CHAPTER 4 Second-Order Linear Differential Equations

CHAPTER 4 Second-Order LinearDifferential Equations

4.1 The Harmonic Oscillator

!!!! The Undamped Oscillator

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

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

x t c t c tx t c t c t( ) = +( ) = − +

1 2

1 2

cos sin! sin cos .

Substituting this expression into the initial conditions x 0 1( ) = , !x 0 0( ) = , gives

x cx c

0 10 0

1

2

( ) = =( ) = =!

so c1 1= , c2 0= . Hence, the IVP has the solution x t t( ) = cos .

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

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

x t c t c tx t c t c t( ) = +( ) = − +

1 2

1 2

cos sin! sin cos .


x cx c

0 10 1

1

2

( ) = =( ) = =!

or c c1 2 1= = . Hence, the IVP has the solution

x t t t( ) = +cos sin .

253

254 CHAPTER 4 Second-Order Linear Differential Equations

In polar form, this would be

x t tb g = −FHGIKJ2

4cos π .

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

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

x t c t c tx t c t c t( ) = +( ) = − +

1 2

1 2

3 33 3 3 3cos sin

! sin cos .


x cx c

0 10 3 1

1

2

( ) = =( ) = =!

so c1 1= , c213

= . Hence, the IVP has the solution

x t t t( ) = +cos sin3 13

3 .


x t tb g b g= −103

3cos δ

where δ = −tan 1 13

. This would be in the first quadrant.

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


x t c t c tx t c t c t( ) = +( ) = − +

1 2

1 2

2 22 2 2 2cos sin

! sin cos .

Substituting this expression into the initial conditions x 0 1( ) = , !x 0 2( ) = − , gives

x cx c

0 10 2 2

1

2

( ) = =( ) = = −!

so c1 1= , c2 1= − . Hence, the IVP has the solution

x t t t( ) = −cos sin2 2 .


x t tb g = +FHGIKJ2 2

4cos π .

SECTION 4.1 The Harmonic Oscillator 255

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


x t c t c tx t c t c t( ) = +( ) = − +

1 2

1 2

4 44 4 4 4cos sin

! sin cos .

Substituting this expression into the initial conditions x 0 1( ) = − , !x 0 0( ) = , gives

x cx c

0 10 4 0

1

2

( ) = = −( ) = =!

so c1 1= − , c2 0= . Hence, the IVP has the solution

x t t( ) = −cos4 .

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


x t c t c tx t c t c t( ) = +( ) = − +

1 2

1 2

4 44 4 4 4cos sin

! sin cos .

Substituting this expression into the initial conditions x 0 0( ) = , !x 0 4( ) = , we get

x cx c

0 00 4 4

1

2

( ) = =( ) = =!

so c1 0= , c2 1= . The IVP has the solution

x t t( ) = sin4 .

!!!! Computer Lab

7. y t t= +cos sin

The equation tells us T = 2π and because

T =2πω

, ω0 1= . We then measure the delay

δω0

08≈ . which we can compute as the phase

angle δ ≈ ( ) =08 1 08. . . The amplitude A can be

measured directly giving A ≈ 14. . Hence,

cos sin . cos .t t t+ ≈ −( )14 08 .

Compare with the algebraically exact form inProblem 13.

π

–1.5

1.5

t

T = 2π

A ≈1.4δω≈ 0.8

y


8. y t t= +2cos sin

The equation tells us T = 2π and because

T =2πω

, ω0 1= . We then measure the delay

δω0

05≈ . , which we can compute as the phase

angle δ ≈ ( ) =05 1 05. . . The amplitude A can be

measured directly giving A ≈ 2 2. . Hence,

2 2 2 05cos sin . cos .t t t+ ≈ −( ).

8–4

–2.5

2.5

4

T = 2πA ≈2.2δ

ω≈ 0.5

y

t

9. y t t= +5 3 3cos sin

The equation tells us that period is T =23π and

because T =2πω

, ω0 3= . We then measure the

delay δω0

005≈ . , which we can compute as the

phase angle δ ≈ ( ) =3 0 05 015. . . The amplitude A

can be measured directly giving A ≈ 51. . Hence,

5 3 3 51 3 015cos sin . cos .t t t+ ≈ −( ) .

–5

5

tπ

T = 2π A ≈5.1

δω≈ 0.05

/3

y

10. y t t= +cos sin3 5 3

The equation tells us the period is T =23π and

because T =2πω


delay δω0

05≈ . , which we can compute as the

phase angle δ ≈ ( ) =05 3 15. . . The amplitude A can

be measured directly giving A ≈ 51. . Hence,

cos sin . cos .3 5 3 51 3 15t t t+ ≈ −( ).

3

–5

5 A ≈5.1

δω≈ 0.5

y

T = 2π /3

t


11. y t t= − +cos sin5 2 5

equation tells us that period is T =25π and

because T =2πω


delay δω

π

0 804≈ or . , which we can compute as

the phase angle δ ≈ ( ) =5 0 4 2. . The amplitude A

can be measured directly giving A ≈ 2 2. . Hence,

− + ≈ −( )cos sin . cos5 2 2 2 5 2t t t .

–2

2

t21

y

A ≈2.2

δω≈ 0.4

T = 2π /5

!!!! Alternate Forms for Sinusoidal Oscillations

12. We have

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

cos cos cos cos cos cos cos sin sincos sin

ω δ ω δ ω δ δ ω δ ωω ω

0 0 0 0 0

1 0 2 0

− = + = ( ) + ( )= +

a f a f

where c A1 = cosδ , c A2 = sinδ .

!!!! Single-Wave Forms of Simple Harmonic Motion

13. cos sint t+

By Equation (5) c1 1= , c2 1= , and ω0 1= . By Equation (6)

A = 2 , δ π=

4

yielding

cos sin cost t t+ = −FH IK24π .

(Compare with solution to Problem 7.)

14. cos sint t−

By Equation (5) c1 1= , c2 1= − , and ω0 1= . By Equation (6)

A = 2 , δ π= −

4

yielding

cos sin cost t t− = +FHIK2

4π .

Because c1 is positive and c2 is negative the phase angle is in the 4th quadrant.


15. − +cos sint t

By Equation (5) c1 1= − , c2 1= , and ω0 1= . By Equation (6)

A = 2 , δ π=

34

yielding

− + = −FHIKcos sin cost t t2 3

4π .

Because c1 is negative and c2 is positive the phase angle is in the 2nd quadrant.

16. − −cos sint t

By Equation (5) c1 1= − , c2 1= − , and ω0 1= . By Equation (6)

A = 2 , δ π=

54

yielding

− − = −FHIKcos sin cost t t2 5

4π .

Because c1 and c2 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:

17. 2 2 2 2 2 2 2cos cos cos sin sin cost t t t−( ) = −( ) − −( ){ } = −π π π

18. cos cos cos sin sin cos sint t t t t+FHGIKJ =

FHGIKJ −

FHGIKJ = −

π π π3 3 3

12

32

19. 34

34 4

3 22

22

3 22

cos cos cos sin sin cos sin cos sint t t t t t t−FHGIKJ = −FHG

IKJ − −FHG

IKJ

RSTUVW = +RST

UVW= +

π π π l q

20. cos cos cos sin sin cos sin36

36

36

32

3 12

3t t t t t−FHGIKJ = −FHG

IKJ − −FHG

IKJ = +

π π π

!!!! Interpreting Oscillator Solutions

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

Because ω0 1= , we know the natural frequency is 12π

and the period is 2π . Using the initial

conditions, we find the solution (see Problem 1)


x t t( ) = cos ,

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

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

Because ω0 1= radians per second, we know the natural frequency is 12π

Hz (cycles per

second), and the periodic function is 2π . Using the initial conditions, we find the solution (seeProblem 2)

x t t( ) = −FH IK24

cos π ,

which tells us the amplitude is 2 and the phase angle is δ π=

4 radians.

23. !!x x+ =9 0, x 0 1( ) = , !x 0 1( ) =


Hz (cycles per

second), and the periodic function is 23π . Using the initial conditions, we find the solution (see

Problem 3)

x t tb g b g= −103

3cos δ

where δ = −tan 1 13

, which tells us the amplitude is

103

and the phase angle is

δ = ≈−tan .1 13

033 radians.

