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296 CHAPTER 4. SECOND ORDER LINEAR EQUATIONS

1 2 3 4

!1.5

!1.0

!0.5

0.5

1.0

1.5

2.0

The solution y oscillates with a growing amplitude as x → 0.

4.4 Mechanical and Electrical Vibrations

1. R cos δ = 3 and R sin δ = 4 implies R =√25 = 5 and δ = arctan(4/3) ≈ 0.9273.

Therefore,

y = 5 cos(2t− 0.9273).

2. R cos δ = −1 and R sin δ =√3 implies R =

√4 = 2 and δ = π + arctan(−3) = 2π/3.

Therefore,

y = 2 cos(t− 2π/3).

3. R cos δ = 4 and R sin δ = −2 implies R =√20 = 2

√5 and δ = arctan(−1/2) ≈ −0.4636.

Therefore,

y = 2√5 cos(3t+ 0.46362).

4. R cos δ = −2 and R sin δ = −3 implies R =√13 and δ = π + arctan(3/2) ≈ 4.1244.

Therefore,

y =√13 cos(πt+ π + arctan(3/2)).

5.

(a) The spring constant is k = 2/(1/2) = 4 lb/ft. The mass m = 2/32 = 1/16 lb-s2/ft.Therefore, the equation of motion is

1

16y�� + 4y = 0

which can be simplified to

y�� + 64y = 0.

The initial conditions are y(0) = 1/4 ft, y�(0) = 0 ft/sec. The general solution of thedifferential equation is y(t) = A cos(8t) + B sin(8t) ft. The initial condition implies

A = 1/4 and B = 0. Therefore, the solution is y(t) =1

4cos(8t) ft.

4.4. MECHANICAL AND ELECTRICAL VIBRATIONS 301

–0.3

–0.2

–0.1

0

0.1

0.2

0.3

x2

–0.3 –0.2 –0.1 0.1 0.2 0.3x1

12. The inductance L = 0.2 henry. The resistance R = 3 × 102 ohms. The capacitanceC = 10−5 farads. Therefore, the equation for charge q is

0.2q�� + 300q� + 105q = 0

which can be rewritten asq�� + 1500q� + 500, 000q = 0.

The initial conditions are q(0) = 10−6 coulombs, q�(0) = 0 couloumbs/sec. The generalsolution of the differential equation is q(t) = Ae−500t+Be−1000t. The initial conditions implyA + B = 10−6 and −500A− 1000B = 0. Therefore, A = 2× 10−6 and B = −10−6 and thespecific solution is

Q(t) = 10−6(2e−500t − e−1000t).

13. The frequency of the undamped motion is ω0 = 1. Therefore, the period of the undamped

motion is T = 2π. The quasi frequency of the damped motion is µ =1

2

�4− γ2. Therefore,

the period of the undamped motion is Td = 4π/2�4− γ2. We want to find γ such that

Td = 1.5T . That is, we want γ to satisfy 4π/2�4− γ2 = 3π. Solving this equation, we have

γ = 2√5/3.

14. The spring constant is k = mg/L. The equation of motion for an undamped system is

my�� +mg

Ly = 0

which simplifies to

y�� +g

Ly = 0.

Therefore, the frequency is ω0 =�g/L, and, therefore, the period is T = 2π/

�g/L =

2π�L/g.

15. Suppose y is a solution of the differential equation

my�� + γy� + ky = 0

and satisfies the initial conditions y(t0) = y0, y�(t0) = y1. Then suppose v is a solution ofthe same differential equation, but satisfies the initial conditions v(t0) = v0 and v�(t0) = 0.302 CHAPTER 4. SECOND ORDER LINEAR EQUATIONS

Let w = y − v. Then w satisfies the same differential equation, since the equation is linear,and, further, w(t0) = 0, w�(t0) = y1. Therefore, y = v + w where v, w satisfy the specifiedconditions.

