x t = −2 sin 2t − 2 cos 2t.

36. 16 0x x+ = 01

0−⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

From Problem 5, x(t) = −cos 4t,

so ( )x t′ = 4 sin 4t.

37. 16 0x x+ = 004⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

From Problem 6, x(t) = sin 4t,

so ( )x t = 4 cos 4t.


38. 216 0x xπ+ = 00π⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

From Problem 7, x(t) = 1 sin 4 ,4

tπ

so ( )x t = cos 4π t.

39. 24 0x xπ+ = 01π⎡ ⎤

= ⎢ ⎥⎣ ⎦

x

From Problem 8, x(t) = cos 2sin ,2 2

t tπ π+

so ( )x t = sin cos .2 2 2

t tπ π ππ− +

Matching Problems

40. B

41. A

42. D

43. C


Changing Frequencies

44. (a) 0 0.5ω = gives tx curve with lowest fre-quency (fewest humps); 0 2ω = gives the

highest frequency (most humps).

(b) 0 0.5ω = gives the innermost phase-plane trajectory; as 0ω increases, the

amplitude of x increases. In Figure 4.1.8

the trajectory that is not totally visible isthe one for 0 2ω = .

3

–4

4x

t

2

–2

2ππ

ω0 05= . ω0 2=

ω0 1=

Detective Work

45. (a) The curve 81.4cos5

y t π⎛ ⎞= −⎜ ⎟⎝ ⎠

is a sinusoidal curve with period 2π , amplitude 1.4A ≈ ,

and phase angle 85πδ ≈ .

(b) From this graph we estimate 0 1ω = , 2.3A ≈ , and 4πδ ≈ . Thus, we have

( ) ( )

( )

0 0cos 2.3cos 2.3 cos cos sin sin4 4 4

2 22.3 cos sin 1.6 cos sin .2 2

x t A t t t t

t t t t

π π πω δ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − = − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎧ ⎫⎪ ⎪= + ≈ +⎨ ⎬⎪ ⎪⎩ ⎭

Pulling a Weight

46. (a) The mass is 2 kgm = . Because a force of 8 nt stretches the spring 0.5 meters, we find

that 8 16 nt m0.5

k = = . If we then release the weight, the IVP describing the motion of

the weight is 2 16 0x x+ = or

8 0x x+ = , ( )0 0.5x = , ( )0 0x = .

The solution of the differential equation is

( ) ( )cos 8x t A t δ= − .

Using the initial conditions, we get the simple oscillation

( ) ( )0.5cos 8x t t= .


(b) 1Amplitude2

= m; 0

2 2 sec8

T π πω

= = , 82

fπ

= cycles per second

(c) Setting ( )cos 8 0t = , we find that the weight will pass through equilibrium at 14

of the

period or after

0.562 8

t π= ≈ seconds.

At that time velocity is

( )0.56 2 sin 1.414m sec2

x π⎛ ⎞= − ≈ −⎜ ⎟⎝ ⎠

moving away from original displacement.

Finding the Differential Equation

47. (a) The mass is 500m = gm, which means the force acting on the spring is 500 980× dynes.

This stretches the spring 50 cm, so the spring constant is

500 980 9800 dynes cm50

k ×= = .

The mass is then pulled down 10 cm from its initial displacement, giving ( )0 10x = (as

long as we measure downward to be the positive direction, which is typical in these problems). The initial velocity of the mass is assumed to be zero, so ( )0 0x = . Thus, the

IVP for the mass is

500 9800 0x x+ =

or

5 98 0x x+ = , ( )0 10x = , ( )0 0x = .

(b) The solution of the differential equation found in part (a) is

( ) 98 98cos 10cos5 5

x t A t tδ⎛ ⎞

= − =⎜ ⎟⎜ ⎟⎝ ⎠

.

(c) In part (b) the amplitude is 10 cm, phase angle is 0, the period is

52 2 1.498

mTk

π π= = ≈ sec,

and the natural frequency is given by the reciprocal 1 0.711.4

f = = oscillations per sec-

ond.


Initial-Value Problems

48. (a) The weight is 16 lbs, so the mass is roughly 16 132 2

= slugs. (See Table 4.1.1 in text.) This

mass stretches the spring 12

foot, hence 12

16 32 lb ftk = = . This yields the equation

( )1 32 02

x x+ = , or

64 0x x+ = .

The initial conditions are that the mass is pulled down 4 inches ( 13

foot) from equilib-

rium and then given an upward velocity of 4 ft sec . This gives the initial conditions of

( ) 103

x = ft, ( )0 4x = − ft/sec, using the engineering convention that for x, down is

positive.

(b) We have the same equation 64 0x x+ = , but the initial conditions are ( ) 106

x = − ft,

( )0 1x = ft/sec.

One More Weight

49. The mass is 12 332 8

m = = slugs. The spring is stretched 12

foot, so the spring constant is

12

12 24 lb ftk = = . The initial position of the mass is 4 inches ( 13

ft) upward so ( ) 103

x = − . The

initial motion is 2 ft sec upward, and thus ( )0 2x = − . Hence, the equation for the motion of the

mass is

64 0x x+ = , ( ) 103

x = − , ( )0 2x = − ,

which has the solution

( ) 1 1cos8 sin83 4

x t t t= − − .


Writing this in polar form, we have

( )

2 22 21 2

1 12

1

1 1 53 4 12

3tan tan4

3.78 radians angle in 3rd quadrant .

A c c

cc

δ − −

⎛ ⎞ ⎛ ⎞= + = − + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

≈

Hence, we have the solution in polar form

( ) ( )5 cos 8 3.7812

x t t= − .

See figure.

1.4

Š0.4

0.4

x

1.210.80.60.40.2t

x t t( ) = −( )512

8 378cos .

Spring oscillation

Comparing Harmonic Motions

50. The period of simple harmonic motion is given by 0

2T πω

= , where 0km

ω = . Notice that this

does not depend at all on our initial conditions. Period is the same so is the frequency, but the amplitude will be twice that in the first case.

Testing Your Intuition

51. 3 0x x x+ + =

Here we have a vibrating spring with no friction, but a nonlinear restoring force 3F x x= − − that

is stronger than a purely linear force –x. For small displacement x the nonlinear F will not be much different (for small x, 3x is very small), but for larger x, the force F will be much stronger

than in a linear spring; as F increases, the frequency of the vibration increases. This equation is called Duffing’s (strong) equation, and the associated springs are called strong springs.

52. 3 0x x x+ − =

Here we have a vibrating spring with no friction, and a nonlinear restoring force 3F x x= − + . For small displacement x the nonlinear term 3x has little effect, but as x increases toward 1, the

restoring force F diminishes (i.e., the spring weakens when it is stretched a lot, and the restoring force becomes zero when 1x = ). The decreasing F causes decreasing frequency (and increasing

period). This equation is called Duffing’s (weak) equation, and the associated springs are called weak springs.


53. 0x x− =

This equation describes a spring with no friction and a negative restoring force. You may wonder if there are such physical systems. In the next two sections we will see that this equation describes the motion of an inverted pendulum (4.3 Problems 58, 59), and it has solutions sinh t and cosh t (4.2 Example 2), in contrast to 0x x+ = , which has solutions sin t and cos t. The restoring force

for the equation under discussion is always directed away from the equilibrium position; hence the solution always moves away from the equilibrium, which us unstable.

54. 1 0x x xt

+ + =

This equation can be interpreted as describing the motion of a vibrating mass that has infinite

friction 1 xt

at 0t = , but friction immediately begins to diminish and approaches zero as t

becomes very large. You may simulate in your mind the motion of such a system. Do you think for large t that the oscillation might behave much like simple harmonic motion? (See 4.3 Problem 68.)

55. ( )2 1 0x x x x+ − + =

This is called van der Pol’s equation and describes oscillations (mostly electrical) where internal friction depends on the value of the dependent variable x. Note that when 1x < , we actually have

negative friction, so for a small displacement x we would expect the system to move away from the zero solution (an unstable equilibrium) in the direction of 1x = . But when 1x > , we will

have positive friction causing damping. We will see in 4.3 Problem 70 and in Chapter 7 that there is a periodic solution between small x and large x that attracts all these other solutions.

56. 0x tx+ =

Here we have a vibrating spring with no friction, but the restoring force –tx gets stronger as time passes. Hence we expect to see no damping, but faster vibrations as t increases.

LR-Circuit

57. (a) Without having a capacitor to store energy, we do not expect the current in the circuit to oscillate. If there had been a constant voltage 0V on in the past, we would expect the

current to be (by Ohm’s law) 0VIR

= . If we then shut off the voltage, we would expect

the current to die off in the presence of a resistance.

(b) If a current I passes through a resistor with resistance R, then the voltage drop is RI; the voltage drop across an inductor of inductance L is LI . We obtain the IVP:

0LI RI+ = , ( ) 00 VIR

=


(c) The solution of the IVP is

( ) ( )0 R L tVI t eR

−= .

(d) If 40R = ohms, 5L = henries, 0 10V = volts, then ( ) 814

tI t e−= ohms.

LC-Circuit

58. (a) With a nonzero initial current and no resistance, we do not expect the current to damp to zero. We would expect an oscillatory current due to the capacitor. Thus the charge on the capacitor would oscillate indefinitely. The exact behavior depends on the initial conditions and the values of the inductance and capacitance.

(b) Kirchoff’s voltage law states that the sum of the voltage drops around the circuit is equal to the impressed voltage source. Hence, we have

1 0LI IC

+ =∫

or, in terms of the charge across the capacitor, we have the IVP

1 0LQ QC

+ = , ( )0 0Q = , ( )0 5Q = .

(c) The solution of the IVP is

( )( )1

1

sin5

LC

LC

tQ t = .

This agrees with the oscillatory behavior predicted in part (a).

(d) With values 10L = henries, 310C −= farads, the charge on the capacitor is

( )( )sin 100 15 sin10

2100

tQ t t= = .

A Pendulum Experiment

59. The pendulum equation is

sin 0gL

θ θ+ = .

For small θ, we can approximate sinθ θ≈ , giving the differential equation

0gL

θ θ+ = .


This is the equation of simple harmonic motion with circular frequency 0gL

ω = , and natural

frequency 01

2gfLπ

= . Hence, the period of motion is 0

1 2 LTf g

π= = .

earth sun

sun earth

400,000 100 40 632T gT g

= = = ≈ .

Changing into Systems

60. 4 2 3 17 cos

1 ( 3 2 17 cos )4

x x x tx y

y x y t

− + = −=

= − + + −

61. 1 ( )

1 1 ( )

Lq Rq q V tc

q I

I q RI V tL c

+ + =

=

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

62. 15 15 5cos310

1 3 cos350

q q q t

q I

I q I t

+ + =

=

= − − +

63. 2

2

2

4 sin 2 04 1 sin 2

4 sin 2

t x tx x t t ttx x x

t ttx y

x ty yt tt

+ + = >

+ + =

=

= − − +

64. 4 16 4sin4 sin

4 sin

x x tx x t

x yy x t

+ =+ =

== − +

Circular Motion

65. Writing the motion in terms of polar coordinates r and θ and using the fact that the angular velocity is constant, we have 0θ ω= (a constant). We also know the particle moves along a circle

of constant radius, which makes r a constant. We then have the relation cosx r θ= , and hence


( )( ) ( ) 2

sin

sin cos .

x r

x r r

θ θ

θ θ θ θ

= −

= − −

Because 0θ = , 0θ ω= , we arrive at the differential equation

20 0x xω+ = .

Another Harmonic Motion

66. For simple harmonic motion the circular frequency 0ω is

2

0 2kR

mR Iω =

+,

so the natural frequency 0f is

2

0 21

2kRf

mR Iπ=

+.

Motion of a Buoy

67. The buoy moves in simple harmonic motion, so the period is

0

22.7 2 mTk

π πω

= = = .

We have one equation in two unknowns, but the buoyancy equation yields the second equation. If we push the buoy down 1 foot, the force upwards will be F V ρ= , where V is the submerged volume and ρ is the density of water. In this case, 2V r hπ= , 9 inches 0.75 ftr = = , 1h = ft, and

62.5 ft secρ = , so the force required to push the buoy down 1 foot is ( )( )9 1 62.5 11016

π ≈ lbs.

But k is the force divided by distance, so 110 110 lbs ft1

k = = . Finally, solving for m in the

equation for T, we get 2

24kTmπ

= , and substituting in all of our numbers, we arrive at 20.4m ≈

slugs (see Table 4.1.1. in the text.) The buoy weighs ( )( )20.4 32.2 657mg = = lbs.

Los Angeles to Tokyo

68. (a) Along the tunnel,

cosmx kr kxθ= − = −

x(0) = d if x is measured positive to the left of the center of the tunnel.

(0)x = 0 means that the train starts from rest (as soon as a brake is released).


(b) The solution to the IVP in part (a) is

x(t) = c1 cos ω0t + c2 sin ω0 t,

where ω0 = km

.

At the surface of the earth mg = kr

where r = R, so ω0 = k qm R

= .

Letting x(0) = d yields c1 = d, while letting (0) 0x = yields c2 = 0.

Hence we have

x(t) = cos .qd tR

For the train to go from L.A. to Tokyo, x(t) goes from d to −d.

and q tR

goes from 0 to π.

Hence,

24000 mi 5280 ft/mi

32 ft/sec2552 sec 42.5 minutes

fRtq

π

π

=

×=

= ≈

(c) The solution tf = Rq

π from part (b) does not depend on the location of the points on the

earth’s surface; π, R, and q are all constant.

Factoring Out Friction

69. (a) Letting ( ) ( ) ( )2b m tx t e X t−= , we have

( ) ( ) ( ) ( ) ( )

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

2 2

22 2 2

2

2

.4

b m t b m t

b m t b m t b m t

bx t e X t e X tmb bx t e X t e X t e X t

mm

− −

− − −

−= +

−= + +


Substituting this into the original equation (1) and dividing through by ( )2b m te− , we arrive

at

[ ]2

2 024

b b bm X X X b X X k Xm mm

⎡ ⎤ ⎡ ⎤− + + − + + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦.

Rearranging terms gives

[ ]2 2

04 2b bmX b b X k Xm m

⎡ ⎤+ − + + − + =⎢ ⎥

⎣ ⎦

or

2

04bmX k Xm

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

.

(b) If we assume 2

04bkm

− > , then divide by m and let

20

1 42

mk bm

ω = − )

we find the solution of this DE in X is

( ) ( )1 0 2 0 0cos sin cosX t c t c t A tω ω ω δ= + = − .

Thus, we have

( ) ( ) ( ) ( ) ( )2 20cosb m t b m tx t e X t Ae tω δ− −= = − .

Suggested Journal Entry

70. Student Project

SECTION 4.2 Real Characteristic Roots 317

4.2 Real Characteristic Roots

Real Characteristic Roots

1. 0y′′ =

The characteristic equation is 2 0r = , so there is double root at 0r = . Thus, the general solution

is

( ) 0 01 2 1 2

t ty t c e c te c c t= + = + .

2. 0y y′′ ′− =

The characteristic equation is 2 0r r− = , which has roots 0, 1. Thus, the general solution is

( ) 1 2ty t c c e= + .

3. 9 0y y′′ − =

The characteristic equation is 2 9 0r − = , which has roots 3, –3. Thus, the general solution is

( ) 3 31 2

t ty t c e c e−= + .

4. 0y y′′ − =

The characteristic equation is 2 1 0r − = , which has roots 1, –1. Thus, the general solution is

( ) 1 2t ty t c e c e−= + .

5. 3 2 0y y y′′ ′− + =

The characteristic equation is 2 3 2 0r r− + = , which factors into ( )( )2 1 0r r− − = , and hence has

roots 1, 2. Thus, the general solution is

( ) 21 2

t ty t c e c e= + .

6. 2 0y y y′′ ′− − =

The characteristic equation is 2 2 0r r− − = , which factors into ( )( )2 1 0r r− + = , and hence has

roots 2, –1. Thus, the general solution is

( ) 21 2


7. 2 0y y y′′ ′+ + =

The characteristic equation is 2 2 1 0r r+ + = , which factors into ( )( )1 1 0r r+ + = , and hence has

the double root –1, –1. Thus, the general solution is

( ) 1 2t ty t c e c te− −= + .


8. 4 4 0y y y′′ ′− + =

The characteristic equation is 24 4 1 0r r− + = , which factors into ( )( )2 1 2 1 0r r− − = , and hence

has the double root 12

, 12

. Thus, the general solution is

( ) 2 21 2

t ty t c e c te= + .

9. 2 3 0y y y′′ ′− + =

The characteristic equation is 22 3 1 0r r− + = , which factors into ( )( )2 1 1 0r r− − = , and hence

has roots 12

, 1. Thus, the general solution is

( ) 21 2

t ty t c e c e= + .

10. 6 9 0y y y′′ ′− + =


the double root 3, 3. Thus, the general solution is

( ) 3 31 2


11. 8 16 0y y y′′ ′− + =

The characteristic equation is 2 8 16 0r r− + = , which factors into ( )( )4 4 0r r− − = , and hence

has the double root 4, 4. Thus, the general solution is

( ) 4 41 2


12. 6 0y y y′′ ′− − =

The characteristic equation is 2 6 0r r− − = , which factors into ( )( )2 3 0r r+ − = , and hence has

roots –2, 3. Thus, the general solution is

( ) 2 31 2


13. 2 0y y y′′ ′+ − =

The characteristic equation is 2 2 1 0r r+ − = , which factors into ( )( )1 2 1 2 0r r+ − + + = , and

hence has roots 1 2− + , 1 2− − . Thus, the general solution is

( ) ( )2 21 2

t t ty t e c e c e− −= + .

14. 9 6 0y y y′′ ′+ + =

The characteristic equation is 29 6 1 0r r+ + = , which factors into ( )23 1 0r + = , and hence has the

double root 13

− , 13

− . Thus, the general solution is

( ) 3 31 2

t ty t c e c te− −= + .


Initial Values Specified

15. 25 0y y′′ − = , ( )0 1y = , ( )0 0y′ =

The characteristic equation of the differential equation is 2 25 0r − = , which factors into ( )( )5 5 0r r− + = , and thus has roots 5, –5. Hence,

( ) 5 51 2


Substituting in the initial conditions ( )0 1y = gives 1 2 1c c+ = . Substituting in ( )0 0y = gives

1 25 5 0c c− = . Solving for 1c , 2c gives 1 212

c c= = . Thus the general solution is

( ) 5 51 12 2

t ty t e e−= + .

16. 2 0y y y′′ ′+ − = , ( )0 1y = , ( )0 0y′ =

The characteristic equation of the differential equation is 2 2 0r r+ − = , which factors into ( )( )2 1 0r r+ − = , and thus has roots 1, –2. Thus, the general solution is

( ) 21 2


Substituting into ( )0 1y = , ( )0 0y′ = yields 113

c = , 223

c = , so

( ) 21 23 3

t ty t e e−= + .

17. 2 0y y y′′ ′+ + = , ( )0 0y = , ( )0 1y′ =

The characteristic equation is 2 2 1 0r r+ + = , which factors into ( )( )1 1 0r r+ + = , and hence has

the double root –1, –1. Thus, the general solution is

( ) 1 2t ty t c e c te− −= + .

Substituting into ( )0 0y = , ( )0 1y′ = yields 1 0c = , 2 1c = , so

( ) ty t te−= .

18. 9 0y y′′ − = , ( )0 1y = − , ( )0 0y′ =

The characteristic equation is 2 9 0r − = , which factors into ( )( )3 3 0r r− + = , and hence has

roots are 3, –3. Thus, the general solution is

( ) 3 31 2


Substituting into ( )0 1y = − , ( )0 0y′ = yields 1 212

c c= = − , so

( ) 3 31 12 2

t ty t e e−= − − .


19. 6 9 0y y y′′ ′− + = , ( )0 0y = , ( )0 1y′ = −


the double root 3, 3. Thus, the general solution is

( ) 3 31 2


Substituting into ( )0 0y = , ( )0 1y′ = − yields 1 0c = , 2 1c = − , so

( ) 3ty t te= − .

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

The characteristic equation is 2 6 0r r+ − = , which factors into ( )( )3 2 0r r+ − = , and hence has

roots –3, 2. Thus, the general solution is

( ) 3 21 2


Substituting into ( )0 1y = , ( )0 1y′ = yields 115

c = , 245

c = , so

( ) 3 21 45 5

t ty t e e−= + .

21. 0y y′′ ′− = y(0) = 2, y(0) = −1

r2 − r = 0 (Characteristic equation)

r(r − 1) = 0 r = 0, 1

y = c1 + c2et ⇒ 2 = c1 + c2

y′ = c2et ⇒ −1 = c2, c1 = 3

y = 3 − et

22. 4 12 0y y y′′ ′− − = y(0) = 1, (0)y′ = −1

r2 − 4r − 12 = 0 (Characteristic equation)

(r + 2)(r − 6) = 0 r = −2, 6

2 6

1 22 6

1 22 6

t t

t t

y c e c e

y c e c e

−

−

= +

′ = − + 1 2

1 2

(0) 1 1(0) 1 2 6 1

y c cy c c

= ⇒ + = ⎫⎬′ = − ⇒ − + = − ⎭

⇒ 1 27 1,8 8

c c= =

y = 2 61 34 4

t te e− −+


Bases and Solution Spaces

23. 4 0y y′′ ′− =

r2 − 4r = 0 (Characteristic equation)

r(r − 4) = 0 ⇒ r = 0, 4

Basis: {1, e4t}

Solution Space: {y ⎜y = c1 + c2e4t; c1, c2 ∈ }

24. 10 25 0y y y′′ ′− + =

r2 − 10r + 25 = 0 (Characteristic equation)

(r −5)2 = 0 ⇒ r = 5, 5

Basis: {e5t, te5t}

Solution Space: {y⎜y = c1e5t + c2te5t; c1, c2 ∈ }

25. 5 10 15 0y y y′′ ′− − =

5r2 − 10r − 15 = 0 (Characteristic equation)

r2 − 2r − 3 = 0 ⇒ (r − 3)(r + 1) = 0 ⇒ r = 3, −1

Basis: {e3t, e−t}

Solution Space: {y ⎜y = c1e3t + c2e−t; c1, c2 ∈ }

26. 2 2 2 0y y′′ ′+ + =

2 2 2 2 0r r+ + = (Characteristic equation)

( 2)( 2) 0r r+ + = r = 2, 2− −

Basis: { }2 2,t te te− −

Solution Space: { }2 21 2 1 2; ,t ty y c e c te c c− −= + ∈

Other Bases

27. 4 0y y′′ − =

r2 − 4 = 0 (Characteristic equation)

r = ±2 ∴ {e2t, e−2t} is a basis

To show {cosh 2t, sinh 2t} is a basis, we need only show that cosh 2t and sinh 2t are linearly

independent solutions:

W = cosh 2 sinh 22sinh 2 2cosh 2

t tt t

= 4 cosh2 2t − 4 sinh2 2t


cosh2 2t = 22 2 4 41

2 2

t t t te e e e− −⎛ ⎞ ⎛ ⎞+ + +=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

sinh2 2t = 22 2 4 41

2 2

t t t te e e e− −⎛ ⎞ ⎛ ⎞− − −=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

so cosh2 2t − sinh2 2t = 1 and W = 4 ≠ 0.

∴ cosh 2t, sinh 2t are linearly independent.

Substitute y = cosh 2t, y′ = 2 sinh 2t, y′′ = 4 cosh 2t

Then y′′ − 4y = 4 cosh 2t − 4 cosh 2t = 0 ∴ y = cosh 2t is a solution.

In similar fashion, we can show that y = sinh 2t is also a solution.

To show that { }2 ,cosh 2te t is a basis, we use the facts that e2t and cosh 2t are solutions.

Then:

W = 2

2

cosh 2

2 2sinh 2

t

t

e t

e t =

2 22

2 2 22

2

t tt

t t t

e ee

e e e

−

−

+

−

= (e4t − 1) − (e4t + 1) = −2 ≠ 0

∴ e2t and cosh 2t are linearly independent

28. 0y′′ =

r2 = 0 (Characteristic equation) so that r = 0, 0. Basis: {1, t}

To show {t + 1, t − 1} is also a basis:

Note that for both y = t + 1 and y = t − 1, y′′ = 0, so both are solutions.

W = 1 1

1 1t t+ −

= (t + 1) − (t − 1) = 2

∴ t + 1, t − 1 are linearly independent

To show {2t, 3t − 1} is another basis:

Note that for both y = 2t and y = 3t − 1, y′′ = 0, so both are solutions.

W = 2 3 12 3t t −

- 6t − 2(3t − 1) = 2

∴ 2t, 3t − 1 are linearly independent


The Wronskian Test

29. W =

2 32 2 3

2

1 11 2 3 1

1 1 2 1 3 ( 1) 0 2 6 1 0 2 60 0 2 6 0 0 6 0 0 60 0 0 6

t t t t tt t t t t t t

t t t t tt

+ − ++ − +

+ = + −

= (t + 1)12 − 12(t − 1) = 24 ≠ 0

Yes, {t + 1, t − 1, t2 + t, t3} is a basis for the solution space for y(4) = 0.

30. W =

5 5 5 5

5 5 5 5

5 5 5

2 1 1 2(5 1) 5 10 5 1 5 10

25 10 25 50(25 10) 25 50

t t t t

t t t t

t t t

te e e t et e e e e t

tt e e e

−− −+ = +

++

= 5 55 10 5 1 10 5 1 51 (2 )

25 50 25 10 50 25 10 25t tt t

e t et t

−⎡ ⎤+ +− + −⎢ ⎥+ +⎣ ⎦

= 25e5t ≠ 0

Yes, {te5t, e5t, 2e5t − 1} is a basis for the solution space for 10 25y y y′′′ ′′ ′− + = 0.

31. The given set has only three solutions, so it cannot be a basis. A basis for the solution space of y(4)

= 0 must have 4 linearly independent solutions.

Sorting Graphs

32.


Relating Graphs

For Problems 33−35, 5 6 0x x x+ + = has (from Example 1) solutions

2 31 2

2 31 2

( ) (1)

( ) 2 3 (2)

t t

t t

x t c e c e

x t c e c e

− −

− −

= +

= − −

33. (a), (b)

x(0) ≈ − 10 (0)x ≈ 0 ⇓ ⇓ c1 + c2 = −10 −2c1 − 3c2 = 0

c1 = −30, c2 = 20

(c) From (1) in box, x(t) = −30e−2t + 20e−3t.

For t > 0, each term diminishes as t increases; the result remains negative, below the

t-axis.

For t < 0, each exponential increases as t decreases; the negative term cancels the positive

term when 30e−2t = 20e−3t or e−t = 1.5,

that is, when t = −ln 1.5 ≈ −.405 which looks about right on the tx-graph.