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


Hz (cycles per second),

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

x t tb g = +FHGIKJ2 2

4cos π ,

which tells us the amplitude is 2 and the phase angle is δ π= −

4 radians.


25. !!x x+ =16 0, x 0 1( ) = − , !x 0 0( ) =



and the periodic function is π2

. Using the initial conditions, we find the solution (see Problem 5)

x t t( ) = −( )cos 4 π ,

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

26. !!x x+ =16 0, x 0 0( ) = , !x 0 4( ) =



and the periodic function is π2

. Using the initial conditions, we find the solution (see Problem 6)

x t tb g = −FHGIKJcos 4

2π ,

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

2 radians.

!!!! Phase Portraits

All of the phase portraits are circles or ellipses. The phase portrait of !!x k x+ =2 0 are ellipses that aremore elongated in the !x direction when k > 1 and elongated in the x direction when k < 1. A curvepassing through x x k, ! , ( ) = 0 2b g will also pass through x x k, ! , ( ) = ( )0 .

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

See Problems 1, 21.

Trajectories are circles centered at the origin,traversed clockwise. x

–1

1

(1, 0)

!x


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

See Problems 2, 22.

Trajectories are circles centered at the origin,traversed clockwise. x

–1

1 (1, 1)

!x

1–1

29. !!x x+ =9 0, x 0 1( ) = , !x 0 1( ) =

See Problems 3, 23.

Trajectories are ellipses centered at the origin,each with height thrice its width. Motion isclockwise.

x

–4

4

(1, 1)

!x

4–4

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

See Problems 4, 24.

Trajectories are ellipses centered at the origin,each with height twice its width. Motion isclockwise.

x

–4

4

(1, –2)

!x

4–4

31. !!x x+ =16 0, x 0 1( ) = − , !x 0 0( ) =

See Problems 5, 25.

Trajectories are ellipses centered at the origin,each with height four times its width. Motion isclockwise.

x

–5

5!x

5–5(–1, 0)


32. !!x x+ =16 0, x 0 0( ) = , !x 0 4( ) =

See Problems 6, 26.

Trajectories are ellipses centered at the origin,each with height four times its width. Motion isclockwise.

x

–5

5!x

5–5

(0, 4)

!!!! Changing Frequencies

33. (a) ω 0 05= . gives tx curve with lowest fre-quency (fewest humps); ω 0 2= gives the

highest frequency (most humps).

(b) ω 0 05= . gives the innermost phase-planetrajectory; as ω 0 increases, the amplitude

of !x increases. In Figure 4.1.7 the trajec-tory that is not totally visible is the onefor ω0 2= .

3

–4

4x

t

2

–2

2ππ

ω0 05= . ω0 2=

ω0 1=

!!!! Detective Work

34. (a) The curve y t= −FHGIKJ14 8

5. cos π is a sinusoidal curve with period 2π , amplitude A ≈ 14. ,

and phase angle roughly 85π .

(b) From the graph we estimate ω0 1= , A ≈ 2 3. , and δ π≈

4. Thus, we have

x t A t t t t

t t t t

b g b g

b g

= − = −FHGIKJ = −FHG

IKJ − −FHG

IKJ

LNM

OQP

= +RST

UVW≈ +

cos . cos . cos cos sin sin

. cos sin . cos sin .

ω δ π π π0 0 2 3

42 3

4 4

2 3 22

22

16

!!!! Pulling a Weight

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

that k = =8

0516

.nt m . If we then release the weight, the IVP describing the motion of

the weight is 2 16 0!!x x+ = or


!!x x+ =8 0, x 0 05( ) = . , !x 0 0( ) = .

The solution of the differential equation is

x t A t( ) = −cos 8 δd h.Using the initial conditions, we get the simple oscillation

x t t( ) = 05 8. cosd h.

(b) Amplitude = 12

m; T = =2 2

80

πω

π sec , f = 82π

cycles per second

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

of the

period or after

t = ≈π

2 80 56. seconds.

At that time velocity is

! . sin . secx 0 56 22

1414b g = − FHGIKJ ≈ −

π m

moving away from original displacement.

!!!! Finding the Differential Equation

36. (a) The mass is m = 500 gm, which means the force acting on the spring is 500 980× dynes.This stretches the spring 50 cm, so the spring constant is

k = ×=

500 98050

9800 dynes cm.

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

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

IVP for the mass is

500 9800 0!!x x+ =

or

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

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

x t A t t( ) = −FHG

IKJ =cos cos98

510 98

5δ .


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

T mk

= = ≈2 2 598

14π π . sec,

and the natural frequency is given by the reciprocal f = =1

14071

.. oscillations per sec-

ond.

!!!! Initial-Value Problems

37. (a) The weight is 16 lbs, so the mass is roughly 1632

12

= slugs. This mass stretches the spring

12

foot, hence k = =16 3212

lb ft . This yields the equation 12

32 0!!x x( ) + = , or

!!x x+ =64 0.

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

x 0 13

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

positive.

(b) We have the same equation !!x x+ =64 0, but the initial conditions are x 0 16

( ) = − ft,

!x 0 1( ) = ft/sec.

!!!! One More Weight

38. The mass is m = =1232

38

slugs. The spring is stretched 12

foot, so the spring constant is

k = =12 2412

lb ft . The initial position of the mass is 4 inches ( 13

ft) upward so x 0 13

( ) = − . The

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

mass is

!!x x+ =64 0, x 0 13

( ) = − , !x 0 2( ) = − ,

which has the solution

x t t t( ) = − −13

8 14

8cos sin .


Writing this in polar form, we have

A c c

cc

= + = −FH IK + −FH IK =

= FHGIKJ =

FH IK≈ ( )

− −

11

22

2 2

1 2

1

1

13

14

512

34

378 radians

δ tan tan

. . angle in 3rd quadrant

Hence, we have the solution in polar form

x t t( ) = −( )5

128 378cos . .

See figure.

1.4

–0.4

0.4

x

1.210.80.60.40.2t

x t t( ) = −( )512

8 378cos .

Spring oscillation

!!!! Comparing Harmonic Motions

39. The period of simple harmonic motion is given by T =2

0

πω

, where ω0 =km

. Notice that this

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

!!!! Testing Your Intuition

40. !!x x x+ + =3 0

Here, we have a vibrating spring with no friction, but a nonlinear restoring force F x x= − − 3 thatis stronger than a purely proportional force F x= − . For small displacement x, the nonlinearrestoring force will not be much different (x3 is very small x), but for larger displacement it willbe much stronger. This equation is called Duffing’s (strong) equation, and you may contemplatethe motion of such springs, called strong springs.

41. !!x x x+ − =3 0

Here, we have a vibrating mass with no friction, but the restoring force is not purely proportionalto the displacement as when F x= − , but has an additional term F x x= − + 3. This means,compared with a linear restoring force −kx , this nonlinear restoring force will not be muchdifferent for small displacement (x3 is very small x), but will have a smaller restoring force forlarger vibration amplitude. (i.e., the spring weakens when it is stretched a lot, and in factbecomes zero when x = 1). This equation is called Duffing’s (weak) equation, and you maycontemplate the motion of such springs, called weak springs.


42. !!x x− = 0

This equation describes a spring with no friction and a negative restoring force. You may wonderif there are such physical systems. Later, we will see that this equation describes the motion of aninverted pendulum, and it has solutions sinh t and cosh t in contrast to !!x x+ = 0, which hassolutions sin t and cos t. The restoring force is always away from the equilibrium position, andhence the solution moves away from it.

43. !! !xt

x x+ + =1 0

This equation can be interpreted as describing the motion of a vibrating mass where the friction1t

x! starts very big when t = 0, but then dies off. Again, you many simulate in your mind the

motion of such a system. For large t, the motion will behave much like simple harmonic motion.

44. !! !x x x x+ − + =2 1 0b gThis is called van der Pol’s equation and describes oscillations (mostly electrical) where internalfriction depends on the value of the dependent variable x. Note that when − < <1 1x , we actuallyhave negative friction, so we would expect the system to move away from the zero solutionx t( ) ≡ 0 (note that this is a solution) in the direction of x t( ) = 1.

45. !!x tx+ = 0

Here, we have a vibrating spring with no friction, but the restoring force –tx gets stronger as timepasses.