16. Using trigonometric identities, we note that r sin(ω0t−θ) = r sin(ω0t) cos θ−r cos(ω0t) sin θ.In order to write A cos(ω0t)+B sin(ω0t) in the form r sin(ω0t− θ), we need A = −r sin θ andB = r cos θ. Therefore, we must have A2 + B2 = r2 and tan θ = −A/B. From the text, weknow that in order to write A cos(ω0t)+B sin(ω0t) = R cos(ω0t−δ), we need R =

√A2 +B2

and tan δ = B/A. Therefore, in order for R cos(ω0t− δ) = r sin(ω0t− θ), we need r = R andtan θ = −1/ tan δ.

17. The spring constant is k = 8/(1.5/12) = 64 lb/ft. The mass is 8/32 = 1/4 lb-s2/ft.Therefore, the equation of motion is

1

4y�� + γy� + 64y = 0.

The motion will experience critical damping when γ = 2√km = 8 lb-s/ft.

18. The inductance L = 0.2 henry. The capacitance C = 0.810−6 farads. Therefore, theequation for charge q is

0.2q�� +Rq� +107

8q = 0.

The motion will experience critical damping when R = 2�

0.2 · 107/8 = 103 ohms.

19. If the system is critically damped or overdamped, then γ ≥ 2√km. If γ = 2

√km

(critically damped), then the solution is given by

y(t) = (A+Bt)e−γt/2m.

In this case, if y = 0, then we must have A+Bt = 0, that is, t = −A/B (assuming B �= 0).If B = 0, the solution is never zero (unless A = 0). If γ > 2

√km (overdamped), then the

solution is given byy(t) = Aeλ1t +Beλ2t

where λ1,λ2 are given by equation (19) in the text. Assume for the moment, that A,B �= 0.Then y = 0 implies Aeλ1t = −Beλ2t which implies e(λ1−λ2)t = −B/A. There is only onesolution to this equation. If A = 0 or B = 0, then there are no solutions to the equationy = 0 (unless they are both zero, in which case, the solution is identically zero).

20. If the system is critically damped, then the general solution is


The initial conditions imply A = y0 and B = v0 + (γy0/2m). If v0 = 0, then B = γy0/2m.In this case, the specific solution is

y(t) = (y0 + (γy0/2m)t)e−γt/2m.

As t → ∞, y → 0. In order for y = 0, we would need y0 + (γy0/2m)t = 0, but there are nopositive times t satisfying this equation. If v0 �= 0, the specific solution is

y(t) = (y0 + (v0 + (γy0/2m))t)e−γt/2m.




√A2 +B2



1

4y�� + γy� + 64y = 0.



0.2q�� +Rq� +107

8q = 0.

The motion will experience critical damping when R = 2�0.2 · 107/8 = 103 ohms.


√km










y(t) = (y0 + (γy0/2m)t)e−γt/2m.


y(t) = (y0 + (v0 + (γy0/2m))t)e−γt/2m.




√A2 +B2



1

4y�� + γy� + 64y = 0.



0.2q�� +Rq� +107

8q = 0.

The motion will experience critical damping when R = 2�0.2 · 107/8 = 103 ohms.


√km










y(t) = (y0 + (γy0/2m)t)e−γt/2m.


y(t) = (y0 + (v0 + (γy0/2m))t)e−γt/2m.


4.5 Nonhomogeneous Equations; Method of Undeter-mined Coefficients

1. The characteristic equation for the homogeneous problem is λ2 − 2λ − 3 = 0, which hasroots λ = 3,−1. Therefore, the solution of the homogeneous problem is yh(t) = c1e3t+c2e−t.To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =Ae2t. Substituting a function of this form into the differential equation, we have

4Ae2t − 4Ae2t − 3Ae2t = 3e2t.

Therefore, we need −3A = 3, or A = −1. Therefore, the solution of the inhomogeneousproblem is

y(t) = c1e3t + c2e

−t − e2t.

2. The characteristic equation for the homogeneous problem is λ2 + 2λ + 5 = 0, whichhas roots λ = −1 ± 2i. Therefore, the solution of the homogeneous problem is yh(t) =e−t(c1 cos(2t) + c2 sin(2t)). To find a solution of the inhomogeneous problem, we look for asolution of the form yp(t) = A cos(2t) + B sin(2t). Substituting a function of this form intothe differential equation, and equating like terms, we have B − 4A = 3 and A + 4B = 0.The solution of these equations is A = −12/17 and B = 3/17. Therefore, the solution of theinhomogeneous problem is

y(t) = e−t(c1 cos(2t) + c2 sin(2t))−12

17cos(2t) +

3

17sin(2t).