(d) From (2), ( )x t = 60e−2t − 60e−3t = 60e−2t(1 − e−t) which is always positive for t > 0,

decreasing as t increases.

( )x t reaches a maximum when 2 3( ) 120 180 0t tx t e e− −= − + =

2 3 0

2 ,3

t

t

e

e

−

−

− + =

=

so 2ln 0.406,3

t = − ≈ which looks about right on the tx -graph.


34. (a)

(b)

(b) x(0) ≈ 5 (0)x ≈ 0 ⇓ ⇓

c1 + c2 = 5 −2c1 − 3c2 = 0

Because all problems for finding ci are of type =Ac b , we solve for 1−=c A b

We have A = 1 12 3

⎡ ⎤⎢ ⎥− −⎣ ⎦

so A−1 = 3 1

,2 1− −⎡ ⎤

− ⎢ ⎥⎣ ⎦

and here 50⎡ ⎤

= ⎢ ⎥⎣ ⎦

b

so

3 1 5 152 1 0 10− − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦c

(c) From (1) in box on previous page, x(t)= 15e−2t − 10e−3t.

As t increases from zero, both exponentials decrease with their sum remaining positive, which

agrees with the tx graph.

(d) From (2), ( )x t = −30e−2t − 30e−3t = −30e−2t (1 + e−t)

For t > 0, this quantity is always negative, and as long as t increases, each term gets

closer to zero, in agreement with tx -graph.


35.

(a) x(0) ≈ 0 and (0)x ≈ −8

(b) From the method of 34(b),

1 1

2

0 3 1 0 88 2 1 8 8

cc

− − − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

A

so from (1), x(t) = −8e−2t + 8e−3t.

(c) For t > 0, 2 3t te e− −> so the sum is always negative and approaches zero as t increases,

in agreement with the tx graph.

(d) From (2) ( )x t = +16e−2t − 24e−3t = 0, so 2e−2t − 3e−3t = 0,

which yields e−t = 23

or t ≈ .406,

which looks about right on the tx -graph.

For t > 0.406, ( ) 0x t > and decreases toward zero as t increases.


For Problems 36−39, 6 0x x x− − = has (from Example 1) solutions

2 31 2

2 31 2

( ) (1)

( ) 2 3 (2)

t t

t t

x t c e c e

x t c e c e

−

−

= +

= − +

36. (a)

(b) From (1) c1 + c2 = 0

From (2) −2c1 + 3c2 = 2 c1 = 2

2 2; 5 5

c− =

x(t) = 2 32 25 5

t te e−− +

2 34 6( )5 5

t tx t e e−= +

(c) For t > 0, e−2t < e3t, so x(t) is always positive, and as t increases, so does x(t).

This result agrees with the tx-graph.

For t < 0, e3t < e−2t so x(t) is always negative, and as t becomes more negative, x(t) becomes more negative.

(d) ( )x t is always positive.

For t > 0, e−2t < e3t, so the second term dominates as t increases, and ( )x t increases as well. These facts are in agreement with the tx -graph.

37. (a)


(b) From (1), c1 + c2 = 2

From (2), −2c1 + 3c2 = 0 c1 = 2

6 4; 5 5

c =

x(t) = 2 36 45 5

t te e− +

2 312 12( )5 5

t tx t e e−= − +

(c) x(t) is always positive.

For t > 0, as t increases, the first term decreases toward 0 and the second term increases ever more rapidly, in agreement with the tx-graph.

(d) For t > 0, e3t > e−2t, so ( )x t is positive, and ( )x t increases as t increases, as shown on the

tx graph.

For t < 0, the first term will dominate and ( )x t will be negative, ever more so as t

becomes more negative, in agreement with the tx -graph.

38. (a)

(b) From (1), c1 + c2 = −3

From (2), −2c1 + 3c2 = 0 c1 = 2

9 6; 5 5

c− = −

x(t) = 2 39 65 5

t te e−− −

2 318 18( )5 5

t tx t e e−= + −

(c) x(t) is always negative, with a maximum at t = 0. (See part (d) and set ( )x t = 0.)

These facts agree with the tx graph.

(d) For t > 0, e3t > e−2t so the negative term dominates in ( )x t and ( )x t is negative, ever

more so as t increases.

For t < 0, e−2t > e3t so the positive term dominates in ( )x t and ( )x t is positive, ever more

so as t becomes more negative. These facts agree with the tx graph.


39. (a)

(b) From (1), c1 + c2 = 0

From (2), −2c1 + 3c2 = −1

c1 = 21 1; 5 5

c = −

x(t) = 2 31 15 5

t te e− −

2 32 3( )5 5

t tx t e e−= − −

(c) For t > 0 the second term dominates, so x(t) is negative, ever more so as t increases.

For t < 0 the first term dominates, so x(t) is positive, ever more so as t becomes more negative.

These facts agree with the tx graph.

(d) ( )x t is always negative. The maximum value will occur when t = 0, as shown on the tx -

graph.

Phase Portraits

Careful inspection shows:

40. (B) 41. (D) 42. (A) 43. (C)


Independent Solutions

44. Letting

1 21 2 0r t r tc e c e+ =

for all t, then by setting 0t = and 1t = we have, respectively

1 2

1 2

1 2

0

0.r r

c c

c e c e

+ =

+ =

When 1 2r r≠ then these equations have the unique solution 1 2 0c c= = , which shows the given functions 1 2, r t r te e are linearly independent for 1 2r r≠ .

Second Solution

45. Substituting ( ) 2bt ay v t e−= into

0ay by cy′′ ′+ + =

gives

2 2

22 2 2

2

2

.4

bt a bt a

bt a bt a bt a

by v e vea

b by v e v e vea a

− −

− − −

′ ′= −

′′ ′′ ′= − +

Substituting , ,v v v′ ′′ into the differential equation gives the new equation (after dividing by 2bt ae− )

2

2 024

b b ba v v v b v v cva aa

⎛ ⎞ ⎛ ⎞′′ ′ ′− + + − + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

.

Simplifying gives

2

04bav c va

⎛ ⎞′′ − − =⎜ ⎟

⎝ ⎠.

Because we have assumed 2 4b ac= , we have the equation 0v′′ = , which was the condition to be

proven.

Independence Again

46. Setting

2 21 2 0bt a bt ac e c te− −+ =

for all t, we set in particular 0t = and then 1t = . These yield, respectively, the equations

12 2

1 2

0

0b a b a

c

c e c e− −

=

+ =

which have the unique solution 1 2 0c c= = . Hence, the given functions are linearly independent.


Repeated Roots, Long-Term Behavior

47. Because 2bt ae− approaches 0 as t →∞ (for a, 0b > ), we know the first term tends toward zero.

For the second term we need only verify that

22 0bt a

bt atte

e− = →

does as well. To use l’Hôpital’s rule, we compute the derivatives of both the numerator and de-nominator of the previous expression, getting

22

1bt ab

a e,

which clearly approaches 0 as t →∞ . Then l’Hôpital’s rule assures us that the given expression ( )2b a tte− approaches 0 as well.

Negative Roots

48. We have 2 4r b b mk= − ± − , so in the overdamped case where 2 4 0b mk− > , these

characteristic roots are real. Because m and k are both nonnegative, 2 24b mk b− < causing

21 4r b b mk= − + − to be a negative sum of negative and positive terms

and

22 4r b b mk= − − − to be a negative sum of two negative terms.

Circuits and Springs

49. (a) The LRC equation is 1 0LQ RQ QC

+ + = , hence the following discriminant conditions

hold:

( )

( )

( )

2

2

2

4 0 underdamped

4 0 critically damped

4 0 overdamped .

LRCLRCLRC

Δ = − <

Δ = − =

Δ = − >


(b) The conditions in part (a) can be written

( )

( )

( )

2 underdamped

2 critically damped

2 overdamped .

LRCLRCLRC

<

=

>

These correspond to the analogy that m, b, and k correspond respectively to L, R, and 1C

.

(see Table 4.1.3 in the textbook.)

A Test of Your Intuition

50. Intuitively, a curve whose rate of increase is proportional to its height will increase very rapidly as the height increases. On the other hand, upward curvature doesn’t necessarily imply that the function is increasing! (The curve te− has upward curvature, yet decreases to 0 as t →∞ .) In this case, the restriction that ( )0 0y′ = will cause the second curve to increase, but probably not nearly as rapidly as the first curve. Solving the equations, the IVP y y′ = , ( )0 1y = has the

solution ty e= , whereas the second curve described by y y′′ = , ( )0 1y = , ( )0 0y′ = has the

solution

( ) 1 12 2

t ty t e e−= + .

The first curve is indeed above the second curve.

An Overdamped Spring

51. (a) The solution of an overdamped equation has the form

( ) 1 21 2

r t r tx t c e c e= + .

Suppose that

1 1 2 11 2 0r t r tc e c e+ =

for some 1t . Because 2 1r te is never zero, we can divide by 2 1r te to get ( )1 2 11 2 0r r tc e c− + = .

Solving for 1t gives

21

1 2 1

1 ln ctr r c

−=

−.

This unique number is the only value for which the curve may pass through 0. If the argument of the logarithm is negative or if the value of 1t is negative, then the solution

does not cross the equilibrium point.

(b) By a similar argument, we can show that the derivative ( )x t also has one zero.


A Critically Damped Spring

52. (a) Suppose

( ) 1 11 2 1 0r tc c t e+ = .

We can divide by the nonnegative quantity 1 1r te getting the equation 1 2 1 0c c t+ = , which

has the unique solution 11

2

ctc

= − . Hence, the solution of a critically damped equation can

pass through the equilibrium at most once. If the value of 1t is negative, then the solution

does not cross the equilibrium point.

(b) By a similar argument, we can show that the derivative ( )x t has one zero.

Linking Graphs

After inspection, we have labeled the yt and y t′ graphs as follows.

53.

–5

y

3t

–5

5

y'

3t

–5

5

y'

5y

–5

12

3

4

1

2

3

44

1 32

t = 0

54.


55.

Damped Vibration

56. The IVP problem is

2 0x x x+ + = , ( ) 10 3 in ft4

x = = , ( )0 0 ft secx = .

The solution is

( ) 1 14 4

t tx t e te− −= + .

This is zero only for 1 1t = − , whereas the physical system does not start before 0t = .

Surge Functions

57. For 0mx bx kx+ + = , let m = 1, find b, k and initial conditions for the solution x = Ate−rt

2

0

0 (characteristic equation)

x bx kx

r br k

+ + =

+ + =

r = 2 4 12

b b k− ± − ⋅ ⋅

b2 − 4k = 0 to obtain repeated roots, r = ,2 2b b

− −

x = c1e−rt + c2te−rt

1 2 ( ( ) )rt rt rtx rc e c t r e e− − −= − + − + ∴ c1 = 0 = x(0) c2 = A = (0)x

(0)x = c2

∴ b = −2r, 4k = b2

and from above we know 4k = 4r2 so that k = ±r, for k > 0


Results: r and A are given, and

b = −2r

k = r

x(0) = 0

(0)x′ = A

LRC-Circuit I

58. (a) 1 2 101 50 0LQ RQ Q Q Q QC

+ + = + + = , ( )0 99Q = , ( )0 0Q =

(b) ( ) 50 2100t tQ t e e− −= − + (c) ( ) ( ) 50 250 50t tI t Q t e e− −= = −

(d) As t →∞ , ( ) 0Q t → and ( ) 0I t →

LRC-Circuit II

59. (a) 1 15 50 0LQ RQ Q Q Q QC

+ + = + + = , ( )0 5Q = , ( )0 0Q =

(b) ( ) 5 1010 5t tQ t e e− −= − (c) ( ) ( ) 5 1050 50t tI t Q t e e− −= = − +

(d) As t →∞ , ( ) 0Q t → and ( ) 0I t →

The Euler-Cauchy Equation 0′′ ′+ + =2at y bty cy

60. Let ( ) ry t t= , so

( )

1

21 .

r

r

y rt

y r r t

−

−

′ =

′′ = −

Hence

( )2 1 0r r rat y bty cy ar r t brt ct′′ ′+ + = − + + = .

Dividing by rt yields the characteristic equation

( )1 0ar r br c− + + = ,

which can be written as

( )2 0ar b a r c+ − + = .

If 1r and 2r are two distinct roots of this equation, we have solutions

( )( )

1

2

1

2 .

r

r

y t t

y t t

=

=


Because these two functions are clearly linearly independent (one not a constant multiple of the other) for 1 2r r≠ , we have

( ) 1 21 2

r ry t c t c t= +

for 0t > .

The Euler-Cauchy Equation with Distinct Roots

For Problems 61–65, see Problem 60 for the form of the characteristic equation for the Euler-Cauchy DE.

61. 2 2 12 0t y ty y′′ ′+ − =

In this case 1a = , 2b = , 12c = − , so the characteristic equation is

( ) ( )( )21 2 12 12 4 3 0r r r r r r r− + − = + − = + − = .

Hence, we have roots 1 4r = − , 2 3r = , and thus

( ) 3 41 2y t c t c t−= + .

62. 24 8 3 0t y ty y′′ ′+ − =


( ) ( )( )24 1 8 3 4 4 3 2 1 2 3 0r r r r r r r− + − = + − = − + = .

Hence, we have roots 112

r = , 232

r = − , and thus

( ) 1 2 3 21 2y t c t c t−= + .

63. 2 4 2 0t y ty y′′ ′+ + =

In this case 1a = , 4b = , 2c = , so the characteristic equation is

( ) ( )( )21 4 2 3 2 1 2 0r r r r r r r− + + = + + = + + = .

Hence, we have roots 1 1r = − , 2 2r = − , and thus

( ) 1 21 2y t c t c t− −= + .


64. 22 3 0t y ty y′′ ′+ − =


( ) ( )( )22 1 3 1 2 1 2 1 1 0r r r r r r r− + − = + − = − + = .

Hence, we have roots 112

r = , 2 1r = − , and thus

( ) 1 2 11 2y t c t c t−= + .

Repeated Euler-Cauchy Roots

65. We are given that the characteristic equation

( )2 0ar b a r c+ − + =

of Euler’s equation

2 0at y bty cy′′ ′+ + =

has a double root of r. Hence, we have one solution 1ry t= . To verify that lnrt t is also a

solution, we differentiate 1 1lnr ry rt t t− −′ = + ,

( ) ( ) ( ) ( )2 2 2 2 21 ln 1 1 ln 2 1r r r r ry r r t t rt r t r r t t r t− − − − −′′ = − + + − = − + − .

By direct substitution we have

( ) ( )

( ) ( )

2 2 2 2 1 11 ln 2 1 ln ln

1 ln 2 1 .

r r r r r

r r

at y bty cy at r r t t r t bt rt t t ct t

ar r br c t t a r b t

− − − −⎡ ⎤ ⎡ ⎤′′ ′+ + = − + − + + +⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤= − + + + − +⎣ ⎦ ⎣ ⎦

We know that ( )1 0ar r br c− + + = , so this last expression becomes simply

( )2 2 1 rat y bty cy a r b t′′ ′ ⎡ ⎤+ + = − +⎣ ⎦ .

Thus the root of the characteristic equation is 2

b ara−

= − , which makes this expression zero.

To verify that rt and lnrt t are linearly independent (where 2

b ara−

= − is the double

root of the characteristic equation), we set

1 2 ln 0r rc t c t t+ =

for specific values 1t = and 2, which give, respectively, the equations

1

1 2

0

2 2 ln 2 0r r

c

c c

=

+ =

and yields the unique solution 1 2 0c c= = . Hence, rt and lnrt t are linearly independent

solutions.


Solutions for Repeated Euler-Cauchy Roots

For Problems 66 and 67 use the result of Problem 60, ( ) 1 2 lnr ry t c t c t t= + .

66. 2 5 4 0t y ty y′′ ′+ + =

In this case, 1a = , 5b = , and 4c = , so our characteristic equation for r is 2 4 4 0r r+ + = , with a

double root at –2. The general solution is

( ) 2 21 2 lny t c t c t t− −= +

for 0t > .

67. 2 3 4 0t y ty y′′ ′− + =

In this case, 1a = , 3b = − , and 4c = , so our characteristic equation for r is 2 4 4 0r r− + = , with

a double root at 2. The general solution is

( ) 2 21 2 lny t c t c t t= +

for 0t > .

68. 9t2y″ + 3ty′ + y = 0 Euler-Cauchy method: y = t m, t > 0 9m(m − 1) + 3m + 1 = 0 (characteristic equation) 9m2 − 6m + 1 = 0

(3m − 1)2 = 0 m = 13

1/3 1/31 2( ) lny t c t c t t= +

69. 24 8 0t y ty y′′ ′+ + = Euler-Cauchy method: y = t m, t > 0

4m(m − 1) + 8m + 1 = 0 4m2 + 4m + 1 = 0

(2m + 1)2 = 0 m = 12

−

1/ 2 1/ 21 2( ) lny t c t c t t− −= +

Computer: Phase-Plane Trajectories

70. (a) ( ) 32 t ty t e e− −= +

The roots of the characteristic equation are –1 and –3, so the characteristic equation is

( )( ) 21 3 4 3 0r r r r+ + = + + = .

( )y t satisfies the differential equation

4 3 0y y y′′ ′+ + = .


(b) To find the IC for the trajectory of ( )y tin yy′ space we differentiate ( )y t ,

getting

( ) 32 3t ty t e e− −′ = − −

The IC of the given trajectory of

( ) ( )( ), y t y t′

in yy′ space is ( ) ( )( ) ( )0 , 0 3, 5y y′ = − .

(c) We plot the trajectory starting at( )3, 5− along with a few other

trajectories in yy′ space.

DE trajectories in yy′ space

71. ( ) 8t ty t e e− −= +

(a) The roots of the characteristic equation are –1 and –8, so the characteristic equation is

( )( ) 21 8 9 8 0r r r r+ + = + + = .


9 8 0y y y′′ ′+ + = .

(b) The derivative is

( ) 88t ty t e e− −′ = − − .

The IC for the given trajectory in yy′

space is

( ) ( )( ) ( )0 , 0 2, 9y y′ = − .

(c) We plot this trajectory in yy′ space.

DE trajectory in yy′ space

72. ( ) t ty t e e−= +

(a) The roots of the characteristic equation are 1 and –1, so the characteristic equation is

( )( ) 21 1 1 0r r r− + = − = .


0y y′′ − = .



( ) t ty t e e−′ = − .


space is

( ) ( )( ) ( )0 , 0 2, 0y y′ = .

(c) We plot this and a few other trajectoriesof this DE in yy′ space.

y

–4

y

4–4 2–2

2

–2

4

(2, 0)


73. ( ) t ty t e te− −= +

(a) The characteristic equation has a double root at –1, so the characteristic equation is

( )2 21 2 1 0r r r+ = + + = .


2 0y y y′′ ′+ + = .


( ) ty t te−′ = − .


space is

( ) ( )0 1, 0 0y y′= = .

(c) See the figure to the right.

–1.5 1 1.5

–1.5

1.5

y(1, 0)

y'

DE trajectory in yy′ space

74. ( ) 23 2 ty t e= +

(a) The roots of the characteristic equation are 0 and 2, so the characteristic equation is

( ) 22 2 0r r r r− = − = .


2 0y y′′ ′− = .



( ) 24 ty t e′ = .


space is

( ) ( )( ) ( )0 , 0 5, 4y y′ = .



Reduction of Order

75. (a) Let 2 1y vy= and

2 1 1

2 1 1 12 .y v y vyy v y v y vy′ ′ ′= +′′ ′′ ′ ′ ′′= + +

Then ( ) ( ) ( )2 2 2 1 1 1 1 1 12 0y p x y q x y v y v y pv y vy pvy qvy′′ ′ ′′ ′ ′ ′ ′′ ′+ + = + + + + + = .

Because 1 1 1 0y py qy′′ ′+ + = , cancel the terms involving v, and arrive at the new equation

( )( )1 1 12 0y v y p x y v′′ ′ ′+ + =

(b) Setting v w′ = and using the fact that 1 1y dx dy′ ′= , we obtain

( )( )( )

( )

( )

( )

( )( )

( )

1 1 1

1 1

1

1 1

1

1

11

21

21

21

2 0

20

2

2ln

2ln

ln ln

p x dx

p x dx

y w y p x y w

y p x yw w

y

y p x ydw dxw y

yw p x dxy

w dy p x dxy

w y p x dx

ew vy

evy

−

−

−

′ ′+ + =

′⎛ ⎞+′ + =⎜ ⎟⎜ ⎟

⎝ ⎠′⎛ ⎞− −

= ⎜ ⎟⎜ ⎟⎝ ⎠

′⎛ ⎞−= −⎜ ⎟

⎝ ⎠−

= −

= −

′= ± =

= ±

∫

∫

∫

∫ ∫

∫

∫

By convention, the positive sign is chosen.


(c) If v is a constant function on I, then 0v′ ≡ and 0w ≡ because v w′ = . The condition 0w ≡ contradicts our work in part (b) as ln w where 0w = is undefined. Because v is

not constant on I, { }1 2, y y is a linearly independent set of I.

Reduction of Order: Second Solution

76. 6 9 0y y y′′ ′− + = , 31

ty e=

We identify ( ) 6p t = − , so

( ) 6p t dt t= −∫ .

Substituting in the formula developed in Problem 75, we have

( )

( ) ( )6

3 32 1 2 231

p t dt tt t

t

e ey y dt e dt tey t e

−

= = =∫

∫ ∫ .

77. 4 4 0y y y′′ ′− + = , 21

ty e=

We won’t use the formula this time. We simply redo the steps in Problem 75. We seek a second solution of the form 2

2 1ty vy ve= = . Differentiating, we have

2 2

22 2 2

2

2

4 4 .

t t

t t t

y v e ve

y v e v e ve

′ ′= +

′′ ′′ ′= + +

Substituting into the equation we obtain

22 2 24 4 0ty y y v e′′ ′ ′′− + = = .

Dividing by 2te gives 0v′′ = or

( ) 1 2v t c t c= + .

Hence, we have found new solutions

2 2 22 1 2

t t ty ve c te c e= = + .

Because 21

ty e= , we let 1 1c = , 2 0c = , yielding a second independent solution

22

ty te= .


78. 2 0t y ty y′′ ′− + = , 1y t=

We won’t use the formula this time. We simply redo the steps in Problem 75. We seek a second solution of the form 2 1y vy tv= = . Differentiating, we have

2

2 2 .y tv vy tv v′ ′= +′′ ′′ ′= +

Substituting into the equation we obtain

2 3 22 2 2 0t y ty y t v t v′′ ′ ′′ ′− + = + = .

Letting w v′= and dividing by 3t yields

1 0w wt

′ + = .

We can solve by integrating the factor method, getting 11w c t−= . Integrating we find

1 2lnv c t c= + ,

so

2 1 2lny tv c t t c t= = + .

Letting 1 1c = , 2 0c = , we get a second linearly independent solution

2 lny t t= .

79. ( )2 1 2 2 0t y ty y′′ ′+ − + = , 1y t=

We won’t use the formula this time. We simply redo the steps in Problem 75. We seek a second solution of the form 2 1y vy tv= = . Differentiating yields

2

2 2 .y tv vy tv v′ ′= +′′ ′′ ′= +

Substituting into the equation we get

( ) ( )2 22 2 21 2 2 1 2 0t y ty y t t v v′′ ′ ′′ ′+ − + = + + = .

Letting w v′= and dividing by ( )2 1t t + , we can solve the new equation using the integrating

factor method, getting

( ) ( )

22

22

2 ln 1 2ln ln11

tdt t ttt t

= − + + =++∫ .


We arrive at

2

21 1 12

1tw c c c tt

−+= = + .

Integrating this, we get

( )11 2v c t t c−= − + ,

so

( )22 1 21y tv c t c t= = − + .

Letting 1 1c = , 2 0c = we get a second linearly independent solution

22 1y t= − .

Classical Equations

80. 2 4 0y ty y′′ ′− + = , ( ) 21 1 2y t t= − (Hermite’s Equation)

Letting ( )22 1 1 2y vy v t= = − , we have

( )( )

22

22

1 2 4

1 2 8 4

y t v tv

y t v tv v

′ ′= − −

′′ ′′ ′= − − −

and perform the long division, yielding the equation

282 0

2 1tv t v

t⎛ ⎞′′ ′+ − + =⎜ ⎟−⎝ ⎠

.

Letting w v′= and solving the first-order equation in w, we get

( )2 221 2 1tw c e t

−= − .

To find 2y we simply let 1 1c = and integrate to get

( )2 222 1tv e t dt−

= −∫ .

Multiplying by ( )21 2t− yields a final answer of

( ) ( ) ( )2 22 22 1 2 2 1ty t t e t dt

−= − −∫ .

81. ( )21 0t y ty y′′ ′− − + = , ( )1y t t= (Chebyshev’s Equation)

Letting 2 1y vy vt= = , we have

2y tv v′ ′= + , 2 2y tv v′′ ′′ ′= + ,


hence we have the equation

( ) ( ) ( )2 2 22 2 21 1 2 3 0t y ty y t t v t v′′ ′ ′′ ′− − + = − + − = .

Dividing by ( )21t t− , and letting w v′= ,

( )

2

2

2 3 01

tw wt t−′ + =−

.

Using partial fractions yields

( )

2

2

2 3 1 12ln ln 1 ln 12 21

t dt t t tt t−

= + − + +−∫ ,

so our integrating factor is 2 21t t− and

1 2 2

1

1w c

t t=

−.

Letting 1 1c = and multiplying by t yields a final answer of

( )2 2 2

1

1y t tv t dt

t t= =

−∫ .

This is a perfect example of a formula that does not tell us much about how the solutions behave. Check out the IDE tool Chebyshev’s Equation to see the value of graphical solutions.

82. ( )1 0ty t y y′′ ′+ − + = , ( )1 1y t t= − (Laguerre’s Equation)

Letting ( )2 1 1y vy v t= = − , we have

( )2 1y v t v′ ′= − + , ( )2 1 2y v t v′′ ′′ ′= − + ,

hence we have the equation

( ) ( ) ( )22 21 1 4 1 0ty t y y t t v t t v′′ ′ ′′ ′+ − + = − + − + − = .

Dividing by ( )1t t − and letting w v′= yields

( )

2 4 1 01

t tw wt t

− + −′ + =−

.

Hence by use of partial fractions, our integrating factor is 1 21

1dt

t tu e− + +

−= ∫ so that

1 2( 1)

tew Ct t

=−

.

Letting 1 1c = and multiplying by 1t − yields a final answer of

( ) ( ) ( )( )2 21 1

1

tey t v t t dtt t

= − = −−∫ .