!!!! LR-Circuit

46. (a) Without having a capacitor to store energy, we do not expect the current in the circuit tooscillate. If there had been a constant voltage V0 on in the past, we would expect the

current to be (by Ohm’s law) I VR

= 0 . If we then shut off the voltage, we would expect

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

(b) If a current I passes through a resistor with resistance R, then the voltage drop is RI; thevoltage drop across an inductor of inductance is LI! . Hence, setting this sum equal to theconstant voltage source V0 gives the first-order differential equation

LI RI! + = 0.

Because the voltage was previously on, we know that the initial current had reached an

equilibrium of I VR

0 0( ) = .


(c) The solution of LI RI! + = 0, I VR

0 0( ) = is

I t VR

e R L t( ) = −0 a f .

(d) If R = 40 ohms, L = 5 henries, V0 10= volts, then I t e t( ) = −14

8 ohms.

!!!! LC-Circuit

47. (a) With a nonzero initial current and no resistance, we do not expect the current to damp tozero. We would expect an oscillatory current due to the capacitor. Thus the charge on thecapacitor would oscillate indefinitely. The exact behavior depends on the initialconditions 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 equalto the impressed voltage source. Hence, we have

LIC

I! + =z1 0

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

LQC

Q!! + =1 0 , Q 0 0( ) = , !Q 0 5( ) = .

(c) The solution of the IVP is

Q ttLC

LC

( ) = 51

1

sine j.

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

(d) With values L = 10 henries, C = −10 3 farads, the charge on the capacitor is

Q tt

t( ) = =5100

10012

10sin

sind h

.

!!!! A Pendulum Experiment

48. The pendulum equation is

!! sinθ θ+ =gL

0 .

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

!!θ θ+ =gL

0 .


This is the equation of simple harmonic motion with circular frequency ω0 =gL

, and natural

frequency f gL0

12

=π

. Hence, the period of motion is Tf

Lg

= =1 20

π .

TT

gg

earth

sun

sun

earth= = = ≈400 000 100 40 632, .

!!!! Circular Motion

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

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

! sin !

!! sin !! cos ! .x rx r r= −( )

= −( ) − ( )

θ θ

θ θ θ θ2

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

!!x x+ =ω0 0.

!!!! Another Harmonic Motion

50. For simple harmonic motion the circular frequency ω0 is

ω 0

2

2=+

kRmR I

,

so the natural frequency f0 is

f kRmR I0

2

21

2=

+π.

!!!! Motion of a Buoy

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

T mk

= = =2 7 2 20

. πω

π .

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, V r h= π 2 , r = =9 0 75 inches ft. , h = 1, and

ρ = 62 5. , so the force required to push the buoy down 1 foot is π 916

1 62 5 110( )( ) ≈. lbs. But k is

the force divided by distance, so k = =110

1110 lbs ft . Finally, solving for m in the equation for


T, we get m kT=

2

24π, and plugging in all of our numbers, we arrive at m ≈ 20 4. slugs (see

Table 4.1.1. in the text.) The buoy weighs mg = ( )( ) =204 32 2 657. . lbs.

!!!! Factoring Out Friction

52. (a) Letting x t e X t X t e x tb m t b m tb g b g b g b gb g b g= ⇒ =− 2 2 , we have

! !

!! ! !! .

X t bm

e x t e x t

X t bm

e x t bm

e x t e x t

b m t b m t

b m t b m t b m t

b g b g b g

b g b g b g b g

b g b g

b g b g b g

= +

= + +

2

4

2 2

2

22 2 2

Plugging this into the original equation, we arrive at

mX k bm

X me x be x ke x

e mx bx kx

b m t b m t b m t

b m t

!! !! !

!! ! .

+ −FHG

IKJ = + +

= + + =

22 2 2

2

4

0

b g b g b g

b g

(b) If we assume k bm

− >2

40 , then the solution of this differential equation (after dividing by

m and letting

ω021

24= −

mmk b )

is

X t c t c t( ) = +1 0 2 0cos sinω ω .

Thus, we have

x t e X t e c t c tb m t b m t( ) = ( ) = +− −2 21 0 2 0

a f a f a fcos sinω ω .

!!!! Suggested Journal Entry

53. Student Project


4.2 Real Characteristic Roots

!!!! Real Characteristic Roots

1. ′′ =y 0

The characteristic equation is r2 0= , so there is double root at r = 0. Thus, the general solutionis

y t c e c te c c tt t( ) = + = +10

20

1 2 .

2. ′′ − ′ =y y 0

The characteristic equation is r r2 0− = , which has roots 0, 1. Thus, the general solution is

y t c c et( ) = +1 2 .

3. ′′ − =y y9 0

The characteristic equation is r2 9 0− = , which has roots 3, –3. Thus, the general solution is

y t c e c et t( ) = + −1

32

3 .

4. ′′ − =y y 0

The characteristic equation is r2 1 0− = , which has roots 1, –1. Thus, the general solution is

y t c e c et t( ) = + −1 2 .

5. ′′ − ′ + =y y y3 2 0

The characteristic equation is r r2 3 2 0− + = , which factors into r r−( ) −( ) =2 1 0, and hence has

roots 1, 2. Thus, the general solution is

y t c e c et t( ) = +1 22 .

6. ′′ − ′ − =y y y2 0

The characteristic equation is r r2 2 0− − = , which factors into r r−( ) +( ) =2 1 0, and hence has

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

y t c e c et t( ) = + −1

22 .

SECTION 4.2 Real Characteristic Roots 271

7. ′′ + ′ + =y y y2 0

The characteristic equation is r r2 2 1 0+ + = , which factors into r r+( ) +( ) =1 1 0, and hence has

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

y t c e c tet t( ) = +− −1 2 .

8. 4 4 0′′ − ′ + =y y y

The characteristic equation is 4 4 1 02r r− + = , which factors into 2 1 2 1 0r r−( ) −( ) = , and hence

has the double root 12

, 12

. Thus, the general solution is

y t c e c tet t( ) = +12

22.

9. 2 3 0′′ − ′ + =y y y

The characteristic equation is 2 3 1 02r r− + = , which factors into 2 1 1 0r r−( ) −( ) = , and hence

has roots 12

, 1. Thus, the general solution is

y t c e c et t( ) = +12

2 .

10. ′′ − ′ + =y y y6 9 0


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

y t c e c tet tb g = +13

23 .

11. ′′ − ′ + =y y y8 16 0



y t c e c tet t( ) = +14

24 .

12. ′′ − ′ − =y y y6 0

The characteristic equation is r r2 6 0− − = , which factors into r r+( ) −( ) =2 3 0, and hence has

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

y t c e c et t( ) = +−1

22

3 .


13. ′′ + ′ − =y y y2 0

The characteristic equation is r r2 2 1 0+ − = , which factors into r r+ − + + =1 2 1 2 0e je j , and

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

y t e c e c et t t( ) = +− −1

22

2d i.14. 9 6 0′′ + ′ + =y y y

The characteristic equation is 9 6 1 02r r+ + = , which factors into 3 1 02r +( ) = , and hence has the

double root − 13

, − 13

. Thus, the general solution is

y t c e c tet t( ) = +− −1

32

3.

!!!! Initial Values Specified

15. ′′ − =y y25 0, y 0 1( ) = , ′( ) =y 0 0

The characteristic equation of the differential equation is r2 25 0− = , which factors intor r−( ) +( ) =5 5 0, and thus has roots 5, –5. Hence,

y t c e c et t( ) = + −1

52

5 .

Substituting in the initial conditions y 0 1( ) = gives c c1 2 1+ = . Plugging in !y 0 0( ) = gives

5 5 01 2c c− = . Solving for c1, c2 gives c c1 212

= = . Thus the general solution is

y t e et t( ) = + −12

12

5 5 .

16. ′′ + ′ − =y y y2 0, y 0 1( ) = , ′( ) =y 0 0

The characteristic equation of the differential equation is r r2 2 0+ − = , which factors intor r+( ) −( ) =2 1 0, and thus has roots 1, –2. Thus, the general solution is

y t c e c et t( ) = +−1

22 .

Substituting into y 0 1( ) = , ′( ) =y 0 0 yields c113

= , c223

= , so

y t e et t( ) = +−13

23

2 .