3. The characteristic equation for the homogeneous problem is λ2 − 2λ − 3 = 0, which hasroots λ = 3,−1. Therefore, the solution of the homogeneous problem is yh(t) = c1e3t +c2e−t. Since y(t) = e−t is a solution of the homogeneous problem, to find a solution ofthe inhomogeneous problem, we look for a solution of the form yp(t) = Ate−t + Bt2e−t.Substituting a function of this form into the differential equation, and equating like terms,we have −4A + 2B = 0 and −8B = −3. The solution of these equations is A = 3/16 andB = 3/8. Therefore, the solution of the inhomogeneous problem is

y(t) = c1e3t + c2e

−t +3

16te−t +

3

8t2e−t.

4. The characteristic equation for the homogeneous problem is λ2 +2λ = 0, which has rootsλ = 0,−2. Therefore, the solution of the homogeneous problem is yh(t) = c1 + c2e−2t. Tofind a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =At+B cos(2t)+C sin(2t). Substituting a function of this form into the differential equation,and equating like terms, we have 2A = 3, −4B + 4C = 0 and −4B − 4C = 4. The solutionof these equations is A = 3/2, B = −1/2 and C = −1/2. Therefore, the solution of theinhomogeneous problem is

y(t) = c1 + c2e−2t +

3

2t− 1

2cos(2t)− 1

2sin(2t).

4.5. NONHOMOGENEOUS EQUATIONS; METHODOF UNDETERMINED COEFFICIENTS311

5. The characteristic equation for the homogeneous problem is λ2 + 9 = 0, which has roots

λ = ±3i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(3t)+c2 sin(3t).To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =Ae3t+Bte3t+Ct2e3t+D. Substituting a function of this form into the differential equation,

and equating like terms, we have 18A + 6B + 2C = 0, 18B + 12C = 0, 18C = 1 and

9D = 6. The solution of these equations is A = 1/162, B = −1/27, C = 1/18 and D = 2/3.Therefore, the solution of the inhomogeneous problem is

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

162e3t − 1

27te3t +

1

18t2e3t +

2

3.

6. The characteristic equation for the homogeneous problem is λ2 + 2λ + 1 = 0, which

has the repeated root λ = −1. Therefore, the solution of the homogeneous problem is

yh(t) = c1e−t + c2te−t. To find a solution of the inhomogeneous problem, we look for a

solution of the form yp(t) = At2e−t . Substituting a function of this form into the differential

equation, and equating like terms, we have 2A = 2. Therefore, A = 1 and the solution of

the inhomogeneous problem is

y(t) = c1e−t

+ c2te−t

+ t2e−t.

7. The characteristic equation for the homogeneous problem is 2λ2 + 3λ + 1 = 0, which

has roots λ = −1,−1/2. Therefore, the solution of the homogeneous problem is yh(t) =

c1e−t + c2e−t/2. To find a solution of the inhomogeneous problem, we will first look for

a solution of the form Y1(t) = A + Bt + Ct2 to account for the inhomogeneous term t2.Substituting a function of this form into the differential equation, and equating like terms,

we have A + 3B + 4C = 0, B + 6C = 0 and C = 1. The solution of these equations is

A = 14, B = −6, C = 1. Therefore, Y1 = 14 − 6t + t2. Next, we look for a solution of the

inhomogeneous problem of the form Y2(t) = D cos t+E sin t. Substituting this function into

the ODE and equating like terms, we find that D = −9/10 and E = −3/10. Therefore, thesolution of the inhomogeneous problem is

y(t) = c1e−t

+ c2e−t/2

+ 14− 6t+ t2 − 9

10cos t− 3

10sin t.


λ = ±i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(t) + c2 sin(t).To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =A cos(2t) + B sin(2t) + Ct cos(2t) + Dt sin(2t). Substituting a function of this form into

the differential equation, and equating like terms, we have 4D − 3A = 0, −3B − 4C = 3,

−3C = 1, and −3D = 0. The solution of these equations is A = 0, B = −5/9, C = −1/3and D = 0. Therefore, the solution of the inhomogeneous problem is

y(t) = c1 cos(t) + c2 sin(t)−5

9sin(2t)− 1

3t cos(2t).