Lagrange’s Adjoint Equation

83. (a)−(b) Differentiating the right side of

[ ] [ ]( ) ( ) ( )dt y y y t y g t ydt

μ μ′′ ′ ′+ + = +

we obtain

y y y y y g y gyμ μ μ μ μ′′ ′ ′ ′ ′ ′+ + = + + +

Setting the coefficients of y, ,y y′ ′′ equal, we find

for y : μ = μ (no information)

for :y′ μ = gμ′ +

for :y′′ gμ ′=

The last equation yields g = dtμ∫ and substituting this into the second equation, and

differentiating, gives a differential equation for the “integrating factor”

μ μ μ′′ ′− + = 0.

(c) We perform the differentiation on the right-hand-side of the given equation, yielding

[ ]( ) ( ) ( ) ( ) ( ) .t y p t y q t y y y g t y g t yμ μ μ′′ ′ ′′ ′ ′ ′ ′+ + = + + +

Multiplying out the left-hand side and subtracting yields

[ ] [ ]( ) ( ) ( ) ( ) 0.p t g t y q t g t yμ μ μ′ ′ ′− − + − =

Setting the first set of coefficients equal to 0 yields p gμ μ′ = − ,

hence ,p p gμ μ μ′′ ′ ′ ′= + − so that .g p pμ μ μ′ ′′ ′ ′= − + +

The second set of coefficients yields q gμ ′− = 0 so that .g qμ′ =

Setting these two equations for g′ equal to each other yields ( ) 0p q pμ μ μ′′ ′ ′− + − =

which was to be shown.


84. Student Project

SECTION 4.3 Complex Characteristic Roots 347

4.3 Complex Characteristic Roots

Solutions in General

1. 9 0y y′′ + =

The characteristic equation is 2 9 0r + = , which has roots 3i, –3i. The general solution is

( ) 1 2cos3 sin 3y t c t c t= + .

2. 0y y y′′ ′+ + =

The characteristic equation is 2 1 0r r+ + = , which has roots 1 32 2

i− ± . The general solution is

( ) 21 2

3 3cos sin2 2

ty t e c t c t− ⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠.

3. 4 5 0y y y′′ ′− + =

The characteristic equation is 2 4 5 0r r− + = , which has roots 2 i± . The general solution is

( ) ( )21 2cos sinty t e c t c t= + .

4. 2 8 0y y y′′ ′+ + =

The characteristic equation is 2 2 8 0r r+ + = , which has roots 1 7i− ± . The general solution is

( ) ( )1 2cos 7 sin 7ty t e c t c t−= + .

5. 2 4 0y y y′′ ′+ + =

The characteristic equation is 2 2 4 0r r+ + = , which has roots 1 3i− ± . The general solution is

( ) ( )1 2cos 3 sin 3ty t e c t c t−= + .

6. 4 7 0y y y′′ ′− + =

The characteristic equation is 2 4 7 0r r− + = , which has roots 2 3i± . The general solution is

( ) ( )21 2cos 3 sin 3ty t e c t c t= + .

7. 10 26 0y y y′′ ′− + =

The characteristic equation is 2 10 26 0r r− + = , which has roots 5 i+ . The general solution is

( ) ( )51 2cos sinty t e c t c t= + .


8. 3 4 9 0y y y′′ ′+ + =

The characteristic equation is 23 4 9 0r r+ + = , which has roots 2 233 3

i− ± . The general solution

is

( ) 2 31 2

23 23cos sin3 3

ty t e c t c t− ⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠.

9. 0y y y′′ ′− + =

The characteristic equation is 2 1 0r r− + = , which has roots 1 32 2

i± . The general solution is

( ) 21 2

3 3cos sin2 2

ty t e c t c t⎛ ⎞

= +⎜ ⎟⎜ ⎟⎝ ⎠

.

10. 2 0y y y′′ ′+ + =

The characteristic equation is 2 2 0r r+ + = , which has roots 1 72 2

i− ± . The general solution is

( ) 21 2

7 7cos sin2 2

ty t e c t c t− ⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠.


11. 4 0y y′′ + = , ( )0 1y = , ( )0 1y′ = −

The characteristic equation is 2 4 0r + = , which has roots 2i± . The general solution is

( ) 1 2cos2 sin 2y t c t c t= + .

Substituting this into the initial conditions gives ( ) 10 1y c= = , ( ) 20 2 1y c′ = = − . Hence, the

solution of the initial-value problem is

( ) 1cos2 sin 22

y t t t= − .

12. 4 13 0y y y′′ ′− + = , ( )0 1y = , ( )0 0y′ =

The characteristic equation is 2 4 13 0r r− + = , which has roots 2 3i± . The general solution is

( ) ( )21 2cos3 sin3ty t e c t c t= + .

Substituting this into the initial conditions yields ( ) 10 1y c= = , ( ) 1 20 2 3 0y c c′ = + = , resulting in

1 1c = , 223

c = − . Hence, the solution of the initial-value problem is

( ) 2 2cos3 sin 33

ty t e t t⎛ ⎞= −⎜ ⎟⎝ ⎠

.


13. 2 2 0y y y′′ ′+ + = , ( )0 1y = , ( )0 0y′ =

The characteristic equation is 2 2 2 0r r+ + = , which has roots 1 i− ± . Hence, the general solution

is

( ) ( )1 2cos sinty t e c t c t−= + .

Substituting this into the initial conditions yields ( ) 10 1y c= = , ( ) 1 20 0y c c′ = − = , resulting in

1 1c = , 2 1c = . Hence, the solution of the initial-value problem is

( ) ( )cos sinty t e t t−= + .

14. 0y y y′′ ′− + = , ( )0 0y = , ( )0 1y′ =

From Problem 6,

( ) ( ) ( )21 2

3 3cos sin2 2

ty t e c t c t⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪= +⎢ ⎥ ⎢ ⎥⎨ ⎬

⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭.

Substituting this into the initial conditions yields ( )0 0y = , ( )0 1y′ = , resulting in 1 0c = ,

22 33

c = . Hence, the solution of the initial-value problem is

( ) 22 33 sin3 2

ty t e t− ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠.

15. 4 7 0y y y′′ ′− + = , ( )0 0y = , ( )0 1y′ = −

From Problem 6,

( ) ( ) ( ){ }21 2cos 3 sin 3ty t e c t c t= + .

Subsituting this into the initial conditions yields ( )0 0y = , ( )0 1y′ = − , resulting in

1 0c = , 21 33

c = − .

Hence, the solution of the initial-value problem is

( ) ( )21 3 sin 33

ty t e t= − .

16. 2 5 0y y y′′ ′+ + = , ( )0 1y = , ( )0 1y′ = −

The characteristic equation is 2 2 5 0r r+ + = , which has roots 1 2i− ± . Hence, the general solu-

tion is

( ) ( )1 2cos2 sin 2ty t e c t c t−= + .


Subsituting this into the initial conditions yields ( )0 1y = , ( )0 1y′ = − , resulting in 1 1c = , 2 0c = .

Hence, the solution of the initial-value problem is

( ) cos2ty t e t−= .

Working Backwards

17. 3 3 2( 1) 3 3 1r r r r− = − + −

3 3 0y y y y′′′ ′′ ′− + − =

18. 3 2( 4)( (1 ))( (1 )) 6 10 8 0r r i r i r r r− − − − + = − + − =

6 10 8 0y y y y′′′ ′′ ′− + − =

19. 3 2( 2)( (2 ))( (2 )) 6 13 10r r i r i r r r− − + − − = − + −

6 13 10 0y y y y′′′ ′′ ′− + − =

20. 2 4 3 2( 4)( (2 ))( (2 )) 4 16 20 0r r i r i r r r r− − + − − = − + + − =

(4) 6 16 20 0y y y y y′′′ ′′ ′− + + − =

Matching Problems

21. 0y y′′ ′− = ⇒ r = 0, 1 y(t) = c1 + c2et Graph D

22. 0y y′′ ′+ = ⇒ r = 0, −1 y(t) = c1 + c2e−t Graph B

23. 3 2 0y y y′′ ′+ + = ⇒ r = −2, −1

y(t) = c1e−2t + c2e−t Graph A

24. 5 6 0y y y′′ ′− + = ⇒ r = 2, 3

y(t) = c1e2t + c2e3t Graph C

25. 0y y y′′ ′+ + = ⇒ r = 1 32

i− ±

y(t) = 12

1 23 3cos sin2 2

t t te c c⎛ ⎞−⎜ ⎟⎝ ⎠

⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠ Graph G

26. 0y y′′ ′+ = ⇒ r = ±i

y(t) = c1 cos t + c2 sin t Graph F

27. 4 4 0 2, 2y y y r′′ ′+ + = ⇒ = − −

21 2( ) ( ) ty t c c t e−= + Graph E


28. 1 302

iy y y r ±′′ ′− + = ⇒ =

12

1 23 3( ) cos sin

2 2

ty t e c t c t

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠ Graph H

Euler’s Formula

29. (a) The Maclaurin series for xe is

2 31 1 112! 3! !

x ne x x x xn

= + + + + + +

(b) ( ) ( ) ( )2 31 1 112! 3! !

nie i i i in

θ θ θ θ θ= + + + + + +

(c) Using the given identities for i, we can write

( ) ( ) ( )2 3

2 4 3 5

1 1 112! 3! !

1 1 1 11 cos sin2! 4! 3! 5!

nie i i i in

i i

θ θ θ θ θ

θ θ θ θ θ θ θ

= + + + + + +

⎛ ⎞ ⎛ ⎞= − + − + + − + − = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(d) Done in part (c). (e) Done in part (c).

Long-Term Behavior of Solutions

30. 1 0r < , 2 0r < . When 1 2r r≠ , the solution is

( ) 1 21 2

r t r ty t c e c e= +

and goes to 0 as t → ∞ . When 1 2 0r r r= = < , the solution has the form

( ) 1 2rt rty t c e c te= + .

In this case using l’Hôpital’s rule we prove the second term rtte goes to zero as t → ∞ when 0r < .

31. 1 0r < , 2 0r = . The solution

( ) 11 2

r ty t c e c= +

approaches the constant 2c as t → ∞ because 1 0r < .

32. r iα β= ± , ( ) ( )1 2cos sinty t e c t c tα β β= + . For 0β ≠ the solution ( )y t oscillates with

decreasing amplitude when 0α < ; oscillates with increasing amplitude when 0α > ; oscillates with constant amplitude when 0α = .


33. 1 0r = , 2 0r = . The solution

( ) 1 2y t c c t= +

approaches ∞ as t → ∞ when 2 0c > and –∞ when 2 0c < .

34. 1 0r > , 2 0r < . The solution

( ) 1 21 2

r t r ty t c e c e= +

approaches ∞ as t → ∞ when 1 0c > and –∞ when 1 0c < .

35. r iβ= ± , ( ) 1 2cos siny t c t c tβ β= + is a periodic function of period 2πβ

, and amplitude

2 21 2c c+

Linear Independence

36. Suppose

1 2cos sin 0t tc e t c e tα αβ β+ =

on an arbitrary interval. Dividing both sides by teα , then differentiating the new equation and

dividing by β, yields

1 2

2 1

cos sin 0cos sin 0.

c t c tc t c t

β ββ β

+ =− =

Hence, 1 20, 0c c= = and we have proven linear independence of the given functions.

Real Coefficients

37. Solution of the differential equation is

( ) ( ) ( )

( ) ( )1 2

1 2 1 2

cos sin cos sin

cos sin .

t t

t t

y t k e t i t k e t i t

e k k t ie k k t

α α

α α

β β β β

β β

= + + −

= + + −

For the solution to be real, there must exist real numbers r and s such that

1 2

1 2

k k rk k si

+ =− =

Solving for 1k and 2k , we get

1

2

1 12 21 1 .2 2

k r si

k r si

= +

= −


Solving =n

nd y 0dt

38. (a)

4

4

3

33

2

3 22

23 2 1

3 23 2 1 0

0

121 13! 2

d ydtd y kdtd y k t kdtdy k t k t kdt

y k t k t k t k

=

=

= +

= + +

= + + +

(b) ( )4 0y = . The characteristic equation is 4 0r = , which has a fourth-order root at

0. Hence, the solution is

( ) 2 30 1 2 3y t c c t c t c t= + + + ,

which is the same as in part (a).

(c) In general we have

( ) ( ) ( )

1 21 2 1 0

1 21 2 1 0

1 11 ! 2 !

n nn n

n nn n

y t k t k t k t kn n

c t c t c t c

− −− −

− −− −

= + + + +− −

= + + + +

because all of the constants are arbitrary.

Higher-Order DEs

39. 5 4 3

5 4 34 4 0d y d y d y

dt dt dt− + =

The characteristic equation is

( ) ( )25 4 3 3 2 34 4 4 4 2 0r r r r r r r r− + = − + = − = ,

which has roots, 0, 0, 0, 2, 2. Hence,

( ) 2 2 21 2 3 4 5

t ty t c c t c t c e c te= + + + + .

40. 3 2

3 24 7 10 0d y d y dy y

dtdt dt+ − − =


3 24 7 10 0r r r+ − − = ,

which has roots, –1, 2, –5. Hence,

( ) 2 51 2 3

t t ty t c e c e c e− −= + + .


41. 5

5 0d y dydtdt

− =


( ) ( )( ) ( )( )( )5 4 2 2 21 1 1 1 1 1 0r r r r r r r r r r r− = − = − + = − + + = ,

which has roots, 0, ±1, ±i. Hence

( ) 1 2 3 4 5cos sint ty t c c e c e c t c t−= + + + + .

42. 4 5 2 0y y y y′′′ ′′ ′− + − =

3 24 5 2 0r r r− + − = (characteristic equation) (1) 1 4 5 2 0f = − + − =

1 is a rootr∴ = By long division, we obtain

2( 1)( 3 2) 0( 1)( 2)( 1) 0 1, 1, 2r r rr r r r

− − + =− − − = =

21 2 3( ) t t ty t c e c te c e= + +

43. 6 12 8 0y y y y′′′ ′′ ′+ + + =

3 26 12 8 0r r r+ + + = (characteristic equation) ( 2) 8 24 24 8 0f − = − + − + =

2 is a rootr∴ = −

By long division, we obtain 2

3

( 2)( 4 4) 0

( 2) 0 2, 2, 2

r r r

r r

+ + + =

+ = = − − −

2 2 2 21 2 3( ) t t ty t c e c te c t e− − −= + +

44. (4) 0y y− =

4 1 0r − = (characteristic equation) 2 2( 1)( 1) 0r r+ − = , 1r i= ± ± 1 2 3 4( ) cos sin t ty t c t c t c e c e−= + + +

Linking Graphs

45.

–5

5y

3t

–5

5y'

1t

–5

5y'

5y

–5

3

2

1

3

1

23

21

t = 0


46.

–5

5y

3t

–5

5y'

1t

–5

5y'

5y

–5

3

2

13

2

1

3

2

1

t = 0

Changing the Damping

47. The curves below show the solution of

0x bx x+ + = , ( )0 4x = , ( )0 0x =

for damping 0b = , 0.5, 1, 2, 4. The larger the damping the faster the curves approach 0. The curve that oscillates has no damping ( )0b = .

–4

4

161284t

x t( )

2

–2

b = 4

b = 2b = 1

b = 0.5b = 0

–2

2

x

4

–4

–2–4 2 4

b = 4b = 2

b = 1

b = 0.5

b = 0

x.

In Figure 4.3.12 (b) in the text the larger the damping b the more directly the trajectory “heads” for the origin. The trajectory that forms a circle corresponds to zero damping. Note that every time a curve in (a) crosses the axis twice the corresponding trajectory in (b) circles the origin.


Changing the Spring

48. (a) The solutions of

0x x kx+ + = , ( )0 4x = , ( )0 0x =

are shown for 14

k = , 12

, 1, 2, 4. For

larger k we have more oscillations.

–4

4

161284t

x t( )

2

–2 k = 4

k = 1k = 0.5

k = 0.25

k = 2

(b) For larger k, since there are more oscilla-tions, the phase-plane trajectory spiralsfurther around the origin.

–2

2

x

4

–4

–2–4 2

k = 4 k = 1

k = 0.5

k = 0.25k = 2

x.

Changing the Mass

49. (a) b = 0 and ωo = km

so that ωo is inversely proportional to m .

(b) If m is doubled, ωo is decreased by a factor of 12

.

(c) If m is doubled, the damping required for critical damping is increased by a factor of 2.

Finding the Maximum

50. (a) 2 3 0x x x+ + = , x(0) = 1, (0) 0x =

r2 + 2r + 3 = 0 (characteristic equation)

r = 1 2i− ±

( )( ) ( )

1 2

1 2 1 2

cos 2 sin 2

2 sin 2 2 cos 2 cos 2 sin 2

t

t t

x e c t c t

x e c t c t e c t c t

−

− −

⎫= + ⎪ ⇒⎬= − + − + ⎪⎭

1 = c1

0 = 22 1c − so that c2 = 12

x = 1cos 2 sin 22

te t t− ⎛ ⎞+⎜ ⎟

⎝ ⎠


To find maximum displacement, set 0x = .

1 12 sin 2 2 cos 2 cos 2 sin 2 02 2

t tx e t t e t t− −⎛ ⎞⎛ ⎞ ⎛ ⎞= − + − + =⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

22 sin 2 02

t⎛ ⎞

− + =⎜ ⎟⎜ ⎟⎝ ⎠

when 2t π= , so that 2

t π= sec

Substituting for t: xmax = 2 1cos 2 sin 22 2 2

eπ

π π− ⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= 2e

π−

−

Max. Amplitude xmax = 2eπ

−

(b) m = 1, b = 2, k = 10 x(0) = 0, (0) 2x =

The DE is 2 10 0x x x′′ ′+ + = for which the characteristic equation

r2 + 2r + 10 = 0 gives r = −1 ± 3i.

1 2

1 2 1 2

( ) ( cos3 sin 3 )

( ) ( 3 sin3 3 cos3 ) ( cos3 sin 3 )

t

t t

x t e c t c t

x t e c t c t e c t c t

−

− −

= +

′ = − + − +

2( ) sin33

tx t e t−= is the solution

To find the maximum displacement xmax , set ′ x (t)=0 and solve for t:

0 =23(3e−t cos3t − e− t sin3t)

so that tan3t = 3 and t = 0.416 radians which gives xmax = 0.4172 .

(c) m = 1, b = 4, k = 4 x(0) = 0, (0) 2x =

The DE is 4 4 0x x′′ ′+ + = for which the characteristic equation

r2 + 4r + 4 = 0 gives r = −2, −2 2 2

1 22 2 2

1 2

( )

( ) 2 ( 2 )

t t

t t t

x t c e c te

x t c e c te e

− −

− − −

⎫= + ⎪⎬

′ = − + − + ⎪⎭ ⇒ 1 20, 2c c= =

The solution is 2( ) 2 tx t te−= .

To find the maximum displacement xmax, we set ( ) 0x t′ = and solve for t:

2 20 2( 2 )t tte e− −= − +

so that t = 1/2 which yields xmax = e−1.


Oscillating Euler-Cauchy

51. We used the substitution ry t= and obtained for 1r iα β= + and 2r iα β= − the solution

( ) ( ) ( )

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

ln ln ln ln ln ln1 2 1 2 1 2

ln1 2 1 2cos ln sin ln cos ln sin ln .

i t i ti i t i t t i t

t

y t k t k t k e k e k e k e

e c t c t t c t c t

α β α βα β α β α β α β

α αβ β β β

+ −+ − + −= + = + = +

= + = +

This is the same process as that used at the start of Case 3 in the text utilizing the Euler’s For-mula (4).

52. 2 2 0t y ty y′′ ′+ + = , ( )1 2 1 0r r r− + + = , 2 1 0r r+ + = , 1 32 2

r i= − ± ,

( ) 1 21 2

3 3cos ln sin ln2 2

y t t c t c t−⎡ ⎤⎛ ⎞ ⎛ ⎞

= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

53. 2 3 5 0t y ty y′′ ′+ + =

Letting ry t= yields

( )

( ){ }

2 2 11 3 5 0

1 3 5 0,

r r r

r

t r r t trt t

t r r r

− −− + + =

− + + =

and gives 2 2 5 0r r+ + = , which has roots 1 2i− ± . Hence, the solution is

( ) ( ) ( )( )11 2cos 2ln sin 2lny t t c t c t−= + .

54. 2 17 16 0t y ty y′′ ′+ + = Euler-Cauchy: y = tm, t > 0

m(m − 1) + 17m + 16 = 0 (characteristic equation)

m2 + 16m + 16 = 0

m = 216 (16) 4(16)

2− ± −

= −8 ± 4 3

( )81 2( ) cos(4 3 ln ) sin(4 3 ln )y t t c t c t−= +


Third-Order Euler-Cauchy

55. The third-order Euler-Cauchy equation has the form 3 2 0at y bt y cty dy′′′ ′′ ′+ + + = . The derivatives of ry t= ( )0t > are

( )( )( )

1

2

3

1

1 2

r

r

r

y rt

y r r t

y r r r t

−

−

−

′ =

′′ = −

′′′ = − −

Substitute these equations into the third-order Euler-Cauchy equation above to obtain:

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

3 3 2 2 11 2 1 0

1 2 1 0

r r r r

r r r r

at r r r t bt r r t ctrt dt

at r r r bt r r ct r dt

− − −− − + − + + =

− − + − + + =

Dividing by rt , we obtain the characteristic equation:

( )( ) ( )1 2 1 0ar r r br r cr d− − + − + + =

Third-Order Euler-Cauchy Problems

56. 3 2 2 2 0t y t y ty y′′′ ′′ ′+ − + = has Euler-Cauchy characteristic equation:

( )( ) ( )3 2 2

3 2

1 2 1 2 2 0

3 2 2 2 0

2 2 0

r r r r r r

r r r r r r

r r r

− − + − − + =

− + + − − + =

− − + =

Note: 1r = is a zero of the polynomial ( ) 3 22 2f r r r r= − − + because

( )1 1 2 1 2 0f = − − + = .

Therefore 1r − is a factor of 3 22 2r r r− − + , which enables us to find the other factors.

( )( )( )3 22 2 1 1 2r r r r r r− − + = − + −

so 1r = , –1, 2. Hence, the general solution to this Euler-Cauchy DE is

( ) 1 21 2 3y t c t c t c t−= + + ,

for 0t > .

57. 3 23 5 0t y t y ty′′′ ′′+ + = Let y = tm, t > 0

3 2 2

3

( 1)( 2) 3 ( 1) 5 0 (characertistic equation)

3 2 3 3 5 0

4 0 0, 2

m m m m m m

m m m m m

m m m i

− − + − + =

− + + − + =

+ = = ±

y(t) = c1 + c2 cos (2 ln t) + c3 sin (2 ln t)


Inverted Pendulum

58. The differential equation 0x x− = has the characteristic equation 2 1 0r − = with roots 1± .

Hence, the general solution is

( ) 1 2t tx t c e c e−= + .

(a) With initial conditions ( )0 0x = , ( )0 1x = , we find 112

c = and 212

c = − . Hence, the solu-

tion of the IVP is

( ) 1 1 sinh2 2

t tx t e e t−= − = .

(b) As t → ∞ , ( ) 0x t → if 1 0c = , and then ( ) 0x t → also. This will happen whenever

( ) ( )0 0x x= − .

Pendulum and Inverted Pendulum

59. (a) The inverted pendulum equation has characteristic equation 2 1 0r − = , which has roots

1± . Hence, the solution

( ) ( ) ( )1 2 1 2 1 2cosh sinh cosh sinh sinh cosht tx t c e c e c t t c t t C t C t−= + = + + − = + ,

where 1 1 2C c c= − , 2 1 2C c c= + .

(b) The characteristic equation of the pendulum equation is 2 1 0r + = , which has roots i± .

Hence, the solution

( ) 1 2cos sinx t c t c t= + .

(c) The reader may think something strange about this because one form (a) appears real and (b) complex, but they are really the same; the difference is taken up by how one chooses the coefficients 1 2, c c in each case. The span of { }, it ite e− is the same as the span of

{ }sin , cost t .


Finding the Damped Oscillation

60. The initial conditions

( )0 1x = , ( )0 1x =

give the constants 1 1c = , 2 2c = . Hence, we have

( ) ( )cos 2sintx t e t t−= + .

5

–0.5

1.5x

t

1

0.5

31

x t e t tt( ) = +( )− cos sin2

Extremes of Damped Oscillations

61. The local maxima and minima of the curve

( ) ( )1 2cos sintx t e c t c tα ω ω= +

have nothing to do with the exponential factor teα ; they depend only on

1 2cos sinc t c tω ω+ ,

which can be rewritten as ( )cosA tω δ− having period 2T πω

= . Hence, consecutive maxima and

minima occur at equidistant values of t, the distance between them being one-half the period, or πω

. (You can note in Problem 32 that the time between the first local maxima and the first local

minima is 1π π= .)

Underdamped Mass-Spring System

62. We are given parameters and initial conditions

0.25m = , 1b = , 4k = , ( )0 1x = , ( )0 0x = .

Hence, the IVP is

0.25 4 0x x x+ + = , ( )0 1x = , ( )0 0x = ,

which has the solution

( ) 2 3cos 2 3 sin 2 33

tx t e t t− ⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠.


Damped Mass-Spring System

63. The IVP is

64 0x bx x+ + = , ( )0 1x = , ( )0 0x = .

(a) 10b = : (underdamped), ( ) 5 5cos 39 sin 3939

tx t e t t− ⎛ ⎞= +⎜ ⎟

⎝ ⎠

(b) 16b = : (critically damped), ( ) ( ) 81 8 tx t t e−= +

(c) 20b = : (overdamped), ( ) ( )4 161 43

t tx t e e− −= −

LRC-Circuit I

64. (a) The IVP is 1 8 25 0LQ RQ Q Q Q QC

+ + = + + = , ( )0 1Q = , ( )0 0Q =

(b) ( ) ( )4 44 5cos3 sin 3 cos 33 3

t tQ t e t t e t δ− −⎛ ⎞= + = −⎜ ⎟⎝ ⎠

where 1 4tan3

δ −=

(c) ( ) ( ) ( ) ( )4 4205 sin 3 cos 33

t tI t Q t e t e tδ δ− −= = − − − − where 1 4tan3

δ −=

(d) Charge on the capacitor and current in the circuit approach zero as t → +∞ .

LRC-Circuit II

65. (a) The IVP is 1 1 1 4 04

LQ RQ Q Q Q QC

+ + = + + = , ( )0 1Q = , ( )0 0Q =

(b) ( ) ( )2 23 2 3cos2 3 sin 2 3 cos 2 33 3

t tQ t e t t e t δ− −⎛ ⎞= + = −⎜ ⎟⎜ ⎟

⎝ ⎠, 3tan

3δ =

(c) ( ) ( ) ( ) ( )2 24 3 cos 2 3 4 sin 2 33

t tI t Q t e t e tδ δ− −= = − − − − , 3tan3

δ =

(d) As t → ∞ , both ( ) 0Q t → and ( ) 0I t →

Computer Lab: Damped Free Vibrations

66. IDE Lab


Effects of Nonconstant Coefficients

67. 1 0x xt

+ =

(a) This ODE describes (among other things) an undamped vibrating spring in which the restoring force is initially very large (when t is near zero), but eventually decays to zero, causing the frequency of vibration to decrease and the solution period to increase as t increases.