17. ′′ + ′ + =y y y2 0, y 0 0b g = , ′( ) =y 0 1

The characteristic equation is r r2 2 1 0+ + = , which factors into r r+( ) +( ) =1 1 0, and hence has

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

y t c e c tet t( ) = +− −1 2 .


Substituting into y 0 0b g = , ′( ) =y 0 1 yields c1 0= , c2 1= , so

y t te t( ) = − .

18. ′′ − =y y9 0, y 0 1( ) = − , ′( ) =y 0 0

The characteristic equation is r2 9 0− = , which factors into r r−( ) +( ) =3 3 0, and hence has roots

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

y t c e c et t( ) = + −1

32

3 .

Substituting into y 0 1( ) = − , ′( ) =y 0 0 yields c c1 212

= = − , so

y t e et t( ) = − − −12

12

3 3 .

19. ′′ − ′ + =y y y6 9 0, y 0 0( ) = , ′( ) = −y 0 1



y t c e c tet t( ) = +13

23 .

Substituting into y 0 0( ) = , ′( ) = −y 0 1 yields c1 0= , c2 1= − , so

y t te t( ) = − 3 .

20. ′′ + ′ − =y y y6 0, y 0 1( ) = , ′( ) =y 1 1

The characteristic equation is r r2 6 0+ − = , which factors into r r+( ) −( ) =3 2 0, and hence has

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

y t c e c et t( ) = +−1

32

2 .

Substituting into y 0 1( ) = , ′( ) =y 1 1 yields c115

= , c245

= , so

y t e et t( ) = +−15

45

3 2 .

!!!! Phase Portraits

Careful inspection shows

21. (B) 22. (D) 23. (A) 24. (C)


!!!! Linking Graphs

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

25.

–5

y

3t

–5

5

y'

3t

–5

5

y'

5y

–5

12

3

4

1

2

3

44

1 32

t = 0

26.

–5

5

y

3t

–5

5

y'

3t

–5

5

y'

5y

–5

4

1

3

2

1

2

34

1

2

3 4

t = 0

27.

–5

5

y

3t

–5

5

y'

3t

3

–5

5

y'

5y

–5

1

2

3

4

32

41

4

1

2

3

t = 0

!!!! Independent Solutions

28. Letting

c e c er t r t1 21 2 0+ =

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

c cc e c er r1 2

1 2

001 2

+ =+ = .


When r r1 2≠ then these equations have the unique solution c c1 2 0= = , which shows the givenfunctions e er t r t1 2, are linearly independent for r r1 2≠ .

!!!! Second Solution I

29. Substituting y v t e bt a= ( ) − 2 into

ay by cy′′ + ′ + = 0

gives

′ = ′ −

′′ = ′′ − ′ +

− −

− − −

y v e ba

ve

y v e ba

v e ba

ve

bt a bt a

bt a bt a bt a

2 2

2 22

22

2

4.

Substituting v v v, ,′ ′′ into the differential equation gives the new equation (after dividing by

e bt a− 2 )

a v ba

v ba

v b v ba

v cv′′ − ′ +FHG

IKJ + ′ −FH

IK + =

2

24 20 .

Simplifying gives

av ba

c v′′ − −FHGIKJ =

2

40 .

Because we have assumed b ac2 4= , we have the equation ′′ =v 0, which was the condition to beproven.

!!!! Independence Again

30. Setting

c e c tebt a bt a1

22

2 0− −+ =

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

cc e c eb a b a

1

12

22

00

=

+ =− −

which have the unique solution c c1 2 0= = . Hence, the given functions are linearly independent.

!!!! Repeated Roots, Long-Term Behavior

31. Because e bt a− 2 approaches 0 as t →∞ (for a, b > 0), we need only verify that

te te

bt abt a

− = →22 0


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

1

22b

abt ae

,

which clearly approaches 0 as t →∞ . Using l’Hopital’s rule, the given expression

12ebt a

approaches 0 as well.

!!!! Negative Roots

32. We have r b b mk= − ± −2 4 , so in the overdamped case where b mk2 4 0− > , these characteristicroots are real. Because m and k are both nonnegative, b mk b2 24− < causing

r b b mk12 4= − + − to be a negative sum of negative and positive terms

and

r b b mk22 4= − − − to be a negative sum of two negative terms.

!!!! Circuits and Springs

33. (a) The LRC equation is LQ RQC

Q!! !+ + =1 0 , hence the following discriminate conditions

hold:

∆

∆

∆

= − < ( )

= − = ( )

= − > ( )

R LC

R LC

R LC

2

2

2

4 0

4 0

4 0

underdamped

critically damped

overdamped .

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

R LC

R LC

R LC

< ( )

= ( )

> ( )

4

4

4

underdamped

critically damped

overdamped .

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.)


!!!! Alternate Solution Form

34. The characteristic equation is r2 1 0− = , which has roots of 1, –1. Hence, the general solution is

y t c e c et t( ) = + −1 2 .

But

sinh

cosh

t e e

t e e

t t

t t

= −

= +

−

−

1212

b g

b g

so we can solve for

e t te t t

t

t

= +

= −−

cosh sinhcosh sinh .

Thus, the general solution can be written

y t c e c e c t t c t tC t C t

t t( ) = + = +( ) + −( )= +

−1 2 1 2

1 2

cosh sinh cosh sinhsinh cosh

where C c c1 1 2= − , C c c2 1 2= + .

!!!! The Euler-Cauchy Equation at at y bty cy2 0′′ + ′ + =

35. Let y t tr( ) = , so

′ =

′′ = −( )

−

−

y rty r r t

r

r

1

21 .

Hence

at y bty cy ar r t brt ctr r r2 1 0′′ + ′ + = −( ) + + = .

Dividing by tr yields the characteristic equation

ar r br c−( ) + + =1 0,

which can be written as

ar b a r c2 0+ −( ) + = .

If r1 and r2 are two distinct roots of this equation, we have solutions

y t ty t t

r

r1

2

1

2

( ) =

( ) = .


Because these two functions are clearly linearly independent (one not a constant multiple of theother) for r r1 2≠ , we have

y t c t c tr r( ) = +1 21 2

for t > 0 .

!!!! The Euler-Cauchy Equation with Distinct Roots

For Problems 36–40, see Problem 35 for the unusual form of the characteristic equation.

36. t y ty y2 2 12 0′′ + ′ − =

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

r r r r r r r−( ) + − = + − = +( ) −( ) =1 2 12 12 4 3 02 .

Hence, we have roots r1 4= − , r2 3= , and thus

y t c t c t( ) = + −1

32

4 .

37. 4 8 3 02t y ty y′′ + ′ − =

In this case a = 4, b = 8, c = −3, so the characteristic equation is

4 1 8 3 4 4 3 2 1 2 3 02r r r r r r r−( ) + − = + − = −( ) +( ) = .

Hence, we have roots r112

= , r232

= − , and thus

y t c t c t( ) = + −1

1 22

3 2 .

38. t y ty y2 4 2 0′′ + ′ + =

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

r r r r r r r−( ) + + = + + = +( ) +( ) =1 4 2 3 2 1 2 02 .

Hence, we have roots r1 1= − , r2 2= − , and thus

y t c t c t( ) = +− −1

12

2 .

39. 2 3 02t y ty y′′ + ′ − =

In this case a = 2, b = 3, c = −1, so the characteristic equation is

2 1 3 1 2 1 2 1 1 02r r r r r r r−( ) + − = + − = −( ) +( ) = .

Hence, we have roots r112

= , r2 1= − , and thus

y t c t c t( ) = + −1

1 22

1.


!!!! Repeated Euler-Cauchy Roots

40. We are given that the characteristic equation

ar b a r c2 0+ −( ) + =

of Euler’s equation

at y bty cy2 0′′ + ′ + =

has a double root of r. Hence, we have one solution y tr1 = . To verify that t tr ln is also a

solution, we differentiate ′ = +− −y rt t tr r1 1ln ,

′′ = −( ) + + −( ) = −( ) + −( )− − − − −y r r t t rt r t r r t t r tr r r r r1 1 1 2 12 2 2 2 2ln ln .

By direct substitution we have

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

r r r r r

r r

2 2 2 2 1 11 2 1

1 2 1

′′ + ′ + = −( ) + −( ) + + +

= −( ) + + + −( ) +

− − − −ln ln ln

ln .

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

at y bty cy a r b tr2 2 1′′ + ′ + = −( ) + .