9. The characteristic equation for the homogeneous problem is λ2 +ω20 = 0, which has roots

λ = ±ω0i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(ω0t) +

4.5. NONHOMOGENEOUS EQUATIONS; METHODOF UNDETERMINED COEFFICIENTS311


λ = ±3i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(3t)+c2 sin(3t).To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =Ae3t+Bte3t+Ct2e3t+D. Substituting a function of this form into the differential equation,

and equating like terms, we have 18A + 6B + 2C = 0, 18B + 12C = 0, 18C = 1 and

9D = 6. The solution of these equations is A = 1/162, B = −1/27, C = 1/18 and D = 2/3.Therefore, the solution of the inhomogeneous problem is

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

162e3t − 1

27te3t +

1

18t2e3t +

2

3.

6. The characteristic equation for the homogeneous problem is λ2 + 2λ + 1 = 0, which

has the repeated root λ = −1. Therefore, the solution of the homogeneous problem is

yh(t) = c1e−t + c2te−t. To find a solution of the inhomogeneous problem, we look for a

solution of the form yp(t) = At2e−t . Substituting a function of this form into the differential

equation, and equating like terms, we have 2A = 2. Therefore, A = 1 and the solution of


y(t) = c1e−t

+ c2te−t

+ t2e−t.

7. The characteristic equation for the homogeneous problem is 2λ2 + 3λ + 1 = 0, which

has roots λ = −1,−1/2. Therefore, the solution of the homogeneous problem is yh(t) =

c1e−t + c2e−t/2. To find a solution of the inhomogeneous problem, we will first look for

a solution of the form Y1(t) = A + Bt + Ct2 to account for the inhomogeneous term t2.Substituting a function of this form into the differential equation, and equating like terms,

we have A + 3B + 4C = 0, B + 6C = 0 and C = 1. The solution of these equations is

A = 14, B = −6, C = 1. Therefore, Y1 = 14 − 6t + t2. Next, we look for a solution of the

inhomogeneous problem of the form Y2(t) = D cos t+E sin t. Substituting this function into

the ODE and equating like terms, we find that D = −9/10 and E = −3/10. Therefore, thesolution of the inhomogeneous problem is

y(t) = c1e−t

+ c2e−t/2

+ 14− 6t+ t2 − 9

10cos t− 3

10sin t.


λ = ±i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(t) + c2 sin(t).To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =A cos(2t) + B sin(2t) + Ct cos(2t) + Dt sin(2t). Substituting a function of this form into

the differential equation, and equating like terms, we have 4D − 3A = 0, −3B − 4C = 3,

−3C = 1, and −3D = 0. The solution of these equations is A = 0, B = −5/9, C = −1/3and D = 0. Therefore, the solution of the inhomogeneous problem is

y(t) = c1 cos(t) + c2 sin(t)−5

9sin(2t)− 1

3t cos(2t).

9. The characteristic equation for the homogeneous problem is λ2 +ω20 = 0, which has roots

λ = ±ω0i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(ω0t) +


c2 sin(ω0t). To find a solution of the inhomogeneous problem, we look for a solution of theform yp(t) = A cos(ωt) +B sin(ωt). Substituting a function of this form into the differentialequation, and equating like terms, we have (ω2

0 − ω2)A = 1 and (ω20 − ω2)B = 0. The

solution of these equations is A = 1/(ω20 − ω2) and B = 0. Therefore, the solution of the

inhomogeneous problem is

y(t) = c1 cos(ω0t) + c2 sin(ω0t) +1

ω20 − ω2

cos(ωt).

10. From problem 9, the solution of the homogeneous problem is yh(t) = c1 cos(ω0t) +c2 sin(ω0t). Since cos(ω0t) is a solution of the homogeneous problem, we look for a solutionof the inhomogeneous problem of the form yp(t) = At cos(ω0t) +Bt sin(ω0t). Substituting afunction of this form into the differential equation, and equating like terms, we have A = 0and B = 1

2ω0. Therefore, the solution of the inhomogeneous problem is


2ω0t sin(ω0t).