(b) We plotted the solution with IC ( )0.1 2x = , ( )0.1 0x = in the tx and xx planes.

10 20–10

–2

–20

–4

1

3

x–1

–3

2

x

(c) As we expected, the tx graph shows that the period of the oscillation increases with t. We see also that the amplitude increases in the absence of friction. The xx phase portrait

shows that as time and amplitude increase, velocity decreases, which is consistent with the previous observations. A good question for further exploration would be whether amplitude increases indefinitely or levels off.

68. 1 0x x xt

+ + =

(a) This ODE describes a damped vibrating spring in which the damping starts very large when t is near zero, but decays to zero. We suspect that initially the amplitude of a solution will rapidly decay, but as time increases the motion could become almost like simple harmonic oscillation, as there will be almost no friction.

(b) We plotted the solution with IC ( )0.1 2x = , ( )0.1 0x = in the tx as well as the xx

planes.


1 2–1

–1

0.5

–2

x–0.5

–1.5

1

x

(c) As first expected, the tx graph shows that the solution is rapidly decaying. However the xx phase portrait, constructed with a longer time interval, shows that our second

expectation is not confirmed. As time increases the oscillations do not become harmonic—the amplitude of the oscillations continues to decrease, gradually and indefinitely.

69. 0tx x+ =

(a) If you divide by t, you will see that this equation is the same as the equation in Problem 67.

70. ( )2 1 0x x x x+ − + =

(a) This ODE shows negative friction for 1x < and positive damping for 1x > . For a small

initial condition near 0x = , we might expect the solution to grow and then oscillate around 1x = .

(b) We plotted the solutions in the tx and xx planes at initial velocity ( )0 0x = for three different initial displacements: ( )0 0.5x = , ( )0 2.0x = , ( )0 4.0x = .

–4

x

x

4

4

–4

.


(c) As expected, the tx graph shows that initially the solution is growing for ( )0 0.5x = and decaying for ( )0 4x = . We also see that all the solutions seem to become periodic with

the same amplitude and period, but we note that the motion is not exactly sinusoidal and that the amplitude is about 2 rather than 1 as we suspected. The xx phase portrait

confirms that the long term trajectories are not circular as in simple harmonic motion, but distorted as we see in the tx graph. This equation is called van der Pol’s equation and describes oscillations (mostly electrical) where internal friction depends on the value of the dependent variable x; further details will be explored in Chapter 7.

71. ( )sin 0x t x x+ + =

(a) In this ODE damping changes periodically from negative to positive, so we can predict oscillation in amplitude as well as periodic vibratory motion.

(b) We plotted the solution with IC ( ) ( )0 2, 0 0x x= = in the tx and xx planes.

(c) The tx graph looks like a superposition of two periodic oscillations. The xx phase

portrait for a longer time interval shows that continued oscillations almost repeat, but never exactly. This is called quasi-periodic motion.

72. 1 0x x txt

+ + =

(a) For this ODE damping is initially large, but vanishes as time increases; the restoring force on the other hand is initially small but increases with time. How will these effects combine?

(b) We plotted the solution with IC ( ) ( )0.1 2, 0.1 0x x= = in the tx and xx planes.


1 2–1

–1

–2

x

1

2x

(c) As we expected, the tx graph shows initially large damping, which rapidly decreases the amplitude of the solution, and increasing frequency, due to the effect of the increasing spring “constant”, which shortens the period. The center of the xx graph will continue to

fill in, very slowly, if you give it a much longer time interval.

73. ( )sin 2 0x t x+ =

(a) In this ODE the restoring force changes periodically from positive to negative with a frequency that is different from the natural frequency of the spring. We expect some complicated but periodic motion.

(b) We plotted the solution with IC ( )0 2x = , ( )0 0x = in the tx and xx planes.

4020

2

4

x

t60

–4

–2

80 100

(c) The tx graph to 100t = indeed looks almost periodic, with period 50. However the xx

phase portrait over a longer time interval shows that continued motion almost repeats, but never exactly. This is another example of quasi-periodic motion, as in Problem 71. Extending the tx graph will be another good way to see that the long term motion is indeed not perfectly repeating.


Boundary-Value Problems

74. 0y y′′ + = , y(0) = 0, 02

y π⎛ ⎞ =⎜ ⎟⎝ ⎠

y(t) = c1 cos t + c2 sin t

y(0) = 0 = c1

2

y π⎛ ⎞⎜ ⎟⎝ ⎠

= 0 = c2, so ( ) 0y t = is the solution.

75. 0y y′′ + = , y(0) = 0, 12

y π⎛ ⎞ =⎜ ⎟⎝ ⎠


y(0) = 0 = c1

2

y π⎛ ⎞⎜ ⎟⎝ ⎠

= 1 = c2, so ( ) siny t t= is the solution.

76. 0y y′′ + = , y(0) = 1, ( ) 1y π =


y(0) = 1 = c1

( ) 1y π = = −c1 *No solutions

77. 0y y′′ + = , 4

y π⎛ ⎞⎜ ⎟⎝ ⎠

= 1, 22

y π⎛ ⎞ =⎜ ⎟⎝ ⎠


1 = 1 21 12 2

c c+ c1 + c2 = 2

2 = c2 c1 = 2 2−

( )( ) 2 2 cos 2siny t t t= − + is the solution.


Exact Second-Order Differential Equations

78. 21 1 0y y yt t

′′ ′+ − = is the same as 1 0y yt

′⎡ ⎤′′ + =⎢ ⎥⎣ ⎦.

Integrating we obtain the linear equation 11 , y y ct

′ + =

for which 1

ln1 so we have .

dt tte e t ty y c tμ ′= = = + =∫

Thus, 1( ) ,d ty c tdt

= so ty = 2122

c t c+ and 1 2( ) .2c cy t t

t= +

Substituting back into the original equation we find c1 = 0, so ( ) cy tt

= is the general solution.

79. 22 2 0y yt t

′′ ′+ − =

2 0y yt

′⎡ ⎤′′ + =⎢ ⎥⎣ ⎦

Integrating and setting c1 = 0, we obtain

2

2ln 22 0 dt tty y e e t

tμ′ + = = = =∫

2 2 0 .t y ty c′ + = =

2( ) 0d t ydt

= , t2y = c, 2( ) cy tt

=

80. 2( 2 ) 4( 1) 2 0t t y t y y′′ ′− + − + = where 2 2 0t t− ≠

Find ( ) :gy ′′

( )( ) ( )

2

gy gy yggy gy yg gy y g yg g y

gy g y yg

′ ′ ′= +′′ ′ ′ ′ ′′ ′ ′ ′′ ′ ′= + = + + +

′′ ′ ′ ′′= + +

Let g = t2 − 2t. Then 2 2, 2g t g′ ′′= − =

2

1

1 2

Then ( ) ( 2 ) 4( 1) 2( ) 0( )

gy t t y t y ygygy c

gy c t c

′′ ′′ ′= − + − +′′ =′ == +

so 1 22( )

2c t cy tt t

+=

−.


81. Student Project

SECTION 4.4 Undetermined Coefficients 369

4.4 Undetermined Coefficients

Inspection First

1. ( )py y t y t t′′ − = ⇒ = − 2. ( )2 2py y y t t′′ ′+ = ⇒ =

3. ( ) 22 py y t t′′ = ⇒ = 4. ( ) 24 pty y t y t t′′ ′+ = ⇒ =

5. ( )2 2 4 2py y y y t′′ ′− + = ⇒ = 6. ( )2cos cospy y t y t t′′ − = − ⇒ =

7. ( )t tpy y y e y t e′′ ′− + = ⇒ = 8. ( )2 2 2py y y t y t t′′′ ′+ + = + ⇒ =

Educated Prediction

The homogeneous equation 2 5 0y y y′′ ′+ + = has characteristic equation 2 2 5 0r r+ + = , which has

complex roots 1 2i− ± . Hence,

( ) 1 2sin 2 cos2t thy t c e t c e t− −= + ,

so for the right-hand sides ( )f t , we try the following:

9. ( ) ( )3 3 22 3 pf t t t y t At Bt Ct D= − ⇒ = + + +

10. ( ) ( ) ( )t tpf t te y t At B e= ⇒ = + 11. ( ) ( )2sin sin cospf t t y t A t B t= ⇒ = +

12. ( ) ( ) ( )2 sin cos sint tpf t e t y t e A t B t− −= ⇒ = +

Guess Again

The homogeneous equation 6 9 0y y y′′ ′− + = has characteristic equation 2 6 9 0r r− + = , which has a

double root 3, 3. Hence,

( ) 3 31 2

t thy t c e c te= + .

We try particular solutions of the form:

13. ( ) ( ) ( ) ( )cos2 sin cospf t t t y t At B t Ct D t= ⇒ = + + +

14. ( ) ( ) ( )3 3 2 3t tpf t te y t At Bt e= ⇒ = +

(We can’t have any terms here dependent on terms in the homogeneous solution.)

15. ( ) ( )sin sin cost tpf t e t y t Ae B t C t− −= + ⇒ = + +

16. ( ) ( )4 2 4 3 21 pf t t t y t At Bt Ct Dt E= − + ⇒ = + + + +


Determining the Undetermined

17. 1y′ = . The homogeneous solution is ( )hy t c= , where c is any constant. By simple inspection we observe that ( )py t t= is a solution of the nonhomogeneous equation. Hence, the general solution

is

( )y t t c= + .

18. 1y y′ + = . The homogeneous solution is ( ) thy t ce−= where c is any constant. By simple inspec-

tion we observe that ( ) 1py t = is a solution of the nonhomogeneous equation. Hence, the general

solution is

( ) 1ty t ce−= + .

19. y y t′ + = . ( ) thy t ce−= , py At B= + , py A′ = . Substituting into the DE gives ( )A At B t+ + = .

Coefficient of t: 1A = . Coefficient of 1: 0A B+ = . Hence, 1A = , 1B = − . 1py t= − ,

1ty ce t−= + − .

20. 1y′′ = . The homogeneous solution of the equation is

( ) 1 2hy t c t c= + ,

where 1c , 2c are arbitrary constants. By inspection, we note that 212py t= is a particular

solution. Hence, the solution of the homogeneous equation is

( ) 21 2

12

y t t c t c= + + .

If you could not find a particular solution by inspection, you could try a solution of the form ( ) 2

py t At Bt C= + + .

21. 4 1y y′′ ′+ = . The characteristic equation is 2 4 0r r+ = , which has roots 0, –4. Hence, the homo-

geneous solution is

( ) 41 2

thy t c c e−= + .

The constant on the right-hand side of the differential equation indicates we seek a particular solution of the form ( )py t A= , except that the homogeneous solution has a constant solution;

thus we seek a solution of the form ( )py t At= . Substituting this expression into the differential

equation yields 4 1A = , or 14

A = . Hence, we have a particular solution

( ) 14py t t= ,


so the general solution is

( ) 41 2

14

ty t c c e t−= + + .

22. 4 1y y′′ + = . The characteristic equation is 2 4 0r + = , which has roots 2i± . Hence, the homoge-

neous solution is

( ) 1 2cos2 sin 2hy t c t c t= + .

The constant on the right-hand side of the differential equation indicates we seek a particular solution of the form ( )py t A= . Substituting this expression into the differential equation yields

4 1A = , or 14

A = . We have a particular solution ( ) 14py t = , so the general solution

( ) 1 21cos 2 sin 24

y t c t c t= + + .

23. 4y y t′′ ′+ = . The characteristic equation is 2 4 0r r+ = , which has roots 0, –4. Hence, the homo-

geneous solution is

( ) 41 2

thy t c c e−= + .

The term on the right-hand side of the differential equation indicates we seek a particular solution of the form

( )py t At B= + .

However, the homogeneous solution has a constant term so we seek a solution of the form

( ) 2py t At Bt= + .

Substituting this expression into the differential equation yields

4 2 8 4y y A At B t′′ ′+ = + + = .

Setting the coefficient of t, 1 equal to each other yields 18

A = , 116

B = − . Thus, the solution

( ) 4 21 2

1 18 16

ty t c c e t t−= + + − .

24. 2 3 6y y y t′′ ′+ − = − . The characteristic equation is 2 2 0r r+ − = , which has roots –2 and 1.

Hence, the homogeneous solution

( ) 21 2

t thy t c e c e−= + .

The linear polynomial on the right-hand side of the equation indicates we seek a particular solution of the form


( )py t At B= + .

(Note that we don’t have any matches with the homogeneous solution.) Substituting this expres-sion into the differential equation yields the equation

2 2 2 3 6y y y A At B t′′ ′+ − = − − = −

so 3A = , 0B = . Hence, we have the general solution

( ) 21 2 3t ty t c e c e t−= + + .

25. 3ty y e′′ + = + . The characteristic equation is 2 1 0r + = , which has roots i± . Hence, the

homogeneous solution is

( ) 1 2cos sinhy t c t c t= + .

The terms on the right-hand side of the differential equation indicates we seek a particular solution of the form

( ) tpy t Ae B= + .


2 3t ty y Ae B e′′ + = + = + .

Setting coefficients of te , 1 equal to each other, we get equations for A, B, which yield 12

A = ,

3B = . Hence, we have the general solution

( ) 1 21cos sin 32

ty t c t c t e= + + + .

26. 2 6 ty y y e′′ ′− − = . The characteristic equation is 2 2 0r r− − = , which has roots –1 and 2. Hence,

the homogeneous solution is

( ) 21 2


The exponential term on the right-hand side of the differential equation indicates we seek a particular solution of the form

( ) tpy t Ae= .

(Note this is not linearly dependent on any of the exponential terms in the homogeneous solution.) Substituting this expression into the differential equation we get

2 2 6t ty y y Ae e′′ ′− − = − = .


Hence, 3A = − , and we have a particular solution

( ) 3 tpy t e= − ,

and hence

( ) 21 2 3t t ty t c e c e e−= + − .

27. 6sin 2y y t′′ ′+ = . The characteristic equation is 2 0r r+ = , which has roots 0 and –1. Hence, the


( ) 1 2t

hy t c c e−= + .

The sine term on the right-hand side of the differential equation indicates we seek a particular solution of the form

( ) cos2 sin 2py t A t B t= + .

Substituting into the differential equation yields

( ) ( )4 2 cos2 4 2 sin 2 6sin 2y y A B t B A t t′′ ′+ = − + + − − = .

Comparing coefficients yields the equations

4 2 04 2 6,

A BB A

− + =− − =

which has the solution 35

A = − , 65

B = − . Hence, we have

( ) 3 6cos 2 sin 25 5py t t t= − − ,

and the general solution is

( ) 1 23 6cos 2 sin 25 5

ty t c c e t t−= + − − .

28. 4 5 2 ty y y e′′ ′+ + = . The characteristic equation of the differential equation is 2 4 5 0r r+ + = ,

which has roots 2 i− ± . Hence, the homogeneous solution is

( ) ( )21 2cos sint

hy t e c t c t−= + .

The exponential on the right-hand side of the differential equation indicates we seek a particular solution of the form

( ) tpy t Ae= .


4 5 10 2t ty y y Ae e′′ ′+ + = = ,


which yields 15

A = . Hence, we have a particular solution ( ) 15

tpy t e= , and the general solution

is given by

( ) ( )21 2

1cos sin5

t ty t e c t c t e−= + + .

29. 4 4 ty y y te−′′ ′+ + = . The characteristic equation is given by 2 4 4 0r r+ + = , which has a double

root of –2, so the homogeneous solution is

( ) 2 21 2

t thy t c e c te− −= + .


( ) t tpy t Ate Be− −= + .


( )4 4 2t t ty y y Ate A B e te− − −′′ ′+ + = + + = .

Comparing coefficients, yields equations, which we solve, getting 1A = , 2B = − . Hence, the general solution is

( ) 2 21 2 2t t t ty t c e c te te e− − − −= + + − .

30. siny y t t′′ − = . The characteristic equation is 2 1 0r − = , which has roots 1± . Hence, the


( ) 1 2t t

hy t c e c e−= + .

The term on the right-hand side of the differential equation indicates we seek a particular solution

( ) ( ) ( )cos sinpy t At B t Ct D t= + + + .

Differentiating this expression two times and substituting it into the differential equation yields the algebraic equation

( ) ( )2 sin 2 cos 2 2 sin 2 2 cos siny y Ct t At t A D t C B t t t′′ − = − − + − − + − = .

Comparing terms in sin t , cos t , sint t , cost t , we get equations that yield

0A = , 12

B = − , 12

C = − , 0D = .

Hence,

( ) ( )1 21 sin cos2

t ty t c e c e t t t−= + − + .


31. 212cosy y t′′ + = . The characteristic equation is 2 1 0r + = , which has roots i± . Hence, the


( ) 1 2cos sinhy t c t c t= + .

Using the trigonometric identity

( )1cos 1 cos22

t t= +

the term on the right-hand side of the differential equation yields

( )212cos 6 1 cos2t t= + .

Hence, we seek a particular solution of the form

( ) cos2 sin 2py t A t B t C= + + .

Substituting this into the differential equation yields

3 cos 2 3 sin 2 6 6cos 2y y A t B t C t′′ + = − − + = + .

Comparing coefficients, we get 2A = − , 0B = , 6C = , so the general solution is

( ) 1 2cos sin 2cos2 6y t c t c t t= + − + .

32. 8 ty y te′′ − = . The characteristic equation is 2 1 0r − = , which has roots 1± . Hence, the homoge-

neous solution is

( ) 1 2t t

hy t c e c e−= + .

The term on the right-hand side of the differential equation indicates we seek a particular solution

( ) t tpy t Ate Be= + ,

but one term in the homogeneous solution is linearly dependent on this term, so we seek

( ) ( )2tpy t e At Bt= + .


( )4 2 2 8t t ty y Ate A B e te′′ − = + + = ,

which gives the two equations 4 8A = , 2 2 0A B+ = , which gives 2A = , 2B = − . Hence, the

general solution is

( ) ( )1 2 2 1t t ty t c e c e te t−= + + − .


33. 24 4 ty y y te′′ ′− + = . The characteristic equation of the differential equation is 2 4 4 0r r− + = ,

which has a double root of 2. Hence, the homogeneous solution is

( ) 2 21 2

t thy t c e c te= + .


( ) 2 2t tpy t Ate Be= + ,

but both terms are linearly dependent with terms in the homogeneous solution, so we choose

( ) 3 2 2 2t tpy t At e Bt e= + .

Differentiating and substituting this expression into the differential equation yields the algebraic equation

( )2 24 4 0 0 6 2t ty y y e At B te′′ ′− + = + + + = .

Comparing coefficients, we get 16

A = , 0B = . Hence, the general solution is

( ) 2 2 3 21 2

16

t t ty t c e c te t e= + + .

34. 4 3 20cosy y y t′′ ′− + = . The characteristic equation of the differential equation is 2 4 3 0r r− + = ,

which has roots 1, 3. Hence,

( ) 31 2

t thy t c e c e= + .

The term on the right-hand side of the differential equation indicates we seek

( ) cos sinpy t A t B t= + .


( ) ( )4 3 4 2 sin 2 4 cos 20cosy y y A B t A B t t′′ ′− + = + + − = .

Comparing coefficients yields 2A = , 4B = − . Hence,

( ) 31 2 2cos 4sint ty t c e c e t t= + + − .

35. 3 2 sinty y y e t′′ ′− + = . The characteristic equation of the differential equation is 2 3 2 0r r− + = ,

which has roots 1, 2. Hence,

( ) 21 2

t thy t c e c e= + .

The term on the right-hand side of the equation indicates we seek a particular solution

( ) cos sint tpy t Ae t Be t= + .


Differentiating and substituting this expression into the equation yields

( ) ( )3 2 sin cos sint t ty y y A B e t A B e t e t′′ ′− + = − + − − = .

Comparing coefficients, we find 12

A = , 12

B = − , yielding the general solution

( ) ( )21 2

1 cos sin2

t t ty t c e c e e t t= + + − .

36. 3 sin 2cosy y t t′′ ′+ = + . The characteristic equation is 2 3 0r r+ = , which has roots 0, –3. Hence,

the homogeneous solution is

( ) 31 2

thy t c c e−= + .

The sine and cosine terms on the right-hand side of the equation indicate we seek a particular solution of the form

( ) cos sinpy t A t B t= + .

Substituting this into the equation yields

( ) ( )3 3 cos 3 sin sin 2cosy y A B t B A t t t′′ ′+ = − + + − − = + .

Comparing terms, we arrive at ( )3 2A B− + = , ( )3 1B A− − = , yielding 12

A = − , 12

B = . From

this, that the general solution is

( ) ( )31 2

1 sin cos2

ty t c c e t t−= + + − .

37. 4 6y y t′′′ ′′− =

(1) Find yh: r3 − 4r2 = 0 ⇒ r2(r − 4) = 0 ⇒ r = 0, 0, 4

∴ yh = c1 + c2t + c3e4t

(2) Find yp: yP = t2(At + B) = At3 + Bt2

py ′ = 3At2 + 2Bt

py ′′ = 6At + 2B

py ′′′ = 6A

4 6 4(6 2 ) 24 6 8 6p py y A At B At A B t′′′ ′′− = − + = − + − =

coefficient of t: −24A = 6, coefficient of 1: 6A − 8B = 0


so that A = 1 ,4

− and 8B = 164

⎛ ⎞−⎜ ⎟⎝ ⎠

= 3 3 so that 2 16

B− = − .

Hence yp = 3 21 34 16

t t− −

(3) 4 3 21 2 3

1 3( )4 16

th py t y y c c t c e t t= + = + + − −

38. (3) 3 3 ty y y y e′′ ′− + − =

(1) Find yh: r3 − 3r2 + 3r − 1 = 0 (characteristic equation)

f(r) =− r3 −3r2 + 3r − 1 = 0

f(1) = 1 − 3 + 3 − 1 = 0 so r = 1 is a root.

By long division, we obtain

r3 − 3r2 + 3r − 1 = (r − 1)(r2 − 2r + 1) = (r − 1)3 Triple root r = 1, 1, 1

yh = c1et + c2tet + c3t2et

(2) Find yp: yp = t3(Aet) = At3et

3 23t tpy At e At e′ = +

3 2 2 3 23 3 6 6 6t t t t t t tpy At e At e At e Ate At e At e Ate′′ = + + + = + +

(3) 3 2 2 23 3 6 3 6 6 6t t t t t t t tpy At e At e At e Ate At e Ate Ate Ae= + + + + + + +

(3) 3 2

3 2

3 2

3 2

3 3 9 18 6

3 18 18 0

3 9 0 0

0 0 0

t t t tp p p p

t t t t

t t t t

t t t t

y y y y At e At e Ate Ae

At e At e Ate e

At e At e te e

At e t e te e

′′ ′− + − = + + +

− − − +

+ + + +

− + + +

16 so that 6

t tAe e A= =

Thus yp = 316

tt e

(3) 2 31 2 3

1( )6

t t t th py t y y c e c te c t e t e= + = + + +


39. (4) 10y y− =

(1) Find yh:

r4 − 1 = 0 2 2 2( 1)( 1) ( 1)( 1)( 1)r r r r r+ − = + − + r = ±i, ± 1

yh = c1 cos t + c2 sin t + c3et + c4e−t

(2) Find yp:

yp = A, so that (4) 0p p p py y y y′ ′′ ′′′= = = =

⇒ (4) 0 10 10 10p p py y A A y− = − = ⇒ = − ⇒ = −

(3) 1 2 3 4( ) cos sin 10t th py t y y c t c t c e c e−= + = + + + −

40. 0y y y y′′′ ′′ ′′′ ′′= ⇒ − =

(1) Find yh: r3 − r2 = 0 ⇒ r2(r − 1) = 0 r = 0, 0, 1

(2) There is no yp because the DE is homogeneous.

(3) y(t) = c1 + c2t + c3et


41. 2 3 6 , (0) 1, (0) 0y y y t y y′′ ′ ′+ − = − = − =

(1) Find yh:

r2 + r − 2 = 0 ⇒ (r − 1)(r + 2) = 0 ⇒ r = 1, −2

∴ yh = c1et + c2e−2t

(2) Find yp: yp = At + B, , 0 2 2( ) 3 6p p p p py A y y y y A At B t′ ′′ ′′ ′= = ⇒ + − = − + = −

coefficient of t: −2A = −6, coefficient of 1: A − 2B = 3 ⇒ A = 3, B = 0

∴ yp = 3t

(3) y(t) = yh + yp = c1et + c2e−2t + 3t; 21 22 3t ty c e c e−′ = − +

y(0) = −1 ⇒ c1 + c2 = −1; 1 2(0) 0 2 3 0y c c′ = ⇒ − + =

1 2 1

1 2 2

513

22 33

c c c

c c c

+ = − = −

− = − =

∴ 25 2 33 3

t ty e e t−= − + +

⇒


42. 4 4 , (0) 1, (0) 1ty y y te y y−′′ ′ ′+ + = = − =

(1) Find yh:

r2 + 4r + 4 = 0 ⇒ (r + 2)2 = 0 ⇒ r = −2, −2

∴ yh = c1e−2t + c2te−2t = (c1 + c2t)e−2t

(2) Find yp:

yp = e−t(At + B) ⇒ ( )t tpy e At B Ae− −′ = − + +

⇒ ( ) ( ) 2t t t t tpy e At B Ae Ae e At B Ae− − − − −′′ = + − − = + −

So 4 4 ( ) 2 4( ( ) ) 4 ( )

( 2 )

t t t t tp p p

t

y y y e At B Ae e At B Ae e At B

e At A B

− − − − −

−

′′ ′+ + = + − + − + + + +

= + +

This gives A = 1, 2A + B = 0 and so A = 1 and B = −2.

Therefore yp = e−t(t − 2).