But the root of the characteristic equation is r b aa

= −−2

, which makes this expression zero.

To verify that tr and t tr ln are linearly independent (where r is the double root of the

characteristic equation, which is r b a= −

−2

), we set

c t c t tr r1 2 0+ =ln

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

cc cr r

1

1 2

02 2 2 0

=

+ =ln

and yields the unique solution c c1 2 0= = . Hence, tr and t tr ln are linearly independent solutions.

!!!! Solutions for Repeated Euler-Cauchy Roots

For Problems 41 and 42 use the result of Problem 40, y t C t C t tr r( ) = +1 2 ln .

41. t y ty y2 5 4 0′′ + ′ + =

In this case, a = 1, b = 5, and c = 4, so our characteristic equation for r is r r2 4 4 0+ + = , andhence it has a double root at –2. The general solution is

y t c t c t t( ) = +− −1

22

2 ln

for t > 0 .


42. t y ty y2 3 4 0′′ − ′ + =

In this case, a = 1, b = −3, and c = 4, so our characteristic equation for r is r r2 4 4 0− + = , andhence it has a double root at 2. The general solution is

y t c t c t t( ) = +12

22 ln

for t > 0 .

!!!! Third-Order Euler-Cauchy

43. (a) The third-order Euler Cauchy equation has the form at y bt y cty dy3 2 0′′′ + ′′ + ′ + = .Substitute y tr= t >( )0 to obtain the characteristic equation

′ =

′′ = −( )

′′′ = −( ) −( )

−

−

−

y rty r r ty r r r t

r

r

r

1

2

3

11 2

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

at r r r t bt r r t ctrt dtat r r r bt r r ct r dt

r r r r

r r r r

3 3 2 2 11 2 1 01 2 1 0

−( ) −( ) + −( ) + + =

−( ) −( ) + −( ) + + =

− − −

Dividing by tr , we obtain the characteristic equation:

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

(b) t y t y ty y3 2 2 2 0′′′ + ′′ − ′ + = has Euler-Cauchy characteristic equation:

r r r r r rr r r r r r

r r r

−( ) −( ) + −( ) − + =

− + + − − + =

− − + =

1 2 1 2 2 03 2 2 2 0

2 2 0

3 2 2

3 2

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

f 1 1 2 1 2 0( ) = − − + = .

Therefore r −1 is a factor of r r r3 22 2− − + . r r r r r r3 22 2 1 1 2− − + = −( ) +( ) −( ) sor = 1, –1, 2. Solution: y t c t c t c t( ) = + +−

1 21

32 for t >( )0 .

!!!! A Test of Your Intuition

44. Intuitively, a curve whose rate of increase is proportional to its height will increase very rapidlyas the height increases. On the other hand, upward curvature doesn’t necessarily imply that thefunction is increasing! (The curve e t− has upward curvature, yet decreases to 0 as t →∞ .) In thiscase, the restriction that ′( ) =y 0 0 will cause the second curve to increase, but probably not nearly


as rapidly as the first curve. Solving the equations, the IVP ′ =y y , y 0 1( ) = has the solutiony et= , whereas the second curve described by ′′ =y y , y 0 1( ) = , ′( ) =y 0 0 has the solution

y t e et t( ) = + −12

12

.

The first curve is indeed above the second curve.

!!!! An Overdamped Spring

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

x t c e c er t r tb g = +1 21 2 .

Suppose that

c e c er t r t1 2

1 1 2 1 0+ =

for some t1 . Because er t2 1 is never zero, we can divide by er t2 1 to get c e cr r t1 21 2 1 0− + =a f .

Solving for t1 gives

tr r

cc1

1 2

2

1

1=

−−ln .

This unique number is the only value for which the curve may pass through 0. If theargument of the logarithm is negative or if the value of t1 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

46. (a) Suppose

c c t er t1 2 1 1 1 0+ =a f .

We can divide by the nonnegative quantity er t1 1 getting the equation c c t1 2 1 0+ = , which

has the unique solution t cc1

1

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

pass through the equilibrium at most once. If the value of t1 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.


!!!! Damped Vibration

47. The IVP problem is

!! !x x x+ + =2 0, x 0 3 14

( ) = = in ft , !x 0 0 ( ) = ft sec .

The solution is

x t e tet t( ) = +− −14

14

.

This is zero only for t1 1= − , whereas the physical system does not start before t = 0.

!!!! LRC-Circuit I

48. (a) LQ RQC

Q Q Q Q!! ! !! !+ + = + + =1 2 101 50 0 , Q 0 99( ) = , !Q 0 0( ) =

(b) Q t e et t( ) = − +− −50 2100 (c) I t Q t e et t( ) = ( ) = −− −! 50 5050 2

(d) As t →∞ , Q t( )→ 0 and I t( )→ 0

!!!! LRC-Circuit II

49. (a) LQ RQC

Q Q Q Q!! ! !! !+ + = + + =1 15 50 0 , Q 0 5( ) = , !Q 0 0( ) =

(b) Q t e et t( ) = −− −10 55 10 (c) I t Q t e et tb g b g= = − +− −! 50 505 10

(d) As t →∞ , Q t( )→ 0 and I t( )→ 0

!!!! Computer: Phase-Plane Trajectories

50. (a) y t e et t( ) = +− −2 3

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

r r r r+( ) +( ) = + + =1 3 4 3 02 .

y t( ) satisfies the differential equation

′′ + ′ + =y y y4 3 0.

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

′( ) = − −− −y t e et t2 3 3 .

The IC of the given trajectory of y t y t( ) ′( )( ), in yy′ space is y y0 0 3 5( ) ′( )( ) = −( ), , .


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

′( ) = − −− −y t e et t2 3 3

The IC of the given trajectory of

y t y t( ) ′( )( ),

in yy′ space is y y0 0 3 5( ) ′( )( ) = −( ), , .

(c) We plot the trajectory starting at 3 5, −( )

along with a few other trajectories in yy′

space.

y

–4

4

!y

4–4 2–2

2

–2

(3, –5)

DE solutions in yy′ space

51. y t e et t( ) = +− −8

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

r r r r+( ) +( ) = + + =1 8 9 8 02 .


′′ + ′ + =y y y9 8 0.

(b) The derivative is

′( ) = − −− −y t e et t8 8 .

The IC for the given trajectory in yy′

space is

y y0 0 2, 9( ) ′( )( ) = −( ), .

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

–10 10

–10

10

y

y'

(2,-9)


52. y t e et t( ) = + −

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

r r r−( ) +( ) = − =1 1 1 02 .


′′ − =y y 0.



′ = − −y t e et tb g .


space is

y y0 0 2, 0( ) ′( )( ) = ( ), .

(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)


53. y t e tet t( ) = +− −

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

r r r+( ) = + + =1 2 1 02 2 .


′′ + ′ + =y y y2 0.


′( ) = − −y t te t .


space is

y y0 1 0 0( ) = ′( ) =, .

(c) See the figure to the right.

–1.5 1 1.5

–1.5

1.5

y(1, 0)

y'


54. y t e t( ) = +3 2 2

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

r r r r−( ) = − =2 2 02 .


′′ − ′ =y y2 0.



′( ) =y t e t4 2 .


space is

y y0 0 5 4( ) ′( )( ) = ( ), , .

(c) See the figure to the right.

y

–8

!y

8–8 4–4

–4

4

8

(5, 4)


!!!! Reduction of Order

55. (a) Let y vy2 1= and

′ = ′ + ′′′ = ′′ + ′ ′ + ′′

y v y vyy v y v y vy

2 1 1

2 1 1 12 .

Then

′′ + ( ) ′ + ( ) = ′′ + ′ ′ + ′ + ′′+ ′ + =y p x y q x y v y v y pv y vy pvy qvy2 2 2 1 1 1 1 1 12 0a f .

Because ′′+ ′ + =y py qy1 1 1 0, cancel the terms involving v, and arrive at the new equationy v y p x y v1 1 12 0′′ + ′ + ( ) ′ =a f

(b) Setting ′ =v w

y w y p x y w

w y p x yy

w

dww

y p x yy

dx

w yy

p x dx

wy

dy p x dx

w y p x dx

w ey

v

v ey

p x dx

p x dx

1 1 1

1 1

1

1 1

1

1

11

12

12

12

2 0

2 0

2

2

2

′ + ′ + ( ) =

′ +′ + ( )F

HGIKJ =

=− ′ − ( )FHG

IKJ

=− ′ − ( )FHG

IKJ

=−

− ( )

= − ( )

= ±z

= ′

= ±z

zz zz

z

−

− ( )

− ( )

a f

ln

ln

ln ln

By convention, the positive sign is chosen.