11. The characteristic equation for the homogeneous problem is λ2 + λ + 4 = 0, whichhas roots λ = (−1 ± i

√15)/2. Therefore, the solution of the homogeneous problem is

yh(t) = e−t/2(c1 cos(√15t/2) + c2 sin(

√15t/2)). To find a solution of the inhomogeneous

problem, we look for a solution of the form yp(t) = Aet + Be−t. Substituting a functionof this form into the differential equation, and equating like terms, we have 6A = 1 and4B = −1. The solution of these equations is A = 1/6 and B = −1/4. Therefore, thesolution of the inhomogeneous problem is

y(t) = e−t/2(c1 cos(√15t/2) + c2 sin(

√15t/2)) +

1

6et − 1

4e−t.

12. The characteristic equation for the homogeneous problem is λ2 − λ − 2 = 0, which hasroots λ = −1, 2. Therefore, the solution of the homogeneous problem is yh(t) = c1e−t+c2e2t.To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =Ate2t+Be−2t. Substituting a function of this form into the differential equation, and equatinglike terms, we have A = 1/6 and B = 1/8. Therefore, the solution of the inhomogeneousproblem is

y(t) = c1e−t + c2e

2t +1

6te2t +

1

8e−2t.

13. The characteristic equation for the homogeneous problem is λ2 + λ − 2 = 0, which hasroots λ = 1,−2. Therefore, the solution of the homogeneous problem is yh(t) = c1et+c2e−2t.To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =At+B. Substituting a function of this form into the differential equation, and equating liketerms, we have −2A = 2 and A − 2B = 0. The solution of these equations is A = −1 andB = −1/2. Therefore, the solution of the inhomogeneous problem is

y(t) = c1et + c2e

−2t − t− 1

2.


c2 sin(ω0t). To find a solution of the inhomogeneous problem, we look for a solution of theform yp(t) = A cos(ωt) +B sin(ωt). Substituting a function of this form into the differentialequation, and equating like terms, we have (ω2

0 − ω2)A = 1 and (ω20 − ω2)B = 0. The

solution of these equations is A = 1/(ω20 − ω2) and B = 0. Therefore, the solution of the

inhomogeneous problem is


ω20 − ω2

cos(ωt).

10. From problem 9, the solution of the homogeneous problem is yh(t) = c1 cos(ω0t) +c2 sin(ω0t). Since cos(ω0t) is a solution of the homogeneous problem, we look for a solutionof the inhomogeneous problem of the form yp(t) = At cos(ω0t) +Bt sin(ω0t). Substituting afunction of this form into the differential equation, and equating like terms, we have A = 0and B = 1

2ω0. Therefore, the solution of the inhomogeneous problem is


2ω0t sin(ω0t).

11. The characteristic equation for the homogeneous problem is λ2 + λ + 4 = 0, whichhas roots λ = (−1 ± i

√15)/2. Therefore, the solution of the homogeneous problem is

yh(t) = e−t/2(c1 cos(√15t/2) + c2 sin(

√15t/2)). To find a solution of the inhomogeneous

problem, we look for a solution of the form yp(t) = Aet + Be−t. Substituting a functionof this form into the differential equation, and equating like terms, we have 6A = 1 and4B = −1. The solution of these equations is A = 1/6 and B = −1/4. Therefore, thesolution of the inhomogeneous problem is

y(t) = e−t/2(c1 cos(√15t/2) + c2 sin(

√15t/2)) +

1

6et − 1

4e−t.

12. The characteristic equation for the homogeneous problem is λ2 − λ − 2 = 0, which hasroots λ = −1, 2. Therefore, the solution of the homogeneous problem is yh(t) = c1e−t+c2e2t.To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =Ate2t+Be−2t. Substituting a function of this form into the differential equation, and equatinglike terms, we have A = 1/6 and B = 1/8. Therefore, the solution of the inhomogeneousproblem is

y(t) = c1e−t + c2e

2t +1

6te2t +

1

8e−2t.