(3) y = yh + yp = c1e−2t + c2te−2t + e−t(t − 2)

2 2 21 2 22 2 ( 2)t t t t ty c e c e c te e t e− − − − −′ = − + − − − +

1 1

1 2 2

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

y c cy c c c

= − ⇒ − = − =′ = ⇒ − + + + = =

∴ y(t) = e−2t + e−t(t − 2)

43. 4 , (0) 1, (0) 1y y t y y′′ ′+ = = = −

(1) Find yh: r2 + 4 = 0 ⇒ r = ± 2i ⇒ yh = c1 cos 2t + c2 sin 2t

(2) Find yp: yp = At + B, , 0p py A y′ ′′= =

∴ 4 4( ) 4 4p py y At b At B t′′ + = + = + =

coefficient of t: 4A = 1, coefficient of 1: 4B = 0 ⇒ A = 1 , 04

B =

∴ yp = 14

t

(3) y = yh + yp = c1 cos 2t + c2 sin 2t + 14

t , 1 212 sin 2 2 cos 24

y c t c t′ = − + +

y(0) = 1 ⇒ c1 = 1; (0)y′ = −1 ⇒ 2c2 + 14

= −1 ⇒ 2c2 = 54

− ⇒ c2 = 58

−

∴ 5 1( ) cos2 sin 28 4

y t t t t= − +

⇒


44. 2 6cos , (0) 1, (0) 1y y y t y y′′ ′ ′+ + = = = −

(1) Find yh: r2 + 2r + 1 = 0 ⇒ (r + 1)2 = 0 ⇒ r = −1, −1

yh = c1e−t + c2te−t

(2) Find yp: yp = A cos t + B sin t , sin cos , cos sinp py A t B t y A t B t′ ′′= − + = − −

2 cos sin

2( cos sin ) cos sin

p p py y y A t B t

B t A tA t B t

′′ ′⇒ + + = − −

+ −+ +

2B cos t − 2A sin t = 6 cost

coefficient of cos t: 2B = 6, coefficient of sin t: −2A = 0 ⇒ A = 0, B = 3

∴ yp = 3 sin t

(3) y = yh + yp = 1 2 1 2 23sin , 3cost t t t tc e c te t y c e c te c e t− − − − −′+ + = − − + +

y(0) = 1 ⇒ c1 = 1; (0) 1y′ = − ⇒ −c1 + c2 + 3 = −1 ⇒ c2 = −3

∴ ( ) 3 3sint ty t e te t− −= − +

45. 4 cos2 , (0) 1, (0) 0y y t y y′′ ′+ = = =

(1) Find yh: 4r2 + 1 = 0 ⇒ r2 = 14

− ⇒ r = 12

i± ⇒ yh = 1 21 1cos sin2 2

c t c t⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(2) Find yp: yp = A cos 2t + B sin 2t, 2 sin 2 2 cos2 , 4 cos 2 4 sin 2p py A t B t y A t B t′ ′′= − + = − −

4 16 cos2 16 sin 2 cos 2 sin 2p py y A t B t A t B t′′⇒ + = − − + +

15 cos2 15 sin 2 cos2A t B t t= − − =

coefficient of cos 2t: −15A = 1, coefficient of sin 2t: −15 B = 0

⇒ A = 1 , 015

B− =

1 cos215py t∴ = −


(3) y = yh + yp = 1 21 1 1cos sin cos22 2 15

c t c t t⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 21 1 1 1 2sin cos sin 22 2 2 2 15

y c t c t t⎛ ⎞ ⎛ ⎞′ = − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

y(0) = 1 ⇒ c1 − 115

= 1 ⇒ c1 = 1615

(0) 0y′ = ⇒ c2 = 0

∴ 16 1 1( ) cos cos 215 2 15

y t t t⎛ ⎞= −⎜ ⎟⎝ ⎠

46. 9 cos3 , (0) 1, (0) 1y y t y y′′ ′+ = = = −

(1) Find yh: r2 + 9 = 0 r = ± 3i

yh = c1 cos 3t + c2 sin 3t

(2) Find yp: yp = t(A cos 3t + B sin 3t)

( 3 sin3 3 cos3 ) ( cos3 sin 3 )py t A t B t A t B t′ = − + + +

( 9 cos3 9 sin3 ) ( 3 sin 3 3 cos3 ) 3 sin3 3 cos3py t A t B t A t B t A t B t′′ = − − + − + − +

9 (9 9 ) cos3 (9 9 ) sin 3 (6 )cos3 ( 6 sin3 )p py y A A t t B B t t B t A t′′ + = − + − + + − = cos 3t

coefficient of cos 3t: 6B = 1, coefficient of sin 3t: −6A = 0

⇒ A = 0, B = 16

so that yp = 1 sin 36

t t

(3) y = yh + yp = c1 cos 3t + c2 sin 3t + 1 sin 36

t t ,

y′ = −3c1 sin 3t + 3c2 cos 3t + 1 1cos3 sin32 6

t t t+

y(0) = 1 ⇒ c1 = 1; (0)y′ = −1 ⇒ 3c2 = −1 ⇒ c2 = 13

−

∴1 1( ) cos3 sin 3 sin33 6

y t t t t t= − +


47. 3 2 4 ,ty y y e−′′ ′− + = y(0) = 1, (0)y′ = 0

(1) Find yh:

r2 − 3r + 2 = 0 ⇒ (r − 1)(r − 2) = 0 ⇒ r = 1, r = 2, so that yh = c1et + c2e2t

(2) Find yp:

yp = Ae−t, ,t tp py Ae y Ae− −′ ′′= − =

23 2 3( ) 2 6 4 so that 3

t t t t tp p py y y Ae Ae Ae Ae e A− − − − −′′ ′− + = − − + = = = ⇒ yp = 2

3te−

(3) y = yh + yp = c1et + c2e2t + 2 ,3

te− 21 2

223

t t ty c e c e e−′ = + −

y(0) = 1 ⇒ c1 + c2 + 23

= 1 c1 + c2 = 13

c1 = 0

1 22(0) 0 2 03

y c c′ = ⇒ + − = c1 + 2c2 = 23

c2 = 13

Thus 21 2( )3 3

t ty t e e−= + .

48. 4 3 ,ty y y e t−′′ ′− + = + y(0) = 0, (0)y′ = 0

(1) Find yh:

r2 − 4r + 3 = 0 ⇒ (r − 1)(r − 3) = 0 so that yh = c1et + c2e3t

(2) Find yp:

yp = Ae−t + Bt + C, py′ ,tAe B−= − + tpy Ae−′′ =

Thus 4 3 4( ) 3( )t t tp p py y y Ae Ae B Ae Bt C− − −′′ ′− + = − − + + + +

4( ) 3( )t t tAe Ae B Ae Bt C− − −= − − + + + +

= 8Ae−t + 3Bt − 4B + 3C = e−t + t

⇒ 8A = 1, 3B = 1, −4B + 3C = 0

⇒A = 1 1 4 4, , 8 3 3 9

B C B= = = .

Thus, yp = 1 1 4 .8 3 9

te t− + +

⇒ ⇒


(3) Find y:

y = yh + yp = c1et + c2e3t + 1 1 48 3 9

te t− + + .

31 2

1 138 3

t t ty c e c e e−′ = + − + .

y(0) = 0 ⇒ c1 + c2 + 1 48 9

+ = 0 c1 + c2 = 4172

− c1 = 34

−

(0)y′ = 0 ⇒ c1 + 3c2 − 1 18 3

+ = 0 c1 + 3c2 = 524

− c2 = 1372

Therefore, y(t) = 33 13 1 1 4 .4 72 8 3 9

t t te e e t−− + + + +

49. 2y y y′′ ′− − = 4 cos 2t, y(0) = 0, (0)y′ = 0

(1) Find yh:

r2 − r − 2 = 0 ⇒ (r + 1)(r − 2) = 0 so r = 2, −1

yh = c1e−t + c2e2t

(2) Find yp:

yp = A cos 2t + B sin 2t, py′ = −2A sin 2t + 2B cos 2t

py′′ = −4A cos 2t − 4B sin 2t

2p p py y y′′ ′− − = −4A cos 2t − 4B sin 2t + 2A sin 2t − 2B cos 2t − 2A cos 2t − 2B sin 2t

= (−6A − 2B) cos 2t + (2A − 6B) sin 2t = 4 cos 2t

∴ coefficient of cos 2t: −6A − 2B = 4, coefficient of sin 2t: 2A − 6B = 0

so A = 3B and B = 15

− ⇒ A = 35

−

Thus yp = 3 1cos 2 sin 25 5

t t− − .

⇒ ⇒


(3) y = yh + yp = c1e−t + c2e2t − 3 1cos2 sin 25 5

t t−

y′ = −c1e−t + 2c2e2t + 6 2sin 2 cos25 5

t t−

y(0) = 0 ⇒ c1 + c2 − 35

= 0 c1 + c2 = 35

c1 = 415

(0)y′ = 0 ⇒ −c1 + 2c2 − 25

= 0 −c1 + 2c2 = 25

c2 = 13

∴ 24 1 3 1( ) cos2 sin 215 3 5 5

t ty t e e t t−= + − −

50. y′′′ − 24 3 ,y y t′′ ′+ = y(0) = 1, (0)y′ = 0, (0)y′′ = 0

(1) Find yh:

r3 − 4r2 + 3r = 0 ⇒ r(r2 − 4r + 3) = 0

r(r − 3)(r − 1) = 0, so r = 0, 1, 3

yh = c1 + c2et + c3e3t

(2) Find yp:

yp = t(At2 + Bt + C) = At3 + Bt2 + Ct

py′ = 3At2 + 2Bt + C, py′′ = 6At + 2B, py′′′ = 6A

4 3p p py y y′′′ ′′ ′− + = 6A − 24At − 8B + 9At2 + 6Bt + 3C = t2

coefficient of t2: 9A = 1, coefficient of t: −24A + 6 B = 0,

coefficient of 1: 6A − 8B + 3C = 0

A = 1 4, ,9 9

B = C = 2627

∴ py = 3 21 4 269 9 27

t t t+ +

(3) y = yh + yp = c1 + c2et + c3e3t + 3 21 4 269 9 27

t t t+ +

Using this general solution and the initial conditions we obtain:

3 3 2161 2 8 1 4 26( )81 3 81 9 9 27

t ty t e e t t t= − − + + +

⇒ ⇒


51. y(4) − y = e2t, y(0) = (0) (0) (0) 0y y y′ ′′ ′′′= = =

(a) Find yh:

r4 − 1 = 0 ⇒ (r2 + 1)(r2 − 1) ⇒ r = ± i, ± 1


(2) Find yp:

yp = Ae2t, py′ = 2Ae2t, py′′ = 4Ae2t, py′′′ = 8Ae2t, (4)py = 16Ae2t

Thus (4)p py y− = 16Ae2t − Ae2t = e2t ⇒ 15Ae2t = e2t ⇒ A = 1

15 so that yp = 21

15te .

(3) y = yh + yp = c1 cos t + c2 sin t + c3et + c4e−t + 2115

te

y′ = −c1 sin t + c2 cos t + c3 et − c4e−t + 2215

te

y′′ = −c1 cos t − c2 sin t + c3 et + c4e−t + 2415

te

y′′′ = c1 sin t − c2 cos t + c3 et − c4e−t + 2815

te

y(0) = 0 ⇒ c1 + c3 + c4 + 115

= 0

2 3 42(0) 0 0

15y c c c′ = ⇒ + − + =

1 3 44(0) 0 0

15y c c c′′ = ⇒ − + + + =

2 3 48(0) 0 0

15y c c c′′′ = ⇒ − + − + =

From these 4 equations in 4 unknowns, we obtain (by the methods of Chapter 3),

c1 = 1 ,10

c2 = 1 ,5

c3 = 14

− and c4 = 112

∴ 21 1 1 1 1( ) cos sin10 5 4 12 15

t t ty t t t e e e−= + − + +

52. y(4) = et, y(0) = 1, (0)y′ = 0, (0) 0y′′ = , (0)y′′′ = 0

(1) Find yh:

r4 = 0 ⇒ r = 0 (multiplicity 4)

yh = c1 + c2t + c3t2 + c4t3


(2) Find yp:

yp = Aet, py′ = Aet, py′′ = Aet, py′′′ = Aet, (4)py = Aet

(4)py = Aet = et ⇒ A = 1 so that yp = et

(3) y = yh + yp = c1 + c2t + c3t2 + c4t3 + et

y′ = c2 + 2c3t + 3c4t2 + et

y′′ = 2c3 + 6c4t + et

y′′′ = 6c4 + et

y(0) = 1 ⇒ c1 + 1 = 1 ⇒ c1 = 0

2 2(0) 0 1 0 1y c c′ = ⇒ + = ⇒ = −

3 31(0) 0 2 1 02

y c c′′ = ⇒ + = ⇒ = −

4 41(0) 0 6 1 06

y c c′′′ = ⇒ + = ⇒ = −

∴ 2 31 1( )2 6

ty t t t t e= − − − +

53. 4 cos2ty y t ⎛ ⎞′′ + = − ⎜ ⎟

⎝ ⎠

Find yh: yh = c1 cos 21 1sin2 2

t c t⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Find yp: 1py = At + B

2py = cos sin

2 2t tt c D

⎛ ⎞⎛ ⎞ ⎛ ⎞+⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

∴ 1 2

( ) cos sin2 2p p pt ty t y y At B Ct Dt⎛ ⎞ ⎛ ⎞= + = + + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

54. 2 ty y t e′′′ ′′− = +

Find yh:

r3 − r2 = 0 ⇒ r2(r − 1) = 0 ⇒ r = 0, 0, 1

∴ yh = c1 + c2t + c3et


Find yp:

1 2

2 2( ), ( )tp py t At Bt C y t De= + + =

∴ 1 2

4 3 2( ) tp p py t y y At Bt Ct Dte= + = + + +

55. 5 6 cos ty y y t te′′ ′− + = −

Find yh:

r2 − 5r + 6 = 0 ⇒ (r − 2)(r − 3) = 0 ⇒ r = 2, 3

∴ yh = c1e2t + c2e3t

Find yp:

1 2

cos sin , ( )tp py A t B t y e Ct D= + = +

∴ 1 2

( ) cos sin ( )tp p py t y y A t B t e Ct D= + = + + +

56. y(4) − y = tet + sin t

r4 − 1 = 0 ⇒ (r2 + 1)(r2 − 1) = 0 ⇒ r = ± i, ±1


1

( )tpy te At B= + ,

2( cos sin )py t C t D t= +

∴ 2( ) ( ) cos sintpy t e At Bt Ct t Dt t= + + +

Judicious Superposition

57. (a) The characteristic equation is 2 6 0r r− − = has roots 3r = , –2, so the general solution is

( ) 3 21 2


(b) (i) Substituting ( ) tpy t Ae= yields

6t t t tAe Ae Ae e− − = ,

which yields 16

A = − . Hence, ( ) 16

tpy t e= − .

(ii) Substituting ( ) tpy t Ae−= yields

6t t t tAe Ae Ae e− − − −+ − =

or 14

A = − . Hence, ( ) 14

tpy t e−= − .


(c) Calling ( ) 6L y y y y′′ ′= − − we found in part (b) that

16

t tL e e⎛ ⎞− =⎜ ⎟⎝ ⎠

, and 14

t tL e e− −⎛ ⎞− =⎜ ⎟⎝ ⎠

.

Multiplying each equation by 12

and using basic properties of derivatives yields

1 112 2

t tL e e⎛ ⎞− =⎜ ⎟⎝ ⎠

, and 1 18 2

t tL e e−⎛ ⎞− =⎜ ⎟⎝ ⎠

and

( )1 1 1 cosh12 8 2

t t t tL e e e e t− −⎛ ⎞− − = + =⎜ ⎟⎝ ⎠

.

Hence, a solution of 6 coshy y y t′′ ′− − = is

( ) 1 112 8

t tpy t e e−= − − .

Wholesale Superposition

58. We first solve the equation

!

nty yn

′ + =

first getting the homogeneous

( ) thy t ce−= .

To find a particular solution, we try

( ) ( ) 11 1 0

n n np n ny t A t A t A t A−

−= + + + +… .

Substituting this into the equation yields

( ) ( )1 2 11 1 1 1 01

!

nn n n n

n n n ntnA t n A t A A t A t A t An

− − −− −⎡ ⎤+ − + + + + + + + =⎣ ⎦… … .

Comparing coefficients, we have


( )

( )

( )

1

2

3

1!

11 !

12 !

1 ,3 !

n

n

n

n

An

An

An

An

−

−

−

=

−=

−

=−

−=

−

and so on. Hence, we have

( )

( ) ( )1 2

! 1 ! 2 !

n n nn

pt t tyn n n

− −

= − + −− −

… .

Further, we have

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

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

0

1

22

3 23

4 3 24

1 2

1

1

12!

13! 2!

14! 3! 2!

! 1 ! 2 !

p

p

p

p

p

n n nn

p

y t

y t t

ty t t

t ty t t

t t ty t t

t t ty tn n n

− −

=

= −

= − +

= − + −

= − + − +

= − + −− −

… … … …

…

By superposition, the sum of these solutions is a solution of ty y e′ + = . (We agree our discussion

is formal in the sense that we have proven superposition for finite sums.) There is a slight problem in adding the preceding functions because the sum changes form depending on whether we add an even or odd number of terms. We have

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

2 4 20 1 2

2

3 5 2 10 1 2 1

2 1

12! 4! 2 !

3! 5! 2 1 !

nn

n p p p

nn

n p p p

t t tS y t y t y tn

t t tS y t y t y t tn

++

+

= + + = + + + +

= + + = + + + ++

… …

… …


However, because the sequence nS converges, it converges to the average of the nth and ( )1n + st terms. That is,

( )2 3

2 2 11 1 112 2 2! 3! 2

tn n

t tS S t e+

⎛ ⎞+ = + + + + =⎜ ⎟

⎝ ⎠…

Hence, we found ( ) 12

tpy t e= .

Discontinuous Forcing Functions

59. y y′′ ′+ = 2 0 4 (0) (0) 01 4

t y yt

′≤ < = =⎧⎨ ≥⎩

Part 1: 1 1 2y y′′ ′+ = y1(0) = 1(0) 0y′ =

Find (y1)h: r2 + r = 0 r(r + 1) = 0 r = 0, −1 (y1)h = c1 + c2e−t

Find (y)p: (y1)p = 2t by inspection, so

y1 = c1 + c2e−t + 2t 0 = c1 + c2

1 2 2ty c e−′ = − + 0 = −c2 + 2

∴ y1 = −2 + 2e−t + 2t

Part 2: 2 2 1y y′′ ′+ =

y2 = c1 + c2e−t + t y2(4) = y1(4) = −2 + 2e−4 + 8 = 6 + 2e−4

2y′ = −c2e−t + 1 42 1(4) (4) 2 2y y e−′ ′= = − +

Thus, when t = 4,

4 4

1 24 4

2

6 2 4

2 2 1

e c c e

e c e

− −

− −

+ = + +

− + = − +

8 = c1 + 5 ⇒ c1 = 3 −2e−4 + 1 = −c2e−4 ⇒ −2 + e4 = −c2 ⇒c2 = 2 − e4 ∴ y2 = 3 + (2e − e4)e−t + t

and y(t) = 4

2 2 2 0 4

3 (2 ) 4

t

t

e t t

e e t t

−

−

⎧− + + ≤ <⎪⎨

+ − + ≥⎪⎩

⇒ ⇒ c2 = 2, c1 = −2


60. cos 0 (0) 1, (0) 0

160

t t y yy y

tπ

π′≤ ≤ = =⎧′′ + = ⎨ >⎩

Part 1: 1 116 cosy y t′′+ = 0 ≤ t ≤ π y1(0) = 1, 1(0)y′ = 0

Find (y1)h: (y1)h = c1 cos 4t + c2 sin 4t

Find (y1)p : (y1)p = A cos t + B sin t

1( ) cos sinpy A t B t′′ = − −

Therefore, y1 = c1 cos 4t + c2 sin 4t + 1 cos15

t

1 1 214 sin 4 4 cos4 sin

15y c t c t t′ = − + −

y1(0) = 1 = c1 + 115

c1 = 1415

−

1 2(0) 0 4y c′ = = c2 = 0

∴ y1(t) = 14 1cos4 cos15 15

t t− +

Part 2: 2 216 0y y′′ + = t > π

y2 = c1 cos 4t + c2 sin 4t

2 1 24 sin 4 4 cos4y c t c t′ = − +

y1(π) = 2 114 1 13 ( )15 15 15

y cπ− + = − = = − c1 = 1315

−

1 2 2( ) 4 ( ) 4y y cπ π′ ′= = = c2 = 1

Thus y2(t) = 13 cos4 sin 415

t t− +

and y(t) =

14 1cos 4 cos 015 1513 cos4 sin 415

t t t

t t t

π

π

⎧− + ≤ ≤⎪⎪⎨⎪− + >⎪⎩

⇒ B = 0, A = 115

⇒


Solutions of Differential Equations Using Complex Functions

61. 2 2siny y y t′′ ′− + =

The homogeneous solution is yh = c1et + c2tet.

For the particular solution we use 2 2 ity y y e′′ ′− + = and seek the imaginary part of the particular

solution.

We let yp = Aeit. Then itpy iAe′ = and py′′ = −Aeit.

By substitution, we obtain

−Aeit − 2iAeit + Aeit = 2eit

−2iA = 2 A = 1 ii

−− =

yp = ieit = i(cos t + i sin t) = i cos t − sin t

Im(yp) = cos t

∴ y(t) = yh + Im yp = c1et + c2tet + cos t

62. 25 6siny y t′′ + = We will use 25 6 ity y e′′ + =

The homogeneous solution is yh = c1 cos 5t + c2 sin 5t

For the particular solution we want Im(yp) where yp = Aeit; and it itp py iAe y Ae′ ′′= = −


−Aeit + 25Aeit = 6eit

24A = 6 so that A = 14

Im(yp) = Im 1 1 sin4 4

ite t⎛ ⎞ =⎜ ⎟⎝ ⎠

∴ y(t) = c1 cos 5t + c2 sin 5t + 1 sin4

t

63. 25 20sin 5y y t′′ + = We will use 525 20 i ty y e′′ + =

The homogeneous solution is yh = c1 cos 5t + c2 sin 5t

For the particular solution we note that e5it is included in yh, so we must use an extra factor of t in yp. We want Im(yp) where yp = Ate5it, so 5 5( 5 )it it

py A t ie e′ = + ,

and 5 5 55 ( 5 ) 5it it itpy A i t ie e Ai e′′ = + + = 5 5(10 25 )it itA ie te− .



5 5 5 5(10 25 ) 25 20i it it itA ie te Ate e− − + = so that 10Ai = 20. Thus A = −2i and yp = −2ite5it.

Im(yp) = ( )2 (cos5 sin5 )Im it t i t− + = −2t cos 5t

y = c1 cos 5t + c2 sin 5t− 2t cos 5t

Complex Exponents

64. 23 2 3 ity y y e′′ ′− + =

The homogeneous solution is yh = c1et + c2e2t.

We seek a particular solution of the form yp = Ae2it.

Then 22 ity iAe′ = and 24 ity Ae′′ = − .


2 2 2 2

2 2 2

4 3(2 ) 2( ) 3

2 6 3(2 6 ) 3

3 2 6 3 92 6 2 6 20 20

it it it it

it it it

Ae iAe Ae e

Ae Aie eA i

iA ii i

− − + =

− − =− + =

− − −= ⋅ = +

+ −

23 9 3 9 (cos 2 sin 220 20 20 20

3 9 9 3cos2 sin 2 cos2 sin 220 20 20 20

itpy i e i t i t

t t i t t

− −⎛ ⎞ ⎛ ⎞= + = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

It can be verified directly by substitution that

Re(yp) = 3 9cos2 sin 220 20

t t− − satisfies

3 2 3cos2y y y t′′ ′− + =

and that Im(yp) = 9 3cos 2 sin 220 20

t t− satisfies

3 2 3sin 2y y y t′′ ′− + =


65. Student Project

SECTION 4.5 Variation of Parameters 395

4.5 Variation of Parameters

Straight Stuff

1. 4y y t′′ + =

The homogeneous solutions to the equation are y t1 1( ) = and y t e t2( ) = − .

To find a particular solution of the form ( ) 1 2t

py t v v e−= + , we solve the equations

1 2 0

0.

t

t

v e v

e v

−

−

′ ′+ =

′− =

for 1 2, v v′ ′ . This gives 1 4v t′ = and 2 4 tv te′ = − .

Integrating yields ( ) 21 2v t t= and ( ) ( )2 4 1tv t e t= − .

Hence, we have a particular solution

( ) ( ) ( )( )2 21 1 2 2 1 2 4 1 2 4 4t t

py t y v y v t e e t t t−= + = + − = − + .

Combining the constant term with the homogeneous solution, we write the general solution as y t c c e t tt( ) = + + −−

1 222 4 .

2. ′′ − ′ = −y y e t

The homogeneous solutions to the equation are y t1 1( ) = and y t et2( ) = .

To find a particular solution of the form ( ) 1 1 2 2 1 2t

py t v y v y v v e= + = + , we solve

1 2

2

0

.

t

t t

v e v

e v e−

′ ′+ =

′ =

This gives 1tv e−′ = − and 2

2tv e−′ = .

Integrating yields ( )1tv t e−= and ( ) 2

212

tv t e−= − .


( ) ( ) 21 1 2 2

1 1 112 2 2

t t t t t tpy t y v y v e e e e e e− − − − −⎛ ⎞= + = + − = − =⎜ ⎟

⎝ ⎠.

The general solution is

( ) 1 212

t ty t c c e e−= + + .


3. ′′ − ′ + =y y yt

et2 1 , t >( )0

The two linear independent solutions y1 and y2 of the homogeneous equation are

y t et1( ) = and y t tet

2( ) = .

Using the method of variation of parameters, we seek the particular solution

( ) ( ) ( )1 2t t

py t v t e v t te= + .

In order for y tp( ) to satisfy the differential equation, υ1 and υ2 must satisfy

1 1 2 2 11 2

1 1 2 2 11 22

0

1( 1)

t t

t t t

y v y v e v te v

y v y v e v e t v et

′ ′ ′ ′+ = =

′ ′ ′ ′+ = + =.

Solving algebraically for 1v′ and 2v′ we obtain ( )1 1v t′ = − and 21vt

′ = .

Integrating gives the values ( )1v t t= − and 2 lnv t= .

Substituting these values into yp yields the particular solution

y t te te tpt t( ) = − + ln .

Hence, the general solution is y t c e c te te tt t t( ) = + +1 2 ln .

4. ′′ + =y y tcsc

The two linearly independent solutions y1 and y2 of the homogeneous equation are

y t t1( ) = cos and y t t2( ) = sin .

Using the method of variation of parameters, we seek the particular solution

( ) ( ) ( )1 2cos sinpy t v t t v t t= + .

In order for y tp( ) to satisfy the differential equation, 1v and 2v must satisfy

1 1 2 2 11 22

1 1 2 2 11 22

(cos ) (sin ) 0

( sin ) (cos ) csc .

y v y v t v t v

y v y v t v t v t

′ ′ ′ ′+ = + =

′ ′ ′ ′+ = − + =

Solving algebraically for 1v′ and 2v′ we obtain ( )1 1v t′ = − and 2 cotv t′ = .

Integrating gives the values ( )1v t t= − and ( )2 ln sinv t= .

Substituting these values into yp yields the particular solution

( ) ( )cos sin ln sinpy t t t t t= − + .

Hence, the general solution is ( ) ( )1 2cos sin cos sin ln siny t c t c t t t t t= + − + .