(c) If v is a constant function on I, then ′ ≡v 0 and w ≡ 0 because ′ =v w . The conditionw ≡ 0 contradicts our work in part (b) as ln w where w = 0 is undefined. Because v isnot constant on I, y y1 2, k p is a linearly independent set of I.

!!!! Second Solution II

56. ′′ − ′ + =y y y6 9 0, y e t1

3=

We identify p t( ) = −6, so

p t dt t( ) = −z 6 .

Plugging in the formula developed in Problem 55, we have

y y ey t

dt e ee

dt tep t dt

tt

tt

2 112

36

3 23=

z( )

= =− ( )z z b g .

57. ′′ − ′ + =y y y4 4 0 , y e t1

2=

We won’t use the formula this time. We simply redo the steps in Problem 55. We seek a secondsolution of the form y vy ve t

2 12= = . Differentiating, we have

′ = ′ +

′′ = ′′ + ′ +

y v e vey v e v e ve

t t

t t t2

2 2

22 2 2

24 4 .

Plugging into the equation we obtain

′′− ′ + = ′′ =y y y v e t2 2 2

24 4 0 .

Dividing by e t2 gives ′′ =v 0 or

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

Hence, we have found new solutions

y ve c te c et t t2

21

22

2= = + .

Because y e t1

2= , we let c1 1= , c2 0= , yielding a second independent solution

y te t2

2= .

58. t y ty y2 0′′ − ′ + = , y t1 =

We won’t use the formula this time. We simply redo the steps in Problem 55. We seek a secondsolution of the form y vy tv2 1= = . Differentiating, we have

′ = ′ +′′ = ′′ + ′

y tv vy tv v

2

2 2 .


Plugging into the equation we obtain

t y ty y t v t v22 2 2

3 2 0′′ − ′ + = ′′ + ′ = .

Letting w v= ′ and dividing by t3 yields

′ + =wt

w1 0 .

We can solve by integrating the factor method, getting w c t= −1

1. Integrating we find

v c t c= +1 2ln ,

so

y tv c t t c t2 1 2= = +ln .

Letting c1 1= , c2 0= , we get a second linearly independent solution

y t t2 = ln .

59. t y ty y2 1 2 2 0+ ′′ − ′ + =b g , y t1 =

We won’t use the formula this time. We simply redo the steps in Problem 55. We seek a secondsolution of the form y vy tv2 1= = . Differentiating yields

′ = ′ +′′ = ′′ + ′

y tv vy tv v

2

2 2 .

Plugging into the equation we get

t y ty y t t v v22 2 2

21 2 2 1 2 0+ ′′ − ′ + = + ′′ + ′ =b g b g .

Letting w v= ′ and dividing by t t2 1+b g, we can solve the new equation using the integrating

factor method, getting

21

1 212

22

2t tdt t t t

t+= − + + =

+z b g b gln ln ln .

We arrive at

w c tt

c c t=+

= + −1

2

2 1 121 .

Integrating this, we get

v c t t c= − +−1

12b g ,

so

y tv c t c t2 12

21= = − +b g .


Letting c1 1= , c2 0= we get a second linearly independent solution

y t22 1= − .

!!!! Classical Equations

60. ′′ − ′ + =y ty y2 4 0 , y t t121 2( ) = − (Hemite’s Equation)

Letting y vy v t2 121 2= = −b g, we have

′ = − ′ −

′′ = − ′′ − ′ −

y t v tv

y t v tv v2

2

22

1 2 4

1 2 8 4

b gb g

and perform the long division, yielding the equation

′′ + − +−

FH

IK ′ =v t t

tv2 8

2 102 .

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

w c e tt= − −1

2 22 2 1b g .

To find y2 we simply let c1 1= and integrate to get

v e t dtt= −z −2 2 12 2b g .

Multiplying by 1 2 2− tb g yields a final answer of

y t t e t dtt2

2 2 21 2 2 1

2b g d i d i= − −z −.

61. 1 02− ′′ − ′ + =t y ty yb g , y t t1( ) = (Chebyshev’s Equation)

Letting y vy vt2 1= = , we have

′ = ′ +y tv v2 , ′′ = ′′ + ′y tv v2 2 ,

hence we have the equation

1 1 2 3 022 2 2

2 2− ′′ − ′ + = − ′′ + − ′ =t y ty y t t v t vb g b g b g .

Dividing by t t1 2−b g and letting w v= ′ , yields

′ +−−

=w tt t

w2 31

02

2b g .

Using partial fractions yields

2 31

2 12

1 12

12

2−−

= + − + +z tt t

dt t t tb g ln ln ln ,


so our integrating factor is t t2 21− or

w ct t

=−

1 2 2

11

.

Letting c1 1= and multiplying by t yields a final answer of

y t tv tt t

dt2 2 2

11

( ) = =−

z .

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.

62. ty t y y′′ + −( ) ′ + =1 0, y t t1 1( ) = − (Laguerre’s Equation)

Letting y vy v t2 1 1= = −( ) , we have

′ = ′ −( ) +y v t v2 1 , ′′ = ′′ −( ) + ′y v t v2 1 2 ,

hence we have the equation

ty t y y t t v t t v′′+ − ′ + = − ′′ + − + − ′ =2 221 1 4 1 0b g b g d i .

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

′ +− + −

−=w t t

t tw

2 4 11

0b g .

Hence

w C et t

t

=−1 21( )

.

Letting c1 1= and multiplying by t −1 yields a final answer of

y t v t t et t

dtt

2 21 11

b g b g b g b g= − = −−z .


63. Student Project


4.3 Complex Characteristic Roots

!!!! Solutions in General

1. ′′ + =y y9 0

The characteristic equation is r2 9 0+ = , which has roots 3i, –3i. The general solution is

y t c t c t( ) = +1 23 3cos sin .

2. ′′ + ′ + =y y y 0

The characteristic equation is r r2 1 0+ + = , which has roots − ±12

32

i . The general solution is

y t e c t c tt( ) = +FHG

IKJ

− 21 2

32

32

cos sin .

3. ′′ − ′ + =y y y4 5 0

The characteristic equation is r r2 4 5 0− + = , which has roots 2 ± i . The general solution is

y t e c t c tt( ) = +21 2cos sina f .

4. ′′ + ′ + =y y y2 8 0

The characteristic equation is r r2 2 8 0+ + = , which has roots − ±1 7i . The general solution is

y t e c t c tt( ) = +−1 27 7cos sind h .

5. ′′ + ′ + =y y y2 4 0

The characteristic equation is r r2 2 4 0+ + = , which has roots − ±1 3i . The general solution is

y t e c t c tt( ) = +−1 23 3cos sind h.

6. ′′ − ′ + =y y y4 7 0

The characteristic equation is r r2 4 7 0− + = , which has roots 2 3± i . The general solution is

y t e c t c tt( ) = +21 23 3cos sind h .

7. ′′ − ′ + =y y y10 26 0

The characteristic equation is r r2 10 26 0− + = , which has roots 5+ i . The general solution is

y t e c t c tt( ) = +51 2cos sina f.

SECTION 4.3 Complex Characteristic Roots 291

8. 3 4 9 0′′ + ′ + =y y y

The characteristic equation is 3 4 9 02r r+ + = , which has roots − ±23

233

i . The general solution

is


IKJ

−2 31 2

233

233

cos sin .

9. ′′ − ′ + =y y y 0

The characteristic equation is r r2 1 0− + = , which has roots 12

32

± i . The general solution is


IKJ

21 2

32

32

cos sin .

10. ′′ + ′ + =y y y2 0

The characteristic equation is r r2 2 0+ + = , which has roots − ±12

72

i . The general solution is


IKJ

− 21 2

72

72

cos sin .

!!!! Initial-Value Problems

11. ′′ + =y y4 0 , y 0 1( ) = , ′( ) = −y 0 1

The characteristic equation is r2 4 0+ = , which has roots ±2i . The general solution is

y t c t c t( ) = +1 22 2cos sin .