13. The characteristic equation for the homogeneous problem is λ2 + λ − 2 = 0, which hasroots λ = 1,−2. Therefore, the solution of the homogeneous problem is yh(t) = c1et+c2e−2t.To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =At+B. Substituting a function of this form into the differential equation, and equating liketerms, we have −2A = 2 and A − 2B = 0. The solution of these equations is A = −1 andB = −1/2. Therefore, the solution of the inhomogeneous problem is

y(t) = c1et + c2e

−2t − t− 1

2.4.5. NONHOMOGENEOUS EQUATIONS; METHODOF UNDETERMINED COEFFICIENTS313

The initial conditions imply

c1 + c2 −1

2= 0

c1 − 2c2 − 1 = 1.

Therefore, c1 = 1 and c2 = −1/2 which implies the particular solution of the IVP is

y(t) = et − 1

2e−2t − t− 1

2.

14. The characteristic equation for the homogeneous problem is λ2 +4 = 0, which has rootsλ = ±2i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(2t)+c2 sin(2t).First, to find a solution of the inhomogeneous problem, we look for a solution of the formY1(t) = A + Bt + Ct2 to correspond with the term t2. Substituting a function of this forminto the differential equation, and equating like terms, we have A = −1/8, B = 0 andC = 1/4. Next, considering the inhomogeneous term 3et, we look for a solution of the formY2(t) = Det. Substituting this function into the equation and equating like terms, we haveD = 3/5. Therefore, the solution of the inhomogeneous problem is

y(t) = c1 cos(2t) + c2 sin(2t)−1

8+

1

4t2 +

3

5et.


19

40+ c1 = 0

3

5+ 2c2 = 2.

Therefore, c1 = −19/40 and c2 = 7/10 which implies the particular solution of the IVP is

y(t) = −19

40cos(2t) +

7

10sin(2t)− 1

8+

1

4t2 +

3

5et.

15. The characteristic equation for the homogeneous problem is λ2 − 2λ+ 1 = 0, which hasthe repeated root λ = 1. Therefore, the solution of the homogeneous problem is yh(t) =c1et + c2tet. First, to find a solution of the inhomogeneous problem, we look for a solutionof the form Y1(t) = At2et + Bt3et to correspond with the term tet. Substituting a functionof this form into the differential equation, and equating like terms, we have Y1(t) = t3et/6.Next, considering the inhomogeneous term 4, we look for a solution of the form Y2(t) =C. Substituting this function into the equation and equating like terms, we have Y2 = 4.Therefore, the solution of the inhomogeneous problem is

y(t) = c1et + c2te

t +1

6t3et + 4.


c1 + 4 = 0

c1 + c2 = 1.


Therefore, c1 = −3 and c2 = 4 which implies the particular solution of the IVP is

y(t) = −3et + 4tet +1

6t3et + 4.

16. The characteristic equation for the homogeneous problem is λ2 − 2λ− 3 = 0, which hasroots λ = 3,−1. Therefore, the solution of the homogeneous problem is yh(t) = c1e3t+c2e−t.To find a solution of the inhomogeneous problem, we look for a solution of the form yp(t) =Ae2t+Bte2t . Substituting a function of this form into the differential equation, and equatinglike terms, we have yp(t) = 2e2t+3te2t. Therefore, the solution of the inhomogeneous problemis

y(t) = c1e3t + c2e

−t + 2e2t + 3te2t.


c1 + c2 + 2 = 1

3c1 − c2 + 4 + 3 = 0.

Therefore, c1 = −2 and c2 = 1 which implies the particular solution of the IVP is

y(t) = −2e3t + e−t + 2e2t + 3te2t.

17. The characteristic equation for the homogeneous problem is λ2 + 4 = 0, which hasroots λ = ±2i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(2t) +c2 sin(2t). To find a solution of the inhomogeneous problem, we look for a solution of theform yp(t) = At cos(2t)+Bt sin(2t). Substituting a function of this form into the differential

equation, and equating like terms, we have yp(t) = −3

4t cos(2t). Therefore, the solution of


y(t) = c1 cos(2t) + c2 sin(2t)−3

4t cos(2t).


c1 = 2

2c2 −3

4= −1.

Therefore, c1 = 2 and c2 = −1/8 which implies the particular solution of the IVP is

y(t) = 2 cos(2t)− 1

8sin(2t)− 3

4t cos(2t).