5. ′′ + =y y t tsec tan

The homogeneous solutions are y t t1( ) = cos and y t t2( ) = sin .

We seek the solution ( ) ( ) ( )1 2cos sinpy y t v t v= + .

We form the system 1 2

1 2

(cos ) (sin ) 0

(sin ) (cos ) sec tan .

t v t v

t v t v t t

′ ′ =

′ ′− + =.

Solving algebraically for 1v′ and 2v′ yields ( ) 21 tanv t t′ = − and 2 tanv t′ = .

Integrating gives the values 1 tanv t t= − and 2 ln secv t= .

The particular solution is y t t t t t t t t t tp = −( ) + = − +tan cos sin ln sec sin cos sin ln sec .

Thus, the general solution is y t c t c t t t t t( ) = + − +1 2cos sin cos sin ln sec .

6. ′′ − ′ + =y y y e tt2 2 sin

The homogeneous solutions are y t e tt1( ) = cos and y t e tt

2( ) = sin .

To find a particular solution of the form ( ) 2 2cos sint tpy t v e t v e t= + ,

we solve the equations

( ) ( )1 2

1 2

cos sin 0

cos sin sin cos sin

t t

t t t t t

e tv e tv

e t e t v e t e t v e t

′ ′+ =

′ ′− + + =

for 1v′ and 2v′ . This yields 211 sinv t′ = − and 2 sin cosv t t′ = .

Integrating yields the functions ( ) ( )11 cos sin2

v t t t t= − + and ( ) 22

1 sin .2

v t t=

Hence, a particular solution

( ) ( ) ( ) ( )21 1 2 2

1 1 1cos cos sin sin sin sin cos2 2 2

t t tpy t y v y v e t t t t e t t e t t t= + = − + + = −


y t c e t c e t te tt t t( ) = + −1 212

cos sin cos .


7. ′′ − ′ + =+ −y y y

e t3 2 11

The homogeneous solutions are y t et1( ) = and y t e t

22( ) = .

Hence we seek the particular solution ( ) 21 2

t tpy y e v e v= + to form the system

21 2

21 2

0

12 .1

t t

t tt

e v e v

e v e ve−

′ ′+ =

′ ′+ =+

Solving algebraically for 1v′ and 2v′ yields 1 1

t

t

eve

−

−

−′ =+

and 2

2 1

t

t

eve

−

−′ =

+.

The first integral is trivial; 1 ln(1 ).tv e−= +

The second one is more difficult. However, if we perform some algebra, we can write

( )2

2

1

1 1 1

t t tt tt

t t t

e e ee ev ee e e

− − −− −−

− − −

+ −′ = = = −

+ + +, which integrates to give ( )2 ln 1t tv e e− −= − + + .

With υ1 and υ2 we have the particular solution

y e e e e ept t t t t= + − + +− −ln ln1 12b g b g


( ) ( ) ( )2 21 2 ln 1t t t t ty t c e c e e e e−= + + + + .

(The term −et in yp was absorbed in the homogeneous solution, giving a better form for the solution.)

8. ′′ + ′ + = −y y y e tt2 ln , t >( )0

The homogeneous solutions are y t e t1( ) = − and y t te t

2( ) = − .

We seek a particular solution ( ) 1 2t t

py y e v te v− −= + to form the system

1 2

1 2

0

( ) ln .

t t

t t t t

e v te v

e v e te v e t

− −

− − − −

′ ′+ =

′ ′− + − =

Solving algebraically for 1v′ and 2v′ , yields 1 lnv t t′ = − and 2 lnv t′ = .

Integrating yields 2 21

1 1ln2 4

v t t t= − + and 2 lnv t t t= − .


2 2 2 2 21 1 1 3ln ln ln .2 4 2 4

t t t t tpy t e t t e t e t t e t e t− − − − − ⎛ ⎞= − + + − = −⎜ ⎟

⎝ ⎠

Thus the general solution is y t c e c te t e tt t t( ) = + + −( )− − −1 2

214

2 3ln .


9. 4 tan 2y y t′′ + =

yh = 1 2

1 2

cos 2 sin 2c t c ty y

+

( )

2 2

1( tan 2 )(sin 2 ) 1 sin 2 1 1 cos 2 1 sec2 cos2

2 2 cos 2 2 cos2 21 ln sec2 tan 2 sin 24

t t t tv dt dt dt t t dtt t

t t t

− −= = − = − = − −

= − + −

∫ ∫ ∫ ∫

2tan 2 cos2 1 cos 2

2 4t tv dt t= = −∫

So yp = y1v1 + y2v2 = ( )1 1cos2 ln sec2 tan 2 sin 2 sin 2 cos24 4

t t t t t t− + − − .

General solution: y(t) = c1 cos 2t + c2 sin 2t + yp.

10. 5 6 cos( )ty y y e′′ ′+ + =

yh = 2 31 2

1 2

t tc e c ey y

− −+

( ) 3

21 5

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

t tt t t t t

t

e ev dt e e dt e e e

e

−

−

−= = = +

−∫ ∫

2

3 22 5

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

t t t t t t tt

e ev dt e e dt e e e e ee

−

−= = − = − =−∫ ∫

So yp = y1v1 + y2v2 = ( ) ( )2 3 2sin( ) cos( ) 2sin( ) sin( ) 2 cos( )t t t t t t t t t te e e e e e e e e e− −+ + − −

= 3 22 sin( ) cos( )t t t te e e e− −−

General solution: y(t) = c1e−2t + c2e−3t + yp.

11. 2secy y t′′ + =

yh = 1 2

1 2

cos sinc t c ty y

+

21 ( sec )sin ( sec tan ) secv t dt t t dt t= − = − = −∫ ∫

22 sec cos sec ln sec tanv t t dt t dt t t= = = +∫ ∫

So, yp = y1v1 + y2v2 = cos sec sin ln sec tant t t t t− + +

= 1 sin ln sec tant t t− + +

General solution: 1 2( ) cos cos sin 1 sin ln sec tany t t c t t t t= + − + +


12. tey yt

′′ − =

yh = 1 2

1 2

t tc e c ey y

−+

11 1 1 1( ) ln2 2 2

ttev e dt dt t

t t−⎛ ⎞

= = =⎜ ⎟⎝ ⎠

∫ ∫

0

2 2

21 1 12 2 2

t t sttt

e e ev e dt dt dst t s

⎛ ⎞= − = − = −⎜ ⎟

⎝ ⎠∫ ∫ ∫

So yp = v1y1 + v2y2 = 0

21 1ln2 2

stt tt

ee t e dss

−− ∫

General solution: y(t) = c1et + c2e−t + yp

Variable Coefficients

13. t y ty y t t2 32 2′′ − ′ + = sin , y t t1( ) = , y t t22( ) =

We begin by dividing the equation by t2 , to get the proper form for using variation of parameters.

′′ − ′ + =yt

yt

y t t2 22 sin .

Substitution verifies that y1 and y2 for a fundamental set of solution to the associated homogeneous equation, so 2

1 2hy c t c t= + ,

we seek a particular solution ( ) 21 2py y v t v t= + ,

where 1v and 2v satisfy the conditions 21 2 0tv t v′ ′+ =

21 22 sinv t v t t′ ′+ =

Solving algebraically for 1v′ and 2v′ , yields 11 sinv t t′ = − and 2 sin .v t′ =

Integrating yields 1 cos sinv t t t= − and 2 cos .v t= −

Thus, y t t t t t t t t tp( ) = − − = −2 2cos sin cos sin .

Hence, the general solution of this equation is y t c t c t t t( ) = + −1 22 sin .

14. t y ty y t t2 2 24 1′′ + ′ − = +b g , y t t12( ) = , y t t2

2( ) = −

We begin by dividing the equation by t2 , to get the proper form for using variation of parameters:

′′ + ′ − = +yt

yt

y t1 4 122.


Substitution verifies that y1 and y2 form a fundamental set of solutions to the associated homogeneous equation, so 2 2

1 2 .hy c t c t−= +

We seek a particular solution ( ) 2 21 2py y v t v t−′= + ,

where υ1 and υ2 satisfy the conditions 2 21 2 0t v t v−′ ′+ =

3 21 22 2 1 .tv t v t−′ ′− = +

Solving algebraically for 1v′ and 2v′ yields 2

11

4tvt

+′ = and ( )3 2

2

1

4

t tv

− +′ = .

Integrating yields 21

1 1ln4 8

v t t= + and 4 62

1 116 24

v t t= − − .

Thus, y t t t t t tp( ) = + − −

14

18 16 24

2 42 4

ln .

Hence, the general solution of this equation is y t c t c t t t tb g = + + +−1

22

2 414

112

ln .

(Notice that the term t2 in yp can be absorbed in the homogeneous solution.)

15. 1 2 1 2−( ) ′′ + ′ − = −( ) −t y ty y t e t , y t t1( ) = , y t et2b g =

We begin by dividing the equation by 1−( )t , to get the proper form for variation of parameters

′′ +−

′ −−

= − −y tt

yt

y t e t

11

12 1b g

Susbtitution verifies that y1 and y2 form a fundamental set of solutions to the associated homogeneous equation, so 1 2

thy c t c e= +

We seek a particular solution ( ) 1 2t

py y v t v e= + ,

where 1v and 2v satisfy the conditions 1 2 0tt v e v′ ′+ =

1 2 2( 1)t tv e v t e−′ ′+ = − −

Solving algebraically for 1v′ and 2v′ yields 1 2 tv e−′ = and 22 2 tv te−′ = − . Integrating

yields 1 2 tv e−= − and 22

1 .2

tv e t− ⎛ ⎞= +⎜ ⎟⎝ ⎠

Thus, ( ) 1 122 2

t t t tpy t te te e e t− − − − ⎛ ⎞= − + + = −⎜ ⎟

⎝ ⎠.

Hence, the general solution of this equation is y t c t c e e tt t( ) = + + −FHIK−

1 212

.


16. ′′ + ′ + −FHIK = −y

ty

ty t1 1 1

4 21 2, ( ) 1/ 2

1 siny t t t−= , ( ) 1/ 22 cosy t t t−=

Substitution verifies that y1 and y2 form a fundamental set of solutions to the associated homogeneous equation, so 1/ 2 1/ 2

1 2sin coshy c t t c t t− −= +

We seek a particular solution ( ) 1/ 2 1/ 21 2sin cospy y v t t v t t− −= + ,

where 1v and 2v satisfy the conditions 1/ 2 1/ 21 2sin cos 0t tv t v− −′ ′+ =

3/ 2 1/ 2 3/ 2 1/ 2 1/ 21 2

1 1sin cos cos sin2 2

t t t t v t t t t v t− − − − −⎛ ⎞ ⎛ ⎞′ ′− + + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

.

Multiplying through by 1/ 2t then solving for 1v′ and 2v′ : 1 cosv t′ = and 2 sin .v t′ = −

1 sinv t= and 2 cos .v t=

Thus, ( ) ( )1/ 2 2 2 1/ 2sin cospy t t t t t− −= + = , and the general solution of this equation is

( ) 1/ 2 1/ 2 1/ 21 2sin cosh py t y y c t t c t t t− − −= + = + + .

Third-Order Theory

17. ( ) ( ) ( ) ( ) ( )L y y p t y q t y r t y f t′′′ ′′ ′= + + + =

Given 1 1 2 2 3 3( )hy t c y c y c y= + + ,

we seek 1 1 2 2 2 3( )py t v y v y v y= + + .

Differentiating yields 1 1 1 1 2 2 2 2 3 3 3 3py v y v y v y v y v y v y′ ′ ′ ′ ′ ′ ′= + + + + +

= 1 1 2 2 3 3v y v y v y′ ′ ′= + + (if we set 1 1 2 2 3 3 0y v y v y v′ ′ ′+ + = ).

Differentiating, again 1 1 1 1 2 2 2 2 3 3 3 3py v y v y v y v y v y v y′′ ′ ′ ′′ ′ ′ ′′ ′ ′ ′′= + + + + +

1 1 2 2 3 3v y v y v y′′ ′′ ′′= + + (if now we set 1 1 2 2 3 3 0y v y v y v′ ′ ′ ′ ′ ′+ + = ).

Differentiating yet again: 1 1 1 1 2 2 2 2 3 3 3 3.py v y v y v y v y v y v y′′′ ′ ′′ ′′′ ′ ′′ ′′′ ′ ′′ ′′′= + + + + +

Substituting yp , ′yp , ′′yp and py′′′ into the L y f( ) = , then regrouping all terms in v1 and v2, we see

that the coefficient of each is 0 because each yi is a solution of L(yi) = 0. Thus we are left with

1 1 2 2 3 3 .py y v y v y v f′′′ ′′ ′ ′′ ′ ′′ ′= + + =

This last equation, together with the two assumptions (in parentheses) that we made while differentiating, gives a system to solve for 1v′ , 2v′ , 3v′ :

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

00

.

y v y v y vy v y v y vy v y v y v f

′ ′ ′+ + =′ ′ ′ ′ ′ ′+ + =′′ ′ ′′ ′ ′′ ′+ + =


We use Cramer’s Rule to solve the system, then integrate to find 1v , 2v , 3v and hence, obtain a

particular solution

( ) 1 1 2 2 3 3py t v y v y v y= + + .

Third-Order DEs

18. ′′′ − ′′ − ′ + =y y y y et2 2

The characteristic equation ( )( )( )1 1 2 0λ λ λ− + − = and has roots 1, –1, and 2. The fundamental set is y et

1 = , y e t2 = − , and y e t

32= . Hence,

y c e c e c eht t t= + +−

1 2 32 .

By variation of parameters, we seek 21 2 3( ) t t t

py t v e v e v e−= + + , as in Problem 17. Hence the

system to solve is

21 2 3

21 2 3

21 2 3

0

2 0

4 .

t t t

t t t

t t t t

e v e v e v

e v e v e v

e v e v e v e

−

−

−

′ ′ ′+ + =

′ ′ ′− + =

′ ′ ′+ + =

Using Cramer’s rule and computing the determinants yields:

2

2 2

2

2 6 ;

4

t t t

t t t t

t t t

e e e

W e e e e

e e e

−

−

−

⎡ ⎤⎢ ⎥

= − = −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

2

2

2 2

1 2

0

0 2

4 3 126

t t

t t

t t t t

t

e e

e e

e e e evW e

−

−

−

⎡ ⎤⎢ ⎥

−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦′ = = = −

−

2

2

2 42

2 2

3 2

0

0 2

4 166

0

0

2 136

t t

t t

t t t tt

t

t t

t t

t t t tt

t

e e

e e

e e e ev eW e

e e

e e

e e e ev eW e

−

−

−−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

−⎢ ⎥⎣ ⎦′ = = =−

⎡ ⎤⎢ ⎥

−⎢ ⎥⎢ ⎥

−⎢ ⎥⎣ ⎦′ = = =−

Hence we obtain 21 2 3

1 1 1 .2 6 3

t tv v e v e−′ ′ ′= − = =

Hence, 21 2 3

1 1 .2 12 3

t ttv v e v e−= − = = −

We get a particular solution of y t te e ept t t( ) = − + −

12

112

13


y t c e c e c e te et t t t t( ) = + + − −−1 2 3

2 12

14

.


19. secy y t′′′ ′+ =

Find yh: yh = r3 + r = 0 r(r2 + 1) = 0 r = 0, ±i

yh = c1 + c2 cos t + c3 sin t

yp = v1 + v2 cos t + v3 sin t

W = 1 cos sin0 sin cos 10 cos sin

t tt tt t

− =− −

1

0 cos sin0 sin cos

sec cos sin cos sinsec sec

sin cos1

t tt t

t t t t tv t t

t t

−− −

′ = = =−

v′ = ln sec tant t+

2

1 0 sin0 0 cos0 sec sin 0 cos cos1 1 1

sec sin1 sec

tt

t t t tvt t t

− −⎛ ⎞′ = = = = −⎜ ⎟− ⎝ ⎠

2v t= −

3

1 cos 00 sin 00 cos sec sin 0 sin1 sin sec

cos sec1 cos

ttt t t tv t t

t t t

−− − −′ = = = − =

−

3 ln cosv t=

y(t) = 1 2 3cos sin ln sec tan cos sin ln cosc c t c t t t t t t t+ + + + − +

20. 9 tan 3y y t′′ ′+ =

Find yh: r3 + 9r = 0 r(r2 + 9) = 0 r = 0 ± 3i

yh = c1 + c2 cos 3t + c3 sin 3t

yp = v1 + v2 cos 3t + v3 sin 3t

W = 1 cos3 sin3

3sin3 3cos30 3sin 3 3cos3 1

9cos3 9sin30 9cos3 9sin3

t tt t

t tt t

t t

−− =

− −− −

= 27


2 21

0 cos3 sin 30 3sin 3 3cos3

tan3 9cos3 9sin 3 cos3 sin 3tan3 tan 3 tan 3(3cos 3 3sin 3 )3sin3 3cos327 27 27 9

t tt t

t t t t tt t tv t tt t

−− −

′ = = = + =−

v1 = ln cos3

27t−

2

1 0 sin30 0 3cos30 tan 3 9sin3 3cos3 tan 3 1 sin 3

27 27 9

tt

t t t tv t− −′ = = = −

v2 = 1 cos327

t

2

3

1 cos3 00 3sin3 00 9cos3 tan 3 3sin3 tan3 1 sin 3

27 27 9 cos3

ttt t t t tv

t

−− −′ = = = −

v3 = ( )21 1 cos 3 1 ln sec3 tan3 sin 3

9 cos3 27t dt t t t

t−

− = − + −∫

y = ( )21 2 3

1 1 sin3cos3 sin 3 ln cos3 cos 3 ln sec3 tan 3 sin 327 27 27

tc c t c t t t t t t+ + − + − + −

Method Choice

21. ( )y y f t′′′ ′− =

We first find the homogeneous solution. The characteristic equation 2( 1)λ λ − = 0 has roots 0, ±1, so the homogeneous solution is 1 2 3

t thy c c e c e−= + + .

(a) 2 ty y e−′′′ ′− = . Because e−t is in yh, we must try yp = a te−t. The method of undetermined

coefficients is straightforward and gives a = 1, so yp = te−t and the general solution can be written 1 2 3( ) t t ty t c c e c e te− −= + + + .

(b) 2siny y t′′′ ′− = . We cannot use undertermined coefficients on sin2 t, so we use variation

of parameters to seek a particular solution of the form yp(t) = v1 + v2et + v3e−t, with the derivatives of v1, v2, and v3 determined from the equations

1 2 3

1 2 3

21 2 3

1 0

0 0

0 sin .

t t

t t

t t

v e v e v

v e v e v

v e v e v t

−

−

−

′ ′ ′+ + =

′ ′ ′+ − =

′ ′ ′+ + =

Discussion continues on next page.


Using Cramer’s rule (as outlined in Problem 18), we obtain

21 sinv t′ = − 2

21 sin2

tv e t−′ = 23

1 sin .2

tv e t′ =

The antiderivative of ′υ1 is easy to find; the other two must be left as integrals

( )11 sin cos2

v t t t= − 22

1 sin2

tv e t dt−= ∫ 23

1 sin2

tv e t dt= ∫ .


( ) ( ) 2 21 2 3

1 1 1sin cos sin sin2 2 2

t t t t t ty t c c e c e t t t e e t dt e e t dt− − −= + + + − + +∫ ∫ .

(c) tany y t′′′ ′− = . As in Part (b) , we must use variation of parameters to find yp, with

1 2 3

1 2 3

1 2 3

1 0

0 0

0 tan .

t t

t t

t t

v e v e v

v e v e v

v e v e v t

−

−

−

′ ′ ′+ + =

′ ′ ′+ − =

′ ′ ′+ + =

Using Cramer’s rule (as outlined in Problem 18), to solve these equations we find

1 tanv t′ = 21 tan2

tv e t−′ = 31 tan .2

tv e t′ =

The antiderivative of ′υ1 is easy to find; the other two must be left as integrals

1 ln cosv t= 21 tan2

tv e t dt−= ∫ 31 tan 2

tv e t dt= ∫ .


( ) 1 2 31 1 1ln cos tan tan .2 2 2

t t t t t ty t c c e c e t e e t dt e e t dt− − −= + + + + +∫ ∫

Parts (b) and (c) demonstrate the power of graphical methods because the algebraic expressions for y(t) are pretty meaningless. It is easier and more informative to use DE software to approximate solutions of this equation in ty space than it is to pursue the analytical formula for the solution. The figures show curves for several initial conditions to show the variety that can occur. For any IVP there would be only one solution.

Note: We used a 3D graphc DE solver with the following equations for ( ) :y y f t′′′ ′− = y x′ = x y′ = y x z′′ ′= = relisted as y x′ = ( )y z x f t′′′ ′= = + ( )z x f t′ = +


(b) f(t) = sin2 t.

The expression for y(t) on the previous page can be further evaluated using the identity

sin2 t = 1 (1 cos2 ),2

t−

but solution behavior is more easily seen on a graph of y(t).

(c) f(t) = tan t

The expression for y(t) on the previous page is even more complicated than that for part (b); again, solution behavior is more readily understood with a graph of y(t).

Green’s Function Representation

22. ( )y y f t′′ + =

We know that y t1 = cos and y t2 = sin are the solutions of the corresponding homogeneous

equation. Their Wronksian is

W y y tt tt t1 2 1,

cos sinsin cos

a f( ) =−

= .

which is makes it easy to use the suggested variation of parameters formulas

( ) ( )

( )( )2

11 2

sin( ) ( ),,

y t f tv t f t

W y y t−

′ = = − ( ) ( )

( )( )1

21 2

cos( ) ( ).,

y t f tv t f t

W y y t′ = =

Integrating yields

( ) ( )1 0sin

tv s f s ds= −∫ ( ) ( )2 0

cos .t

v s f s ds= ∫

Hence, ( ) 1 1 2 2py t y v y v′ ′= +

[ ]0 0

0

0

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

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

sin( ) ( ) .

t t

t

t

t s f s ds t s f s ds

t s t s f s ds

t s f s ds

= − +

= − +

= −

∫ ∫

∫

∫

Green Variation

23. The homogeneous solutions are y et1 = and y e t

2 = − . We seek a particular solution of the form

1 1 1 2py v y v y= + , where 1v′ and 2v′ satisfy

( )1 2

1 2

0

.

t t

t t

e v e v

e v e v f t

−

−

′ ′+ =

′ ′− =


Adding and subtracting the equations and solving yields

( )112

tv e f t−′ = ( )212

tv e f t′ = −

Integrating gives ( )1 0

12

t sv e f s ds−= ∫ ( )2 0

12

t sv e f s ds= − ∫ .

Hence, 1 1 2 2py v y v y= +

( ) ( )

( )

( ) ( )

0 0

0

0

1 12 2

2

sinh .

t tt s t s

t s t st

t

e e f s e e f s ds

e ef s ds

t s f s ds

− −

− − +

= −

⎛ ⎞−= ⎜ ⎟

⎝ ⎠

= −

∫ ∫

∫

∫

Green’s Follow-Up

24. From the Leibniz Rule in multivariable calculus we have the following result:

For a continuous function g(t,s),

( )0 0

, lim ( , ) ( , ) .t r

r t

d gg t s ds g r r t s dsdt t→

∂⎡ ⎤= +⎢ ⎥∂⎣ ⎦∫ ∫

In Problem 22, the solution of the equation ′′ + = ( )y y f t is

y t t s f s dst

( ) = −( ) ( )z sin0

Differentiating yields

( ) ( ) ( ) ( ) ( ) ( )0 0

sin cos cost t

y t t f t t s f s ds t s f s ds′ = − + − = −∫ ∫

and

( ) ( ) ( ) ( ) ( ) ( ) ( )0 0

cos sin sin .t t

y t t f t t s f s ds f t t s f s ds′′ = − − − = − −∫ ∫

Hence,

′′ + = ( ) − −( ) ( ) + −( ) ( ) = ( )z zy y f t t s f s ds t s f s ds f tt tsin sin

0 0.

Suggested Journal Entry I

25. Student Project

Suggested Journal Entry II

26. Student Project

SECTION 4.6 Forced Oscillations 409

4.6 Forced Oscillations

Mass-Spring Problems

1. 2 6cosx x x t′′ ′+ + =

Find xh: r2 + 2n + 1 = 0 ⇒ (r + 1)2 = 0 ⇒ r = −1

xh = c1e−t + c2te−t.