Plugging this into the initial conditions gives y c0 11( ) = = , ′( ) = = −y c0 2 12 . Hence, the solution

of the initial-value problem is

y t t t( ) = −cos sin2 12

2 .

12. ′′ − ′ + =y y y4 13 0, y 0 1( ) = , ′( ) =y 0 0

The characteristic equation is r r2 4 13 0− + = , which has roots 2 3± i . The general solution is

y t e c t c tt( ) = +21 23 3cos sina f .

Plugging this into the initial conditions yields y c0 11( ) = = , ′( ) = + =y c c0 2 3 01 2 , resulting in

c1 1= , c223

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

y t e t tt( ) = −FH IK2 3 23

3cos sin .


13. ′′ + ′ + =y y y2 2 0 , y 0 1( ) = , ′( ) =y 0 0

The characteristic equation is r r2 2 2 0+ + = , which has roots − ±1 i . Hence, the general solutionis

y t e c t c tt( ) = +−1 2cos sina f .

Plugging this into the initial conditions yields y c0 11( ) = = , ′( ) = − =y c c0 01 2 , resulting in c1 1= ,c2 1= . Hence, the solution of the initial-value problem is

y t e t tt( ) = +( )− cos sin .

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

From Problem 6,

y t e c t c tt( ) = ( )LNMOQP + ( )

LNMOQP

RSTUVW

21 2

32

32

cos sin .

Plugging this into the initial conditions yields y 0 0( ) = , ′( ) =y 0 1, resulting in c1 0= , c223

3= .

Hence, the solution of the initial-value problem is

y t e tt( ) =FHGIKJ

−23

3 32

2 sin .

15. ′′ − ′ + =y y y4 7 0 , y 0 0( ) = , ′( ) = −y 0 1

From Problem 6,

y t e c t c tt( ) = +21 23 3cos sind h d hn s.

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

c1 0= , c213

3= − .


y t e tt( ) = −13

3 32 sind h .

16. ′′ + ′ + =y y y2 5 0 , y 0 1( ) = , ′( ) = −y 0 1

The characteristic equation is r r2 2 5 0+ + = , which has roots − ±1 2i . Hence, the general solu-tion is

y t e c t c tt( ) = +−1 22 2cos sina f .


Plugging this into the initial conditions yields y 0 1( ) = , ′( ) = −y 0 1, resulting in c1 1= , c2 0= .


y t e tt( ) = − cos2 .

!!!! Euler’s Formula

17. (a) The Maclaurin series for ex is

e x x xn

xx n= + + + + + +1 12

13

12 3! ! !

" "

(b) e i i in

ii nθ θ θ θ θ= + + ( ) + ( ) + + ( ) +1 12

13

12 3

! ! !" "

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

e i i in

i

i i

i nθ θ θ θ θ

θ θ θ θ θ θ θ

= + + ( ) + ( ) + + ( ) +

= − + − +FH IK + − + −FH IK = +

1 12

13

1

1 12

14

13

15

2 3

2 4 3 5

! ! !

! ! ! !cos sin

" "

" "

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

!!!! Changing the Damping

18. The curves below show the solution of

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

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

–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 4.3.6 (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 everytime a curve in (a) crosses the axis twice the corresponding trajectory in (b) circles the origin.


!!!! Changing the Spring

19. (a) The solutions of

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

are shown for k = 14

, 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.

!!!! Linking Graphs

20.

–5

5

y

3t

–5

5

y'

1t

–5

5

y'

5y

–5

3

2

1

3

1

23

21

t = 0

21.

–5

5

y

3t

–5

5

y'

1t

–5

5

y'

5y

–5

3

2

13

2

1

3

2

1

t = 0


!!!! Long-Term Behavior of Solutions

22. r1 0< , r2 0< . When r r1 2≠ , the solution is

y t c e c er t r t( ) = +1 21 2

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

y t c e c tert rt( ) = +1 2 .

In this case using l’Hopital’s rule we prove the second term tert goes to zero as t →∞ whenr < 0 .

23. r1 0< , r2 0= . The solution

y t c e cr t( ) = +1 21

approaches the constant c2 as t →∞ because r1 0< .

24. r i= ±α β , y t e c t c tt( ) = +α β β1 2cos sina f . For β ≠ 0 the solution y t( ) oscillates with decreasing

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

25. r1 0= , r2 0= . The solution

y t c c t( ) = +1 2

approaches plus ∞ as t →∞ when c2 0> and minus ∞ when c2 0< .

26. r1 0> , r2 0< . The solution

y t c e c er t r t( ) = +1 21 2

approaches plus ∞ as t →∞ when c1 0> and minus ∞ when c1 0< .

27. r i= ±β , y t c t c t( ) = +1 2cos sinβ β is a periodic function of 2πβ

and amplitude c c12

22+

!!!! Linear Independence

28. Suppose

c e t c e tt t1 2 0α αβ βcos sin+ =

on an arbitrary interval. Dividing both sides by e tα then differentiating the new equation anddividing by β, yields

c t c tc t c t

1 2

2 1

00

cos sincos sin .

β ββ β+ =− =

Hence, c c1 20 0= =, and we have proven linear independence of the given functions.


!!!! Real Coefficients

29. Solution of the differential equation is

y t k e t i t k e t i te k k t ie k k t

t t

t t

( ) = + + −

= + + −1 2

1 2 1 2

α α

α α

β β β β

β β

cos sin cos sincos sin .

a f a fa f a f

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

k k rk k si

1 2

1 2

+ =− =

Solving for k1 and k2 , we get

k r si

k r si

1

2

12

12

12

12

= +

= − .

!!!! Solving d ydt

n

n = 0

30. (a)d ydtd ydt

k

d ydt

k t k

dydt

k t k t k

y k t k t k t k

4

4

3

3 3

2

2 3 2

32

2 1

33

22

1 0

0

1213

12

=

=

= +

= + +

= + + +!

(b) y 4 0( ) = . The characteristic equation is

r4 0= , which has a fourth-order root at0. Hence, the solution is

y t c c t c t c t( ) = + + +0 1 22

33,

which is the same as in part (a).

(c) In general we have

y t kn

t kn

t k t k

c t c t c t c

nn

nn

nn

nn

( ) =−( )

+−( )

+ + +

= + + + +

−−

−−

−−

−−

11

22

1 0

11

22

1 0

11

12! !

"

"

because all of the constants are arbitrary.

!!!! Higher-Order DEs

31. d ydt

d ydt

d ydt

5

5

4

4

3

34 4 0− + =

The characteristic equation is

r r r r r r r r5 4 3 3 2 3 24 4 4 4 2 0− + = − + = −( ) =b g ,


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

y t c c t c t c e c tet t( ) = + + + +1 2 32

42

52 .

32. d ydt

d ydt

dydt

y3

3

2

24 7 10 0+ − − =


r r r3 24 7 10 0+ − − = ,

which has roots, –1, 2, –5. Hence,

y t c e c e c et t t( ) = + +− −1 2

23

5 .

33. d ydt

dydt

5

5 0− =


r r r r r r r r r r r5 4 2 2 21 1 1 1 1 1 0− = − = − + = −( ) +( ) + =b g b gb g b g ,

which has roots, 0, ±1, ±i. Hence

y t c c e c e c t c tt t( ) = + + + +−1 2 3 4 5cos sin .

!!!! Oscillating Euler-Cauchy

34. We used the substitution y tr= and obtained for r i1 = +α β and r i2 = −α β the solution

y t k t k t k e k e k e k ee c t c t t c t c t

i i i t i t t i t t i t

t

( ) = + = + = +

= ( ) + ( ) = ( ) + ( )

+ − +( ) −( ) + −1 2 1 2 1 2

1 2 1 2

α β α β α β α β α β α β

α αβ β β β

ln ln ln ln ln ln

ln cos ln sin ln cos ln sin ln .a f a fThis is the same process as that used at the start of Case 3 in the text utilizing the Euler’s For-mula (4).

35. t y ty y2 2 0′′ + ′ + = , r r r−( ) + + =1 2 1 0, r r2 1 0+ + = , r i= − ±12

32

,

y t t c t c t( ) =FHGIKJ +

FHGIKJ

LNM

OQP

−1 21 2

32

32

cos ln sin ln

36. t y ty y2 3 5 0′′ + ′ + =

Letting y tr= yields

′ =

′′ = −( )

−

−

y rty r r t

r

r

1

21 .