18. The characteristic equation for the homogeneous problem is λ2 + 2λ + 5 = 0, whichhas roots λ = −1 ± 2i. Therefore, the solution of the homogeneous problem is yh(t) =e−t(c1 cos(2t) + c2 sin(2t)). To find a solution of the inhomogeneous problem, we look for asolution of the form yp(t) = Ate−t cos(2t) + Bte−t sin(2t). Substituting a function of this4.5. NONHOMOGENEOUS EQUATIONS; METHODOF UNDETERMINED COEFFICIENTS315

form into the differential equation, and equating like terms, we have yp(t) = te−t sin(2t).Therefore, the solution of the inhomogeneous problem is

y(t) = e−t(c1 cos(2t) + c2 sin(2t)) + te−t sin(2t).


c1 = 1

− c1 + 2c2 = 0.

Therefore, c1 = 1 and c2 = 1/2 which implies the particular solution of the IVP is

y(t) = e−t

�cos(2t) +

1

2sin(2t)

�+ te−t sin(2t).

19.

(a) The characteristic equation for the homogeneous problem is λ2+3λ = 0, which has rootsλ = 0,−3. Therefore, the solution of the homogeneous problem is yh(t) = c1 + c2e−3t.Therefore, to find a solution of the inhomogeneous problem, we look for a solution of theform Y (t) = t(A0t4+A1t3+A2t2+A3t+A4)+t(B0t2+B1t+B2)e−3t+D sin 3t+E cos 3t.

(b) Substituting a function of this form into the differential equation, and equating liketerms, we have A0 = 2/15, A1 = −2/9, A2 = 8/27, A3 = −8/27, A4 = 16/81, B0 =−1/9, B1 = −1/9, B2 = −2/27, D = −1/18, E = −1/18 Therefore, the solution of theinhomogeneous problem is

y(t) = c1 + c2e−3t + t((2/15)t4 − (2/9)t3 + (8/27)t2 − (8/27)t+ (16/81))+

t((−1/9)t2 − (1/9)t− 2/27)e−3t − 1

18sin 3t− 1

18cos 3t

20.

(a) The characteristic equation for the homogeneous problem is λ2 +1 = 0, which has rootsλ = ±i. Therefore, the solution of the homogeneous problem is yh(t) = c1 cos(t) +c2 sin(t). Therefore, to find a solution of the inhomogeneous problem, we look for asolution of the form Y (t) = A0t+ A1 + t(B0t+B1) sin t+ t(D0t+D1) cos t.

(b) Substituting a function of this form into the differential equation, and equating liketerms, we have A0 = 1, A1 = 0, B0 = 0, B1 = 1/4, D0 = −1/4, D1 = 0. Therefore, thesolution of the inhomogeneous problem is

y(t) = c1 cos(t) + c2 sin(t) + t+1

4t sin t− 1

4t2 cos t.

21.


(b)

0.5

1

1.5

2

2.5

0.6 0.8 1 1.2 1.4 1.6 1.8 2x

(c) The amplitude for a similar system with a linear spring is given by

R =5√

25− 49ω2 + 25ω4.

Amplitude

1

2

3

4

5

0.6 0.8 1 1.2 1.4 1.6 1.8 2w

4.7 Variation of Parameters

1.

(a)

Xu =

�x11(t) x12(t)x21(t) x22(t)

��u1(t)u2(t)

�=

�x11(t)u1(t) + x12(t)u2(t)x21(t)u1(t) + x22(t)u2(t)

�.

Therefore,

(Xu)� =

�x�11u1 + x11u�

1 + x�12u2 + x12u�

2

x�21u1 + x21u�

1 + x�22u2 + x22u�

2

�.

Now

X� =

�x�11 x�

12

x�21 x�

22

�u� =

�u�1

u�2

�.

4.7. VARIATION OF PARAMETERS 341

4. Let X(t) be the fundamental matrix,

X(t) =

�0 −e−t

e−t te−t

�.

Then

X−1(t) =

�tet et

−et 0

�.

Therefore,

X−1(t)g(t) =

�tet et

−et 0

��−1t

�

=

�0et

�.

Therefore,�

X−1(t)g(t) dt =

� �0et

�dt

=

�0et

�.