Find xp: xp = A cos t + B sin t, sin cos ,px A t B t′ = − + cos sinpx A t B t′′ = − −

2 cos sinp p px x x A t B t′′ ′+ + = − −

2( cos sin ) cos sin

2 cos 2 sin 6cos

B t A tA t B t

B t A t t

+ −+ +

− =

⇒ 2B = 6, −2A = 0 ⇒ A = 0, B = 3

xp = 3 sin t

x(t) = xh + xp = c1e−t + c2te−t + 3 sin t

xss = 3 sin t = 3cos2

t π⎛ ⎞−⎜ ⎟⎝ ⎠

Amplitude C = 3; phase shift 2

δ πβ= radians

2. 2 3 cos3x x x t′′ ′+ + =

Find xh: r2 + 2r + 3 = 0 ⇒ r = −1 ± 2i

xh = ( )1 2cos 2 sin 2te c t c t− +

Find xp: xp = A cos 3t + B sin 3t, 3 sin 3 3 cos3 ,px A t B t′ = − + 9 cos3 9 sin3px A t B t′′ = − −

2 3p p px x x′′ ′+ + = 9Acos3 9 sin 3 ) 2(3 cos3 3 sin 3 ) 3( cos3 sin3 )

( 6 6 )cos3 + ( 6 6 )sin3 cos3

t B tB t A t

A t B t

A B t A B t t

− −+ −+ +

− + − − =

⇒ −6A + 6B = 1, −6A − 6B = 0

⇒ A = 1 1,12 12

B− = , so


xp = 1 1cos3 sin 312 12

t t− +

x(t) = xh + xp = ( )1 21 1cos 2 sin 2 cos3 sin3

12 12te c t c t t t− + − +

xss = 1 1cos3 sin 312 12

t t− + = 2 3cos 312 4

t π⎛ ⎞−⎜ ⎟⎝ ⎠

Amplitude C = 212

; phase shift 4

δ πβ= radians

3. 2 3 4cos8x x t′′ + =

Find xh: 2r2 + 3 = 0 ⇒ r2 32

− ⇒ r = 32

i±

⇒ xh = 1 23 3cos sin2 2

c t c t+

Find xp: xp = A cos 8t, 8 sin8 ,px A t′ = − 64 cos8px A t′′ = −

2 3 2( 64 cos8 )

3( cos8 )

125 cos8 4cos8

p px x A t

A t

A t t

′′ + = −

+

− =

⇒ −125A = 4

⇒ A = 4125

−

⇒ xp = 4 cos8125

t−

x(t) = xh + xp = 1 23 3 4cos sin cos82 2 125

c t c t t+ −

xss = 4 4cos8 cos(8 )125 125

t t π− = −

Amplitude C = 4125

; phase shift 8

δ πβ= radians


4. 1 52 2 cos2 2

x x x t′′ ′+ + =

Find xh: 2r2 + 2r + 12

= 0 ⇒ r = 12

−

xh = (1/ 2) (1/ 2)1 2

t tc e c te− −+

Find xp: xp = A cos t + B sin t, sin cospx A t B t′ = − + , cos sinpx A t B t′′ = − −

12 2 2( cos sin )2

2( cos sin ) 1 ( cos sin )

23 3 5 2 cos 2 sin cos2 2 2

p p px x x A t B t

B t A t

A t B t

A B t A B t t

′′ ′+ + = − −

+ −

+ +

⎛ ⎞ ⎛ ⎞− + + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ 3 5 32 , 2 02 2 2

A B A B− + = − − =

⇒ A = 3 4, 5 5

B− =

xp = 3 4cos sin5 5

t t− +

x(t) = xh + xp = (1/ 2) (1/ 2)1 2

3 4cos sin5 5

t tc e c te t t− −+ − +

xss = 3 4cos sin5 5

t t− + = cos( 2.2)t −

Amplitude C = 1; phase shift 2.2δβ≈ radians

5. 2 2 2cosx x x t′′ ′+ + =

Find xh: r2 + 2r + 2 = 0 ⇒ r = −1 ± i

xh = ( )1 2cos sinte c t c t− +

Find xp: xp = A cos t + B sin t, sin cospx A t B t′ = − + cos sinpx A t B t′′ = − −

2 cos sin

2( cos sin )

2( cos sin )

( 2 )cos ( 2 )sin 2cos

p p px x x A t B t

B t A tA t B t

A B t A B t t

′′ ′+ + = − −

+ −+ +

+ + − + =

⇒ A + 2B = 2, −2A + B = 0


⇒ A = 2 4, 5 5

B =

xp = 2 4cos sin5 5

t t+

x(t) = xh + xp = 1 22 4( cos sin ) cos sin5 5

te c t c t t t− + + +

xss = 2 4cos sin5 5

t t+ = 2 cos( 1.1)5

t −

Amplitude C = 25

; phase shift 1.1δβ≈ radians

6. 4 5 2cos2x x x t′′ ′+ + =

Find xh: r2 + 4r + 5 = 0 ⇒ r = −2 ± i

xh = 21 2( cos sin )te c t c t− +

Find xp: xp = A cos 2t + B sin 2t, 2 sin 2 2 cos 2 ,px A t B t′ = − + 4 cos2 4 sin 2px A t B t′′ = − −

4 5 4 cos2 4 sin 2 )

4(2 cos2 2 sin 2 )

5( cos2 sin 2 )

( 8 )cos 2 ( 8 )sin 2 2cos2

p p px x x A t B t

B t A tA t B t

A B t A B t t

′′ ′+ + = − −

+ −+ +

+ + − + =

⇒ A + 8B = 2, −8A + B = 0

⇒ A = 2 ,65

B = 1665

xp = 2 cos265

t + 16 sin 265

t

x(t) = xh + xp = 21 2

2 16( cos sin ) cos 2 sin 265 65

te c t c t t t− + + +

xss = 2 16cos2 sin 265 65

t t+ = 2 cos(2 1.4)65

t −

Amplitude C = 265

; phase shift 0.73δβ≈ radians


Pushing Up

7. m = 1 , 2.5, 64

b k= =

1 2.5 6 2cos24

x x x t+ + =

The IVP is 10 24 8cos 2 , (0) 2, (0) 0.x x x t x x+ + = = − =

Find xh: r2 + 10r + 24 = 0

(r + 4)(r + 6) = 0 r = −4, −6

xh = c1e−4t + c2e−6t

Find xp: xp = A cos 2t + B sin 2t

px = −2A sin 2t + 2B cos 2t

4 cos2 4 sin 2px A t B t= − −

10 24 4 cos 2 4 sin 2 10( 2 sin 2 2 cos2 ) 24( cos2 sin 2 )p p px x x A t B t A t B t A t B t+ + = − − + − + + +

coeff. of cos 2t: −4A + 24A + 20B = 8 20A + 20 B = 8

coeff. of sin 2t: −4B + 24B − 20A = 0 −20A + 20B = 0

∴ A = B = 15

xp = 1 1cos2 sin 25 5

t t+

Therefore

x(t) = xh + xp = 4 61 2

1 1cos2 sin 25 5

t tc e c e t t− −+ + +

4 61 2

2 24 6 sin 2 cos 25 5

t tx c e c e t t− −= − − − +

Substituting initial conditions gives

1 2

0 2

1(0) 25

2(0) 0 4 65

x c c

x c c

⎫= − = + + ⎪⎪⎬⎪= = − − +⎪⎭

⇒ 1 234 23, 5 5

c c= − =

Thus x(t) = 4 634 23 1 1cos2 sin 25 5 5 5

t te e t t− −− + + + .


Pulling Down

8. m = 16 1 , 6, 1632 2

b k= = =

1 6 16 4cos 42

x x x t+ + =

The IVP is 12 32 8cos 4 , (0) 1, (0) 0x x x t x x+ + = = = .

Find xh: r2 + 12r + 32 = 0

(r + 4)(r + 8) = 0 r = −4, −8

xh = c1e−4t + c2e−8t

Find xp: xp = A cos 4t + B sin 4t

px = −4A sin 4t + 4B cos 4t

16 cos4 16 sin 4px A t B t= − −

12 32 16 cos4 16 sin 4 12( 4 sin 4 4 cos 4 ) 32( cos4 sin 4 )

8cos 4p px x x A t B t A t B t A t B t

t

+ + = − − + − + + +

=

Coeff. of cos 4t: −16A + 32A + 48B = 8 16A + 48B = 8

Coeff. of sin 4t: −16B + 32B − 48A = 0 −48A + 16B = 0

∴ A = 120

B = 320

xp = 1 3cos4 sin 420 20

t t+

Therefore

x(t) = xh + xp = 4 81 2

1 3cos 4 sin 420 20

t tc e c e t t− −+ + +

4 81 2

1 34 8 sin 4 cos45 5

t tx c e c e t t− −= − − +

Substituting intial conditions,

1 2

1 2

1120

30 4 85

c c

c c

= + +

= − − +

1 2

1 2

1920

3220

c c

c c

⎫+ = ⎪⎪⎬⎪+ =⎪⎭

⇒ 1 27 4, 4 5

c c= = −

Thus x(t) = 4 87 4 1 3cos4 sin 44 5 20 20

t te e t t− −− − + + .


Mass-Spring Again

9. (a) The mass is 100 kgm = ; gravitational force (weight) acting on the spring is ( )100 9.8 980mg = = newtons. Because the weight stretches the spring by

20 cm 0.2 m= , we have

980 4900 nt m0.20

k = = .

(b) The initial-value problem for this mass is

49 0x x+ = , ( )0 0.40x = , ( )0 0x = .

Solving we write the transient solution in polar form

( ) ( ) ( )0cos cos 7x t C t C tω δ δ= − = −

where the circular frequency is 0 7ω = radians per second. Using the initial conditions

gives

( )( )0 cos 0.4

0 7 sin 0

x C

x C

δ

δ

= =

= − =

or 0δ = , 0.4C = . Hence,

( ) ( )0.4cos 7x t t= .

(c) Amplitude: 0.4C = meter; period: 2 seconds7

T π= .

(d) If 500b = , then

( )( )2 4 250,000 4 100 4900 0b mk− = − < .

The system is underdamped.

(e) 100 500 4900 0x x x+ + = has characteristic equation

2 5 49 0r r+ + = ,

which has roots 1,25 1 1712 2

x i= − ± . Hence, the general solution is

( ) 5 21 2

171 171cos sin2 2

tx t e c t c t−⎡ ⎤⎛ ⎞ ⎛ ⎞

= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦.

Using the initial conditions ( )0 0.4x = , ( )0 0x = gives

( )

( )

1

2 1

0 0.4

171 50 02 2

x c

x c c

= =

= − =


which implies 1 0.4c = , 22 171

171c = . Hence, the solution is

( ) 5 2 171 2 171 1710.4cos sin2 171 2

tx t e t t− ⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠.

Adding Forcing

10. We change the unforced equation in Problem 9 to the forced equation 100 500 4900 100cos fx x x tω+ + = .

(a) ( )( )2 24 500 4 100 4900 0b mk− = − < , so the system is underdamped. From Equation (21)

in the text, the amplitude is a maximum when the forcing frequency is

( )

2 2

2 24900 500 6.04 rad sec1002 2 100

fk bm m

ω = − = − = .

(b) Given 7fω = , we have seen that 0 7ω = , 100m = , 500b = , and hence tanδ becomes

infinite, so that 2πδ = radians. Hence, by Equation (17)

( )( ) ( )

( )

( ) ( ) ( )

02 22 2 2

0

2 2 2

cos

100 cos 7 0.029cos 7 .2 2100 49 49 500 7

ss f

f f

Fx t tm b

t t

ω δω ω ω

π π

= −− +

⎛ ⎞ ⎛ ⎞= − ≈ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− + ⋅

See the graph for the solution of the IVPproblem. You can see that the steadystate appears to be

( ) 0.029cos 72ssx t t π⎛ ⎞≈ −⎜ ⎟

⎝ ⎠.

(c) The undamped equation is

100 4900 100cos7x x t+ = .

The particular solution now has the form ( ) ( )cos 7ssx t Ct t δ= −

or ( ) ( )cos7 sin 7ssx t t A t B t= + .


Electric Analog

11. From Problem 10 we found the differential equation for the mechanical system to be

100 500 4900 100cos fx x x tω+ + =

or

5 49 cos fx x x tω+ + = .

So if 4 ohmsR = , then the equivalent electrical equation (making one equation a constant multi-

ple of the other equation)

01 cos fLQ RQ Q V tC

ω+ + =

would be:

( ) ( )4 4 4 45 49 cos5 5 5 5 fQ Q Q tω+ + = .

This means we have

( )

0.80 henries1 139.2 0.025 farads

39.20.80cos .f

L

CC

V t tω

=

= ⇒ = ≈

=

Damped Forced Motion I

12. 8 36 72cos6x x x t+ + =

The given characteristic equation has roots 4 2 5i− ± , hence in the long run the homogeneous

equation solution always decays to zero. We are only interested in a particular solution, and in this case that solution is

cos6 sin 6x A t B t= + .

Differentiating and substituting into the differen-tial equation gives

0A = , 32

B = .

Hence, the steady-state solution is given by

( ) 3 sin 62ssx t t= .

The graph of the steady-state solution is shown.

2 31

–2

2x

t

1

–1


Damped Forced Motion II

13. The initial-value problem

4 20 20cos2x x x t+ + = , ( ) ( )0 0 0x x= = ,

has

( ) ( )21 2cos4 sin 4t

hx t e c t c t−= + ,

and

( ) cos2 sin 2px t A t B t= + .

Substituting px into the differential equation we find 1A = , 12

B = , so

1cos2 sin 22px t t= + .

Substituting the general solution into the initialconditions ( ) ( )0 0 0x x= = , we find

( ) 2 3 1cos4 sin 4 cos 2 sin 24 2

tx t e t t t t− ⎛ ⎞= − + + +⎜ ⎟⎝ ⎠

.

The steady-state portion of the solution is shown.(See figure.)

4 632

Š1.5

1.5

1 5t

0.5

Š0.5

x t( )

Calculating Charge

14. 4 100 10cos4 , (0) 0, (0) 0 Q Q t Q Q′+ = = =

Find Qh : 1 24 100 0 25 0 cos5 sin 5hQ Q Q Q Q c t c t+ = ⇒ + = ⇒ = +

Find Qp : Qp has the form A cos 4t + B sin 4t.

Substitution in 4 100 10cos 4p pQ Q t+ = leads to A = 518

, B = 0 and so Qp = 5 cos 4 .18

t

Thus Q = Qh + Qp = c1 cos 5t + c2 sin 5t + 5 cos418

t .

The initial conditions Q(0) = 0 and (0) 0Q′ = give us c1 = 5 ,18

− c2 = 0

and the solution of the IVP is Q(t) = 5 5cos5 cos 418 18

t t− + .


Charge and Current

15. 12 100 12cos10 , (0) 0, (0) 0 Q Q Q t Q Q′+ + = = =

(a) Find Qh: 2 61 212 100 0 6 8 ( cos8 sin8 ).t

hr r r i Q e c t c t−+ + = ⇒ = − ± ⇒ = +

Find Qp: Qp has the form A cos 10t + B sin 10t.

Substitution in 4 12 100 12cos10p p pQ Q Q t+ + = leads to A = 0, B = 110

and so Qp = 1 sin10 .10

t

Thus Q = Qh + Qp = 61 2

1( cos8 sin8 ) sin10 .10

te c t c t t− + +

The initial conditions Q(0) = 0 and (0) 0Q = give us c1 = 0 c2 = 18

−

and the solution of the IVP is Q(t) = 61 1sin8 sin108 10

te t t−− + .

(b) I(t) = 6 3( ) sin8 cos8 cos10 .4

tQ t e t t t− ⎛ ⎞= − +⎜ ⎟⎝ ⎠

True/False Questions

16. True

The steady-state solution, being a particular solution, has the form A cos ωf t + B sin ωf t, where the forcing function is Fo cos ωf t. The steady-state solution can be written in the form

xss = C cos(ωf t − δ).

Hence, the frequency of the steady-state is the same as that of the forcing function.

17. False

The amplitude of the steady-state is a function of the frequency of the forcing function.

In fact, 02 2 2 2 2

0

( )( ) ( )

ff f

FAm b

ωω ω ω

=− +

.

We can see that A(ωf) → 0 as ωf → ∞ and that

0Fk

as ωf → 0.


Beats

18. The identity

( ) ( )cos cos 2sin sinA B A B A B+ − − = −

may be used here. In this case, if 2A t= , B t= , and we have 3A B t+ = , A B t− = . Hence,

cos3 cos 2sin 2 sint t t t− = − .

The Beat Goes On

19. The trigonometric identity

( ) ( )sin sin 2sin cosA B A B B A+ − − =

may be used here. Let 3A t= and B t= . From this we get 3A B t+ = and 2A B− = . Hence,

3 3sin 3 sin 2sin cos

2 22sin cos2 .

t t t tt t

t t

− +⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

Steady State

Note: We must be careful in finding the phase angle using the formula

1tan BA

δ −=

because we don’t know in which quadrant δ lies using 1tan BA

δ −= . The value of δ you get

might be π units different from the correct value. Unless you know by some other means in which quadrant δ lies, it is best to use the two equations

cosC Aδ = , sinC Bδ = .

A good rule of thumb is to think of the AB plane; when both A, B are positive δ will be in the first

quadrant (i.e., δ between 0 and 2π ), but when both A and B are negative δ will be in the third

quadrant, and so on.


20. 4 4 cosx x x t+ + =

The homogeneous solution to the equation is

( ) 2 21 2

t thx t c e c te− −= + .

We use the method of undetermined coefficients to find

( ) cos sinpx t A t B t= + .

Differentiating, we get

( )( )

sin cos

cos sin .p

p

x t A t B t

x t A t B t

′ = − +

′′ = − −

Substituting into the equation gives the equation

( ) ( )4 4 4 4 cos 4 4 sin cosp p px x x A B A t B A B t t′′ ′+ + = − + + + − − + = .

Hence 325

A = , 425

B = with the particular solution

( ) 3 4cos sin25 25px t t t= + .

Nothing in xp dies off with time; this is our steady-state solution. Putting this in polar form

2 2

1

3 4 5 125 25 25 5

4tan 0.93 radians.3

C

δ −

⎛ ⎞ ⎛ ⎞= + = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ≈

Hence, the steady-state solution is

( ) ( )( ) 0.20cos 0.93p ssx t x t t= = − .

21. 2 2 2cosx x x t+ + =

The roots of the characteristic equation are 1 i− ± . We use for the particular solution:

( ) ( )cos sin cospx t A t B t C t δ= + = − .


Another approach to px is to note that 0 2F = ,

0 2ω = , 1fω = , 1m = , 2b = and simply

substitute these numbers into the text solution tofind

( ) ( )2 cos5ssx t t δ= −

where the phase angle is ( )1tan 2 1.1δ −= ≈ radi-

ans.

22. 4cos3x x x t+ + =

The roots of the characteristic equation are 1 32 2

i− ± . Notice that, as in previous problems, there

will be an 2te− involved in all of the terms in the homogeneous equation, so hx is transient, and none of these terms will be involved in px . We move on to find a particular solution, using the

method of undetermined coefficients. Let

( ) cos3 sin3px t A t B t= + .

Then we have

3 sin3 3 cos3px A t B t′ = − + , ( ) 9 cos3 9 sin3px t A t B t′′ = − − .

Hence,

( ) ( )8 3 cos3 3 8 sin3 4cos3p p px x x A B t A B t t′′ ′+ + = − + + − − = .

Solving we get

3273

A = − , 1273

B = .

In polar coordinates we have

2 2 1168 473 73

12arctan 2.78 radians.32

C A B

δ

= + = =

⎛ ⎞= ≈⎜ ⎟−⎝ ⎠

Hence, the steady-state solution is

( ) ( )4( ) cos 3 2.7873p ssx t x t t= = − .


Resonance

23. The differential equation is given by

12 16cosx x tω+ = .

The circular frequency is

0 12 2 3km

ω = = = radians per second,

the frequency is

01 3fπ

= oscillations per second,

and the period of oscillations is

3T π= seconds.

24. If resonance exists, the input frequency fω is the same as the natural frequency 0 2 3ω = (see

Problem 23). Hence, we have the initial-value problem

( )12 16cos 2 3x x t+ = , ( ) ( )0 0 0x x= = .

This equation has homogeneous solution

( ) ( ) ( )1 2cos 2 3 sin 2 3hx t c t c t= + .

To find a particular solution we seek a function of the form

( ) ( )cos 2 3 sin 2 3px At t Bt t= + .

Differentiating and substituting into the differential equation yields

0A = , 16 3434 3

B = = ,

so the general solution is

( ) ( ) ( ) ( )1 24 3cos 2 3 sin 2 3 sin 2 3

3x t c t c t t t= + + .

Substituting this into ( ) ( )0 0 0x x= = yields 1 0c = , 2 0c = . Hence, the solution to the IVP is

( ) ( )4 3 sin 2 33

x t t t= .


Ed’s Buoy

25. (a) Simple harmonic motion with

0

2000 125 62.5 slugs32 2

2 5 seconds,

m

T πω

= = =

= =

hence

025

km

πω = =

or

2

2 20

125 4 102 25

k m πω π= = × = .

We measure the displacement of the buoy ( )x t from the water level with ( ) 0x t =

corresponding to the position of the buoy with 4 2 32+

= feet being above water.

Because the forced equation in rough seas has an amplitude of 3 feet and a period

of 7 seconds, the frequency of the forced response is 27π . We therefore get the equation

2 262.5 10 3cos7

tx x ππ+ = .

We are interested in the steady-state solution of this equation, hence we use the method of undetermined coefficients to find a particular solution. In this case we let

2 2cos sin7 7p

t tx A Bπ π= + .

We now differentiate and substitute into the equation yielding 249

80A

π= , 0B = . Hence,

we have

( ) 249 2 2( ) cos 0.06cos

7 780p sst tx t x t π π

π= = ≈ .

[Although no friction term has been included in the preceding DE, there will in reality be such a term, so the homogeneous solution would go to zero leaving only the oscillation

( )ssx t .]


(b) The steady-state solution never varies more than 249 0.06

80π≈ feet from its equilibrium

position 3 feet above the level water line. The steady-state solution has the buoy moving in phase with the waves so when a 3-foot wave crest hits, the buoy’s height above sea level is approximately 3.06 feet. Thus the buoy is always at least 0.06 feet above the water and is never submerged.

General Solution of the Damped Forced System

26. (a) We know the form of the particular solution is

( ) cos sinss f fx t A t B tω ω= + .

Substituting this into the equation 0 cos fmx bx kx F tω+ + =

and simplifying, we find

( ) ( )2 20cos sin cosf f f f f f fk m A b B t k m B b A t F tω ω ω ω ω ω ω⎡ ⎤ ⎡ ⎤− + + − − =⎣ ⎦ ⎣ ⎦ .

Setting the coefficients of the sine and cosine terms equal, yields the two equations

( )( )

20

2 0.

f f

f f

k m A b B F

k m B b A

ω ω

ω ω

− + =

− − =

Solving, we obtain

( )( )

( )

20

22 2 2

022 2 2

.

f

f f

f

f f

F k mA

k m b

F bB

k m b

ω

ω ω

ω

ω ω

−=

− +

=− +

(b) From part (a) we have

( )( )

( )2022 2 2

cos sinss f f f f

f f

Fx t k m t b tk m b

ω ω ω ωω ω

⎡ ⎤= − +⎣ ⎦− +.

Rewriting this in polar form, yields

( )( )

( )022 2 2

cosss f

f f

Fx t tk m b

ω δω ω

= −− +

,

with ( )2 2

0

tan f

f

b

m

ωδ

ω ω=

−. From this equation it can be seen that the long-term response

of the system is oscillatory with the same frequency fω as the forcing term, but with a

phase lag.


Phase Portrait Recognition

27. 0.3 cosx x x t+ + =

(C) We have damping but we also have a sinusoidal forcing term. Hence, the homogeneous solution goes to zero and particular solutions consist of sines and cosines, which give rise to circles in the phase plane. Therefore, starting from the origin ( ) ( )0 0 0x x= = we get a curve that

approaches a circle from the inside.

28. 0x x+ =

(A) The equation models the undamped harmonic oscillator, which has circular trajectories.

29. cosx x t+ =

(D) This equation has resonance so the trajectories in phase space spiral to infinity.

30. 0.3 0x x x+ + =

(B) The system is unforced but damped, and hence trajectories must approach ( ) ( )0 0 0x x= = .

Matching 3D Graphs

31. (a) E (b) B (c) C, D (d) A, C (e) B, D, E (f) C, D (g) A

Mass-Spring Analysis I

32. (a) 4cos 4 3sin 4hx t t= − (b) The amplitude of 5hx = .

(c) The amplitude (time-varying) of px is 5t .

(d) 5 sin 4px t t= ; px will be unchanged.

(e) Because 0fω ω= , 0 4 k km

ω = = = , 16k = .

(f) The system is in a state of pure resonance because 0 fω ω= . The mass will oscillate with

increasing amplitude.

Electrical Version

33. (a) 4cos4 5sin 4hQ t t= −

(b) The amplitude of the transient solution is 2 24 5 41A = + = (c) 6 cos 4s pQ Q t t= = (d) 6 cos4pQ t t=

(e) 1 14LC C

= = , 116

C =

(f) The charge on the capacitor will oscillate with ever-increasing amplitude due to pure resonance.


Mass-Spring Analysis II

34. (a) 2 23 cos 2 sint thx e t e t− −= −

(b) From the exponential function 2te− we see that 2

bm− = −2. Hence if m = 1, 4b = .

(c) Underdamped (d) The amplitude (time-varying) of 213 t

hx e−= .

(e) ( )2 cos 5ss px x t δ= = −

(f) 5 rad secfω = , 4 1612

kβ −= = so 5 Nt mk = , and 0 5 rad secω = . From For-

mula (19), we obtain ( ) ( )

0 02 22

28005 5 4 5

F F= =

− + ⋅. Therefore 0 40 NtF = .

Perfect Aim

35. (a) The dart is fired straight at the target with initial velocity vo.

Let yD denote the vertical position of the dart at time t. Dy g′′ = − Dy gt c′ = − +

(0) sinD oy v θ′ = so 0 sinDy gt v θ′ = − +

Integrating: yD = 20

1 ( sin )2

gt v t cθ− + +

(0) 0Dy = so 20

1 ( sin )2Dy gt v tθ= − +

Now consider the target. Let yT denote the vertical position of the target at time T. The

initial conditions are yT(0) = y0 and (0)Ty′ = 0. By similar calculations, yT = y0 − 12

gt2.

(b) To find the time t1 when the heights of dart and target are equal, set yT(t) = yD(t).

Then yo − 2 20

1 1sin2 2

gt v t gt= − so that T1 = 0

0 siny

v t and x1 = 0 0

00

( cos )sin tany yv

vθ

θ θ⎛ ⎞

=⎜ ⎟⎝ ⎠

However, tan θ = 0 0

0 1

y yx x

= so that x1 = x0 (i.e., the dart hits the target).

(c) Substituting t1 into either equation for the height of the dart or the target at impact yields yT.

yT = y0 − 20

202( sin )gy

v θ

Simplifying by using the diagram so that sin θ = 0yd

, we obtain

yT = y0 − 2

0

12

dgv

⎛ ⎞⎜ ⎟⎝ ⎠

.

2 20 0d x y= +


Extrema of the Amplitude Response

36. We write

( )( )

0 02 2 22 2 2

2 2

/f

f ff f

F F mAk bk m bm m

ωω ω ω ω

= =⎡ ⎤⎛ ⎞ ⎛ ⎞− + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

.

Differentiating A with respect to fω , we find

( )( )

( ) ( )

222

03 22 22 2

22

ff

f

f f

k bmm m

A Fk bm m

ωω

ω

ω ω

⎛ ⎞ ⎡ ⎤− − −⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠′ = ×⎡ ⎤

− +⎢ ⎥⎣ ⎦

from which it follows that ( ) 0fA ω′ = if and only if 0fω = or

2

22fk bm m

ω = − .

When 2 2b mk> , fω is not real. Hence ( ) 0fA ω′ = only when 0fω = . In this case

( )fA ω damps to zero as fω goes from 0 to ∞. It is clear then that the maximum of ( )fA ω

occurs when 0fω = and has the value

( ) 10Ak

= .

When 2 2b mk< , then fω is real and positive. It is easy using the sign of the derivative to see

that the maximum of ( )fA ω occurs at

2

22fk bm m

ω = − .

Evaluating the amplitude response at this value of fω yields the expression

0max 2

24

FAk bb m m

=−

.


37. Student Project

SECTION 4.7 Conservation and Conversion 429

4.7 Conservation and Conversion

Total Energy of a Mass-Spring

1. 0x x+ = , ( )0 1x = , ( )0 4x = −

The total energy of the system is 2 21 12 2

E mx kx= + . Here 1m = , 1k = , so

2 21 12 2

E x x= + .

Because the system is conservative it does not change over time. Initially we have ( )0 1x = ,

( )0 4x = − , so the initial energy of this system is

( ) ( )21 1 174 12 2 2

E = − + = ,

which remains constant in time.

Nonconservative Mass-Spring System

2. 2 26 0x x x+ + = , ( )0 1x = , ( )0 4x =

(a) The solution of the IVP is

( ) ( ) ( )sin 5 cos 5tx t e t t− ⎡ ⎤= +⎣ ⎦ .