Plugging this into the differential equation yields

t r r t trt tt r r r

r r r

r

2 2 11 3 5 01 3 5 0

−( ) + + =

−( ) + +{ } =

− −

,

and gives r r2 2 5 0+ + = , which has roots − ±1 2i . Hence, the solution is

y t t c t c t( ) = ( ) + ( )−11 22 2cos ln sin lna f .

!!!! Inverted Pendulum

37. The differential equation !!x x− = 0 has the characteristic equation r2 1 0− = with roots ±1.Hence, the general solution is

x t c e c et t( ) = + −1 2 .

(a) With initial conditions x 0 0( ) = , !x 0 1( ) = , we find c112

= and c212

= − . Hence, the solu-

tion of the IVP is

x t e e tt t( ) = − =−12

12

sinh .

(b) As t →∞ , x t( )→ 0 if c1 0= , and then !x t( )→ 0 also. This will happen whenever

x x0 0( ) = − ( )! .

!!!! Pendulum and Inverted Pendulum

38. (a) The inverted pendulum equation has characteristic equation r2 1 0− = , which has roots±1. Hence, the solution

x t c e c e c t t c t t C t C tt t( ) = + = +( ) + −( ) = +−1 2 1 2 1 2cosh sinh cosh sinh sinh cosh ,

where C c c1 1 2= − , C c c2 1 2= + .

(b) The characteristic equation of the pendulum equation is r2 1 0+ = , which has roots ±i .Hence, the solution

x t c t c t( ) = +1 2cos sin .

(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 choosesthe coefficients c c1 2, in each case. The span of e eit it, −l q is the same as the span of

sin , cost t { } .


!!!! Finding the Damped Oscillation

39. The initial conditions

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

give the constants c1 1= , c2 2= . Hence, we have

x t e t tt( ) = +( )− cos sin2 .

5

–0.5

1.5x

t

1

0.5

31

x t e t tt( ) = +( )− cos sin2

!!!! Extremes of Damped Oscillations

40. The local maxima and minima of the curve

x t e c t c tt( ) = +α ω ω1 2cos sina fhave nothing to do with the exponential factor e t−α ; they depend only on

c t c t1 2cos sinω ω+ ,

which can be rewritten as A tcos ω δ−a f having periodic function T =2πω

. 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 24 that the time between the first local maxima and

the first local minima is π π1= .)

!!!! Underdamped Mass-Spring System

41. We are given parameters and initial conditions

m = 0 25. , b = 1, k = 4 , x 0 1( ) = , !x 0 0( ) = .

Hence, the IVP is

0 25 4 0. !! !x x x+ + = , x 0 1( ) = , !x 0 0( ) = ,

which has the solution

x t e t ttb g = +FHG

IKJ

−2 2 3 33

2 3cos sin .


!!!! Damped Mass-Spring System

42. The IVP is

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

(a) b = 10: (underdamped), x t e t tt( ) = +FH IK−5 39 539

39cos sin

(b) b = 16: (critically damped), x t t e t( ) = +( ) −1 8 8

(c) b = 20: (overdamped), x t e et t( ) = −− −13

4 4 16b g

!!!! LRC-Circuit I

43. (a) The IVP is LQ RQC

Q Q Q Q!! ! !! !+ + = + + =1 8 25 0 , Q 0 1( ) = , ′( ) =Q 0 0

(b) Q t e t tt( ) = +FH

IK−4 3 4

33cos sin . Q t e tt( ) = −( )−5

334 cos δ where δ = −tan 1 4

3

(c) I t Q t e t e tt t( ) = ′( ) = − −( ) − −( )− −5 3 203

34 4sin cosδ δ where δ = −tan 1 43

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

!!!! LRC-Circuit II

44. (a) The IVP is LQ RQC

Q Q Q Q!! ! !! !+ + = + + =1 1

41 4 0 , Q 0 1( ) = , ′( ) =Q 0 0

(b) Q t e t t e tt t( ) = +FHG

IKJ = −− −2 22 3 3

32 3 2 3

32 3cos sin cos δd h , tanδ = 3

3

(c) I t Q t e t e tt t( ) = ( ) = − − − −− −! cos sin4 33

2 3 4 2 32 2δ δd h d h

(d) As t →∞ , Q t( )→ 0 and I t( )→ 0

!!!! Computer Lab: Damped Free Vibrations

45. IDE Lab

!!!! Reading Nonautonomous Differential Equations

46. (a) !!xt

x+ =1 0 . 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 eventuallydecays to zero. One would expect that the solution amplitude would increase as the


restoring force decays to zero. It would be interesting to investigate the solutions oft hisequation using an open-ended graphic solver.

(b) Solution was plotted with IC x 01 2.( ) = , ! .x 01 0( ) = in the xt as well as the !xx planes.

200100

5

10

x

t300

15

–15

–10

–5400 500

10 20–10

–2

–20

–4

1

3

x–1

–3

2

!x

(c) As we expected the oscillations amplitude increases with t.

47. (a) !! !xt

x x+ + =1 0 . 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 theamplitude of a solution will rapidly decay, but as time increases the motion will becomealmost periodic oscillations, as there will be almost no friction.

(b) Solution was plotted for IC x 01 2.( ) = , ! .x 01 0( ) = in the xt as well as the !xx planes.

2010

0.5

x

t30

1

–0.5

40 50

1 2–1

–1

0.5

–2

x–0.5

–1.5

1

!x

(c) As first expected, initially the solution is rapidly decaying; however, our second expecta-tion is not confirmed! As time increases the oscillations do not become harmonic—theamplitude of the oscillations decrease gradually.

48. tx x!!+ = 0. If you divide by t, you will see that this equation is the same as the equation inProblem 46.


49. (a) !! !x x x x+ − + =2 1 0b g . Negative friction for 0 1< <x , positive damping for x ≥1. For a

small initial condition near x = 0, we might expect the solution to grow and thenoscillate around x = 1. This would be a good DE to investigate with an open-endedgraphical DE solver.

(b) Solution was plotted with different IC:

x 0 05( )( ) = . ,x 0 2( ) = ,

and

x 0 4( ) = ; !x 0 0( ) = ).

(c) As expected, the solution grows for smallinitial conditions, whereas for x >1 we

see damping.

105

23

x

t15

4

–2–1

1

20

50. (a) !! sin !x t x x+ ( ) + = 0. Damping changes periodically from negative to positive. It would be

fascinating to investigate the solutions of this equation using an open-ended graphicssolver. (See figure.)

(b) Solution was plotted with IC x x0 2, 0 0( ) = ( ) =! in the xt as well as in the !xx planes.

4020

1

2

x

t60

–2

–1

80 100 1 2–1

–1

–2

–2

x

1

2!x

(c) The solution is a superposition of two periodic oscillations.

51. (a) !! !xt

x tx+ + =1 0 . Damping is initially large, but it vanishes as time increases; the restoring

force, on the other hand, is initially small, but increases with time. The question is:which effect will dominate in hte long run? It is rather difficult to predict; you mightexplore the problem with an open-ended graphic ODE solver.


(b) Solution was plotted with IC x x01 2, 01 0. ! .( ) = ( ) = in the xt as well as in the !xx planes.

105

1

1.5

x

t15–0.5

0.5

20 25 30

2

–1

1 2–1

–1

–2

x

1

2!x

(c) Note that initially there is large damping, so the amplitude of the solution is decreasing.As time progresses, however, the damping keeps on decreasinga nd the spring constantkeeps on increasing. As a result, the spring vibration frequency keeps on growing and thespring vibration amplitude keeps on declining.

52. (a) !! sinx t x+ ( ) =2 0. Restoring force changes periodically from positive to negative with a

frequency different from the natural frequency of the spring. The behavior seems to beperiodic but rather complicated as the phase plane trajectory shows (see figure).

(b) Solution was plotted with IC x 0 2( ) = , !x 0 0( ) = in the xt as well as in the !xx planes.

4020

2

4

x

t60

–4

–2

80 100 2 4–2

–1

–4

–2

x

2

4

–6 6

!x

(c) The solution is, indeed, periodic.


53. Student Project

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CHAPTER 4 Second-Order Linear Differential Equations