At time 5

t π= we have

5

5x e ππ −⎛ ⎞ = −⎜ ⎟⎝ ⎠

.

Also

( ) ( ) ( ) ( ) ( )sin 5 cos 5 5cos 5 5sin 5t tx t e t t e t t− −⎡ ⎤ ⎡ ⎤= − + + −⎣ ⎦ ⎣ ⎦ ,

so

5 5 55 45

x e e eπ π ππ − − −⎛ ⎞ = − = −⎜ ⎟⎝ ⎠

.

(b) Because 1m = , 26k = , we have

( ) ( )2 2

2 5 2 5 2 51 1 1 116 26 215 2 5 2 5 2 2

E m x k x e e eπ π ππ π π − − −⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

.


(c) Because the initial energy of the system was

( ) ( ) ( ) ( ) ( ) ( )2 2 2 21 1 1 10 0 0 1 4 26 1 21 joules or ergs2 2 2 2

E m x k x⎡ ⎤ ⎡ ⎤= + = ⋅ + ⋅ =⎣ ⎦ ⎣ ⎦

the energy loss ( )2 5 2 521 21 21 1e eπ π− −= − = − .

General Formula for Total Energy in an LC-Circuit

3. 1 0LQ QC

+ = , ( ) 00Q Q= , ( ) 00I I=

The total energy of this LC system is the constant value

( ) 2 2 2 20 0

1 1 1 12 2 2 2

E t LQ Q LI QC C

= + = + .

Energy in an LC-Circuit

4. 1 0LQ QC

+ = , ( ) 00Q Q= , ( ) 00I I=

The total energy of this LC system is the constant value

( ) ( ) ( )2 2 2 2 2 20 0

1 1 1 1 1 164 1 4 1302 2 2 2 2 2

E t LQ Q LI QC C

= + = + = + = .

Energy Loss in LRC-Circuit

5. 0LQ RQ CQ+ + = , ( ) 00Q Q= , ( ) 00I I=

We are given 1L = henry, 1R = ohm, 4C = farads, 0 0Q = coulomb, 0 2I = amps. Hence, the

IVP is

0.25 0Q Q Q+ + = , ( )0 0Q = , ( )0 2I = ,

whose solution is given by

( )( ) ( ) ( )

2

2 2 2

2

2 2 .

t

t t t

Q t te

I t Q t e te e t

−

− − −

=

= = − = − −

Hence, the initial energy is

( ) ( )( ) ( )2 2 20 0

1 1 1 10 1 2 0 22 2 2 8

E LI QC

= + = + = joules.

At time t the energy is

( ) ( ) ( ) ( ) ( ) ( ) ( )2 22 2 2 2 21 1 1 1 12 4 2 2 22 2 2 8 2

t t t tE t LI t Q t e t t e e t t e t tC

− − − −⎡ ⎤= + = − + = − + = − +⎣ ⎦ .

Hence, after time t the energy loss is ( )22 2 2te t t−− − + joules.


Questions of Energy

6. 3 0x x x− + =

(a) 21KE2

x= , ( )3 2 41 12 4

V x x dx x x= − + = − +∫

( ) 2 2 41 1 1, KE2 2 4

E x x V x x x= + = − +

(b) To find the equilibrium points, we seek the solutions of

3 0E x xx

∂= − + =

∂, 0E x

x∂

= =∂

.

Solving these equations, we find three equilibrium points at ( )1, 0− , (0, 0) and (1, 0).

Because 0x = for all these points, we determine which points are stable (local maxima) by simply drawing the graph of ( )V x (shown in part (c)).

(c) Graph of the potential energy ( )V x is shown. Note that ( )V x has local minima

at 1x = ± and a local maxima at 0x = .

Hence,

( )1, 0− and ( )1, 0

are stable points, and

( )0, 0

is an unstable point.

1

2

x

3

–0.5–1–2 1 2

V x( )

Potential energy of 3 0x x x− + =

7. 3 0x x x− − =

(a) 21KE2

x= , ( )3 2 41 12 4

V x x dx x x= − − = − −∫

( ) 2 2 41 1 1, KE2 2 4

E x x V x x x= + = − −

(b) To find the equilibrium points, we seek the solutions of the two equations

3 0E x xx

∂= − − =

∂, 0E x

x∂

= =∂

.

Solving these equations, we find one equilibrium point at ( )0, 0 . Because 0x = we

determine if it is a stable point (local minima) or unstable point (local maxima) by simply drawing the graph of ( )V x shown in part (c).


The graph of potential energy ( )V x is shown. Note that ( )V x has local maximaat 0x = , and hence ( )0, 0 is an unstable

equilibrium point.


–2

–1

x

1

–1–2 1 2

V x( )

Potential energy of 3 0x x x− − =

8. 2 0x x x− + =

(a) 21KE2

x= , ( )2 2 31 12 3

V x x dx x x= − + = − +∫

( ) 2 2 31 1 1, KE2 2 3

E x x V x x x= + = − +


2 0E x xx

∂= − + =

∂, 0E x

x∂

= =∂

.

Solving these equations, we find two equilibrium points at ( )0, 0 , ( )1, 0 . Because 0x =

for both these points, we determine which points are stable (local minima) and which are unstable (local maxima) by simply drawing the graph of ( )V x shown in part (c).

The graph of potential energy ( )V x is shown. Note that ( )V x has local minima

at 1x = + and a local maxima at 0x = . Hence, ( )0, 0 is an unstable point and

( )1, 0 is a stable point.

(c) See figure.

–2

–1

x

1

–1–2 1 2

V x( )

Potential energy of 2 0x x x− + =

9. 2 0x x+ =

(a) 21KE2

x= , 2 313

V x dx x= =∫


( ) 2 31 1, KE2 3

E x x V x x= + = +

(b) To find the equilibrium points, we seek the solutions of the equations

2 0E xx

∂= =

∂, 0E x

x∂

= =∂

.

Solving these equations, we find one equilibrium point at

( )0, 0 .

To determine if the point is stable, wenote that the potential energy

( ) 313

V x x=

does not have a local maxima or minimaat ( )0, 0 , so ( )0, 0 is an unstable (or

semistable) equilibrium point.

(c) The graph of ( )V x is a simple cubic.

–1

2

x

3

–2

–1–2 1 2

V x( )

Potential energy of 2 0x x+ =

10. 1 0xx e− − =

(a) 21KE2

x= , ( )1x xV e dx e x= − − = − −∫

( ) 21, KE2

xE x x V x e x= + = − −


1 0xE ex

∂= − − =

∂, 0E x

x∂

= =∂

.

It is clear that the first of these equationshas no root, so the equation has no equi-librium points.

(c) Note that the graph of the potential energy( )V x does not have any local maxima or

minima points, which corresponds to thelack of equilibrium points found in part (b).

–1

1

x

2

–2

–1–2 1 2

V x( )

Potential energy of 1 0xx e− − =


11. ( )21 0x x+ − =

(a) 21KE2

x= , ( )2 3 2113

V x dx x x x= − = − +∫

( ) 2 3 21 1, KE2 3

E x x V x x x x= + = + − +


2 2 1 0E x xx

∂= − + =

∂, 0E x

x∂

= =∂

.

Solving these equations, yields only one real equilibrium point ( )1, 0 . Because 0x = ,

we determine if it is stable (local minima) or unstable (local maxima) by simply drawing the graph of ( )V x show in part (c). Thus, we find that ( )1, 0 is an unstable (or

semistable) equilibrium point.

(c) The graph of the potential energy ( )V xis shown. Note that ( )V x has neither a

maximum nor a minimum at 1x = , and hence ( )1, 0 is an unstable (or semista-

ble) equilibrium point.

–1

1

x

2

–2

–1–2 1 2

V x( )

Potential energy of ( )21 0x x+ − =

12. 21xx

=

(a) 21KE2

x= , 21 1V dx

xx= − =∫

( ) 21 1, KE2

E x x V xx

= + = +


21 0E

x x∂ −

= =∂

, 0E xx

∂= =

∂.

Because the first equation does not have a solution, there is no equilibrium point.


(c) The graph of the potential energy ( )V xis shown. Note that ( )V x does not have

any local maxima or minima, which cor-responds to the absence of equilibriumpoints noted in part (b).

–2

2

x

4

–4

–1–2 1 2

V x( )

Potential energy of 21x x=

13. ( )( )1 2x x x= − −

(a) 21KE2

x= , ( )( ) 3 21 31 2 23 2

V x x dx x x x= − − − = + −∫

( ) 2 3 21 1 3, KE 22 3 2

E x x V x x x x= + = − + −


2 3 2 0E x xx

∂= − + − =

∂, 0E x

x∂

= =∂

.

Solving these equations, we find two equilibrium points at ( )1, 0 and ( )2, 0 . Because

0x = , we determine which points are stable (local minima) and which are unstable (local maxima) by simply drawing the graph of ( )V x show in part (c).

(c) The graph of the potential energy ( )V xis shown. Note that ( )V x has local

minima at 1x = and a local maxima at2x = . Hence, ( )1, 0 is a stable point and

( )2, 0 is an unstable point.

–1

1

x

2

–2

–2–4 2 4

V x( )

Potential energy of ( )( )1 2x x x= − −


Conservative or Nonconservative?

14. 2 0x x+ =

Conservative because it is of the form

( ) 0mx F x+ = .

The total energy of this conservative system is

( ) ( ) 2 31 1 1,2 2 3

E x x mx F x dx x x= + = +∫ .

–2

2

x

4

–4

–2–4 2 4

x

We draw contour curves for this surface over the xx -plane to view the trajectories of the differential equation in the xx plane.

15. 0x kx+ =

Conservative because it has the form

( ) 0mx F x+ = .

The total energy of this conservative system is

( ) ( ) 2 21 1 1,2 2 2

E x x mx F x dx x kx= + = +∫ .

–2

2

x

4

–4

–2–4 2 4

x

We draw contour curves for this surface over the xx -plane to view the trajectories of the differen-tial equation in the xx plane. The trajectories of 0x kx+ = are ellipses each with height k

times its width.

16. 2 1x x x+ + =

Not conservative due to the x term. The spiral

trajectories in its phase plane cannot be levelcurves of any surface.

x

5

–5

–5 5

x


17. sin 0θ θ+ =

Conservative because it is of the form

( ) 0m Fθ θ+ = .

The total energy of this nonconservative systemis

( ) ( ) 21 1, cos2 2

E m F dθ θ θ θ θ θ θ= + = −∫ .

–2

2

4

–4

θ−π π

θ

We can draw contour curves for this surface over the θθ -plane to view the trajectories of the differential equation in the θθ plane.

18. sin 1θ θ+ =

Conservative because it can be written in theform

( ) 0m Fθ θ+ = ,

where ( ) sin 1F θ θ= − . The total energy is

( ) 21, cos2

E θ θ θ θ θ= − − .

–1

1

θ

2

–2

–4–8 4 8

θ

We can draw contour curves for this surface over the θθ -plane to view the trajectories of the differential equation in the θθ plane.

19. sin 1θ θ θ+ + =

Not conservative due to the θ term. The

following phase plane portrait shows equilibriaalong the axis. Trajectory cannot be level curvesfor any surface.

–2

2

4

–4

θ−π π−2π 2π

θ

Trajectories of a nonconservative system


Time-Reversible Systems

20. (a) ( )mx F x= . If we introduce backwards time tτ = − , then taking the derivatives, yields

( ) ( )

2 2 2

2 2 21

dx dx d dxdt d dt d

d x d d dx d d x d xxdt d d dtdt d d

ττ τ

ττ τ τ τ

= = −

⎛ ⎞= = − = − − =⎜ ⎟⎝ ⎠

The conservative system ( ) 0x F x+ = is transformed into exactly the same equation

( )2

2 0d x F xdτ

+ =

in backwards time τ .

(b) The solution of the IVP

0x x+ = , ( )0 1x = , ( )0 0x =

is ( ) cosx t t= . If we replace t by –t, it yields the solution ( ) ( )cos cosx t t t− = − = . Hence,

running the system backwards looks exactly like running the system forward.

(c) The solution of the IVP x mg= − , ( )0 0x = , ( )0 100x = is

( ) 21 1002

x t mgt t= − + .

If we replace t by –t, we get ( ) 21 1002

x t mgt t− = − − . Hence, the solution is not the same,

and the system is not time reversible.

(d) If we think of a time-reversible system as a system where equations of motion are the same when we replace t by –t, we might make the following conclusions.

(i) yes (ii) no (iii) no

(iv) no (v) yes

Computer Lab: Undamped Spring

21. IDE Lab

Computer Lab: Damped Spring

22. IDE Lab


Conversion of Equations

23. ( )20x x f tω+ =

Letting 1x x= , 2x x= , we have

( )

1 22

2 0 1 .

x x

x x f tω

=

= − +

In matrix form, this becomes

( )1 1

22 20

0 1 00

x xf tx xω

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

.

24. sin 0gL

θ θ+ =

Letting 1x θ= , 2x θ= , we have

1 2

2 1sin .

x xgx xL

=

= −

This system is not linear, so there is no matrix form.

25. 0ay by cy′′ ′+ + =

Letting 1x y= , 2x y′= , we have

1 2

2 1 2 .

x xc bx cx xa a

=

= − −


1 1

2 2

0 1x xc bx xa a

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

.

26. 1 0LQ RQ QC

+ + =

Letting 1x Q= , 2dQxdt

= , we have

1 2

2 1 21 .

x xRx x x

LC L

=

= − −



1 1

2 2

0 11

x xRx x

LC L

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

.

27. ( )2 2 2 0t x tx t n x+ + − =


( )1 2

2 2

2 1 221 .

x x

t nx x x

tt

=

−= − −


1 12 2

2 22

0 1

1x x

t nx xtt

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

.

28. ( )1 sin 0x t xω+ + =


( )

1 2

2 11 sin .x xx t xω=

= − +


( )1 1

2 2

0 11 sin 0

x xtx xω

⎡ ⎤⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦⎣ ⎦

.

29. ( ) ( )21 2 1 0t y ty n n y′′ ′− − + + =

Letting 1x y= , 2x y′= , we have

( )1 2

2 1 22 2

1 2 .1 1

x xn n tx x x

t t

=

+= − +

− −


1 1

2 22 2

0 11 2

1 1

x xn tx xn

t t

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥− −⎣ ⎦

.


30. 4 3 2

4 3 23 2 4 1d y d y d y dy ydtdt dt dt

+ + + + =

If we introduce

1

2

2

3 2

3

4 3

x ydyxdtd yxdtd yxdt

=

=

=

=

we have the differential equations

1 2

2 3

3 4

4 1 2 3 44 2 3 1

x xx xx xx x x x x

===

= − − − − +

or in matrix form

1 1

2 2

3 3

4 4

0 1 0 0 00 0 1 0 00 0 0 1 04 1 2 3 1

x xx xx xx x

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

− − − −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

.

Conversion of IVPs

31. 2 siny y y t′′ ′− + = , ( )0 1y = , ( )0 1y′ =

Letting 1x y= , 2x y′= yields

( )( )

1 2 1

2 1 2 2

0 12 sin 0 1

x x xx x x t x′ = =′ = − + + =

.

In matrix form this becomes

1 1

2 2

0 1 02 1 sin

x xx x t′⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥′ −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

; ( )( )

1

2

0 10 1

xx⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

.

32. 1y ty y′′′ ′+ + = , ( )0 0y = , ( )0 1y′ = , ( )0 2y′′ =

Letting 1x y= , 2x y′= , 3x y′′= yields

( )( )( )

1 2 1

2 3 2

3 1 2 3

0 10 1

1 0 2

x x xx x xx x tx x

′ = =′ = =′ = − − + =

.



1 1

2 2

3 3

0 1 0 00 0 1 01 0 1

x xx xx t x

′⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥′ = +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥′ − −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

, ( )( )( )

1

2

3

0 00 10 2

xxx

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

.

33. 3 2 ty y z e−′′ ′+ + = , ( )0 0y = , ( )0 1y′ =

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

Letting 1x y= , 2x y′= , 3x z= , 4x z′= yields

( )( )( )( )

1 2 1

2 2 3 2

3 4 3

4 1 3 4

0 0

3 2 0 10 1

2 1 0 0

t

x x x

x x x e xx x xx x x x

−

′ = =

′ = − − + =′ = =′ = − − + =

.


1 1

2 2

3 3

4 4

00 1 0 00 3 2 00 0 0 1 01 0 2 0 1

t

x xx x ex xx x

−

′ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′ − − ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥= + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′ − − ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

,

( )( )( )( )

1

2

3

4

0 00 10 10 0

xxxx

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

.

34. 2 1y y z′′′ ′+ + = , ( )0 1y = , ( )0 0y′ = , ( )0 1y′′ =

2 sinz y z t′ + + = , ( )0 1z =

Letting 1x y= , 2x y′= , 3x y′′= , 4x z= yields

( )( )( )( )

1 2 1

2 3 2

3 2 4 3

4 1 4 4

0 00 0

2 1 0 12 sin 0 1

x x xx x xx x x xx x x t x

′ = =′ = =′ = − − + =′ = − − + =

.


( )( )( )( )

11 1

22 2

33 3

44 4

00 1 0 0 0 000 0 1 0 0 000 1 0 2 1 101 0 0 2 sin 1

xx xxx xxx xxx x t

′ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥′ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥′ − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥′ − − ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

.


Conversion of Systems

35. 1 1 2

2 2

22 0

tx x x ex x

−+ + =+ =

Letting 1 1z x= , 2 1z x= , 3 2z x= , 4 2z x= yields the system

1 2

2 1 3

3 4

4 3

2

2 .

t

z z

z z z ez zz z

−

=

= − − +

=

= −


1 1

2 2

3 3

4 4

00 1 0 01 0 2 00 0 0 1 00 0 2 0 0

t

z zz z ez zz z

−

′ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′ − − ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥= + ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥′ − ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦

.

36. ( )( )

, , , , ,

, , , , ,

y f t y y y z z

z f t y y y z z

′′′ ′ ′′ ′=

′′ ′ ′′ ′=

Letting 1x y= , 2x y′= , 3x y′′= , 4x z= , 5x z′= yields

( )

( )

1 2

2 3

3 1 2 3 4 5

4 5

5 1 2 3 4 5

, , , , ,

, , , , , .

x xx xx f t x x x x xx xx g t x x x x x

′ =′ =

′ =

′ =

′ =

37. 1 11 1 12 2 13 3

2 21 1 22 2 23 3

3 31 1 32 2 33 3

x a x a x a xx a x a x a xx a x a x a x

= + +

= + +

= + +

If we let 1 1z x= , 2 1z x= , 3 2z x= , 4 2z x= , 5 3z x= , 6 3z x= , we get

1 2

2 11 1 12 3 13 5

3 4

4 21 1 22 3 23 5

5 6

6 31 1 32 3 33 5 .

z zz a z a z a zz zz a z a z a zz zz a z a z a z

== + +

=

= + +

=

= + +


In matrix form =z Az , this becomes

1 1

2 11 12 13 2

3 3

4 21 22 23 4

5 5

6 31 32 33 6

0 1 0 0 0 00 0 0

0 0 0 1 0 00 0 0

0 0 0 0 0 10 0 0

z zz a a a zz zz a a a zz zz a a a z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

.

Solving Linear Systems

38. 1 2

2 1 22 3x xx x x′ =′ = − −

From first DE: 2 1x x′= . Substituting in second DE gives

( ) ( )1 1 12 3x x x′′ ′= − −

or the second order DE

1 1 13 2 0x x x′′ ′+ + = .

Solving the second order DE gives

21 1 2

t tx c e c e− −= + .

Substituting this result in first DE gives

22 1 1 22 t tx x c e c e− −′= = − − .

39. 1 1 2

2 1 2

3 22 2

x x xx x x′ = −′ = −

From first DE: 2 1 11 32 2

x x x′= − + . Substituting in second DE yields a second order DE to solve for

1x .

1 1 1 1 1

1 1 1 1 1

1 1 12

1 1 2

1 3 1 32 22 2 2 2

1 3 2 32 2

2 0t t

x x x x x

x x x x x

x x x

x c e c e−

′⎛ ⎞ ⎛ ⎞′ ′− + = − − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′′ ′ ′− + = + −

′′ ′− − =

= +

To find 2x , substitute the solution for 1x back into the first DE.

( ) ( )2 2 22 1 1 1 2 1 2 1 2

1 3 1 3 12 22 2 2 2 2

t t t t t tx x x c e c e c e c e c e c e− − −′= − + = − − + + = + .


40. 1 1 2

2 1 24x x xx x x′ = +′ = +

From first DE: 2 1 1x x x′= − . Substituting in second DE yields a second order DE to solve for 1x .

( ) ( )1 1 1 1 1

1 1 1 1 1

1 1 13

1 1 2

4

42 3 0

t t

x x x x x

x x x x xx x x

x c e c e−

′′ ′− = + −

′′ ′ ′− = + −′′ ′− − =

= +

From first calculation, 2 1 1x x x′= − , so ( ) ( )3 3 32 1 2 1 2 1 23 2 2t t t t t tx c e c e c e c e c e c e− − −= − − + = − .

41. 1 2

2 1 22 3 5x x tx x x′ = +′ = − + +

From first DE: 2 1x x t′= − . Substituting in second DE yields a second order DE to solve for 1x .

( ) ( )1 1 1

1 1 1

1 1 1

21 1 2

2 3 51 2 3 3 5

3 2 3 6.t t

h

x t x x tx x x t

x x x t

x c e c e

′′ ′− = − + − +

′′ ′− = − + − +′′ ′− + = − +

= +

To find 1px by the method of undetermined coefficients, substitute 1px at b= + , 1px a′ = , 1 0px′′ =

to obtain

0 3 2 2 3 6a at b t− + + = − + .

Comparing like terms,

Coefficients of t: 2 3a = − so 32

a = − .

Constants: 3 2 6a b− + = so 34

b = . Hence 13 32 4px t= − + . Therefore,

21 1 2

3 32 4

t tx c e c e t= + − + .

From first calculation 2 1x x t′= − , so

22 1 2

322

t tx c e c e t= + − − .


Solving IVPs for Systems

42. 1 1 2

2 1 2

6 32

x x xx x x′ = −′ = +

From first DE: 2 1 1123

x x x′= − . Substituting in second DE yields a second order DE to solve for 1x .

1 1 1 1 1

1 1 13 4

1 1 2

1 12 2 23 3

7 12 0

.t t

x x x x x

x x x

x c e c e

′⎛ ⎞ ⎛ ⎞′ ′− = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠′′ ′− + =

= +

From first calculation, 2 1 1123

x x x′= − , so

( ) ( )3 4 3 4 3 42 1 2 1 2 1 2

1 22 3 43 3

t t t t t tx c e c e c e c e c e c e= + − + = + .

Applying initial conditions:

( )

( )1 1 2

2 1 2

0 2 220 3 33

x c c

x c c

= ⇒ + =

= ⇒ + =

so

2 3c = − and 1 5c = .

The solution to the IVP is 3 41 5 3t tx e e= − , 3 4

2 5 2t tx e e= − .

43. 1 1 2

2 1 2

3 42

x x xx x x′ = +′ = +

From first DE: 2 1 11 34 4

x x x′= − . Substituting in second DE yields a second order DE to solve for

1x .

1 1 1 1 1

1 1 1 1 1

1 1 15

1 1 2

1 3 1 324 4 4 4

3 8 34 5 0

t t

x x x x x

x x x x xx x x

x c e c e−

′⎛ ⎞ ⎛ ⎞′ ′− = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′′ ′ ′− = + −′′ ′− − =

= +

From first calculation, 2 1 11 34 4

x x x′= − , so

( ) ( )5 5 52 1 2 1 2 1 2

1 3 154 4 2

t t t t t tx c e c e c e c e c e c e− − −= − − + = −


Applying initial conditions:

( )

( )1 1 2

2 1 2

0 1 110 1 12

x c c

x c c

= ⇒ + =

= − ⇒ − = −

so

1 0c = and 2 1c = .

The solution to the IVP is 1tx e−= , 2

tx e−= − .

Counterexample

44. An example: The degenerate system

1 2 1

1 2 1

00

x x xx x x+ + =+ + =

where both equations are exactly the same clearly cannot be written as a second-order equation in either 1x or 2x . The reader might contemplate finding all the solutions of such an undetermined

system.

Another approach: Note that when we write an nth-order equation such as

0ay by cy′′ ′+ + =

as a system of first-order equations by letting 1x y= , 2x y′= , the system has the form

1 1

2 2

0 1x xc bx xa a

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

.

This shows we cannot obtain a second-order equation in 1x with 2 1x x= unless the coefficient

matrix has the preceding form in which the first row contains a 0 and 1. Hence, a system such as

1 1

2 2

1 14 1

x xx x⎡ ⎤ ⎡ ⎤⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

cannot be transformed into a second-order equation in 1x with 2 1x x= .

Coupled Mass-Spring System

45. Given the linear system

( ) ( )

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

2 2 2 1 2 1 2 2 ,

mx k x k x x k k x k x

mx k x x k x k x

= − + − = − + +

= − − = −


we let

1 1 3 2

2 1 4 2.z x z xz x z x= == =

We then have the first-order system

1 2

1 2 22 1 3

3 4

2 24 1 3 .

z zk k kz z z

m mz z

k kz z zm m

=+

= − +

=

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


( )1 11 2 2

2 2

3 3

4 42 2

0 1 0 0

0 0

0 0 0 1

0 0

z zk k kz zm mz zz zk k

m m

⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤+⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦−⎢ ⎥⎣ ⎦

.

Satellite Problem

46. ( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( )

212

22 1

kr r t t u tr t

t r tu t

r t r t

θ

θθ

= − +

= +

Letting

1 3

2 4 ,x r xx r x

θθ

= == =

we have the system

( )

( )

1 2

22 1 4 12

1

3 4

2 44 2

1 1

2 1 .

x xkx x x u tx

x xx xx u tx x

=

= − +

=

= +


Two Inverted Pendulums

47. ( ) ( )( ) ( )

1 1 2

2 1 2

1

1

mg mg u t

mg mg u t

θ θ θ

θ θ θ

= + + −

= + + −

Letting

1 1

2 1

3 2

4 2

x

xx

x

θ

θθ

θ

=

==

=

we have first-order linear system

( ) ( )

( ) ( )

1 2

2 1 3

3 4

4 1 3

1

1 .

x xx mg x mgx u tx xx mgx mg x u t

=

= + + −

=

= + + −


1 1

2 2

3 3

4 4

0 1 0 0 01 0 0 ( )

0 0 0 1 00 1 0 ( )

x xx xmg mg u tx xx xmg mg u t

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥+ −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

+ −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

.


48. Student Project

Diff EQ Chapter-4

Documents

Transcript of Diff EQ Chapter-4