Therefore,

xp(t) = X(t)

�X−1(t)g(t) dt

=

�0 −e−t

e−t te−t

��0et

�

=

�−1t

�.

Therefore,

xp(t) =

�−1t

�.

5. Let X(t) be the fundamental matrix,

X(t) =

�cos t sin t− sin t cos t

�.

Then

X−1(t) =

�cos t − sin tsin t cos t

�.

Therefore,

X−1(t)g(t) =

�cos t − sin tsin t cos t

��cos t− sin t

�

=

�10

�.


Therefore, c1 − c2 = −1 and c2 − 1 = 1. Solving this system of equations, we have c1 = 1and c2 = 2. Therefore, the solution of the IVP is

x(t) =

�e−t − 2et + te−t + 2t

2et − t− 1

�.

8. The general solution of the equation in problem 4 is

x(t) = c1

�0e−t

�+ c2

�−e−t

te−t

�+

�−1t

�.

The initial condition x(0) =�1 0

�timplies

x(0) = c1

�01

�+ c2

�−10

�+

�−10

�=

�10

�.

Therefore, −c2 − 1 = 1 and c1 = 0. Solving this system of equations, we have c1 = 0 andc2 = −2. Therefore, the solution of the IVP is

x(t) = −2

�−e−t

te−t

�+

�−1t

�.

9. The general solution of the equation in problem 5 is

x(t) = c1

�cos t− sin t

�+ c2

�sin tcos t

�+

�t cos t−t sin t

�.

The initial condition x(0) =�1 1

�timplies

x(0) = c1

�10

�+ c2

�01

�+

�00

�=

�11

�.

Therefore, c1 = 1 and c2 = 1. Therefore, the solution of the IVP is

x(t) =

�cos t− sin t

�+

�sin tcos t

�+

�t cos t−t sin t

�.

10. The solution of the homogeneous equation is yh(t) = c1e2t + c2e3t. The functionsy1(t) = e2t and y2(t) = e3t form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = e5t. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

e3t(2et)

W (t)dt = 2e−t

u2(t) =

�e2t(2et)

W (t)dt = −e−2t.


Therefore, the particular solution is Y (t) = 2et − et = et.

11. The solution of the homogeneous equation is yh(t) = c1e2t + c2e−t. The functionsy1(t) = e2t and y2(t) = e−t form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = −3et. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

e−t(2e−t)

W (t)dt = −2

9e−3t

u2(t) =

�e2t(2e−t)

W (t)dt = −2t

3.

Therefore, a particular solution is Y (t) = −2

9e−t − 2

3te−t. As −2e−t/9 is a solution of the

homogeneous equation, we can omit it and conclude that yp(t) = −2

3te−t is a particular

solution of the inhomogeneous problem.

12. The solution of the homogeneous equation is yh(t) = c1e−t + c2te−t. The functionsy1(t) = e−t and y2(t) = te−t form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = e−2t. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

te−t(3e−t)

W (t)dt = −3

2t2

u2(t) =

�e−t(3e−t)

W (t)dt = 3t.

Therefore, a particular solution is Y (t) = −32t

2e−t + 3t2e−t = 32t

2e−t.

13. The solution of the homogeneous equation is yh(t) = c1et/2 + c2tet/2. The functionsy1(t) = et/2 and y2(t) = tet/2 form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = et. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

tet/2(4et/2)

W (t)dt = −2t2

u2(t) =

�et/2(4et/2)

W (t)dt = 4t.

Therefore, a particular solution is Y (t) = −2t2et/2 + 4t2et/2 = 2t2et/2.

14. The solution of the homogeneous equation is yh(t) = c1 cos(t) + c2 sin(t). The functionsy1(t) = cos(t) and y2(t) = sin(t) form a fundamental set of solutions. The Wronskian ofthese functions is W (y1, y2) = 1. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

sin(t) tan(t)

W (t)dt = sin(t)− ln(sec(t) + tan(t))


Therefore, the particular solution is Y (t) = 2et − et = et.

11. The solution of the homogeneous equation is yh(t) = c1e2t + c2e−t. The functionsy1(t) = e2t and y2(t) = e−t form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = −3et. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

e−t(2e−t)

W (t)dt = −2

9e−3t

u2(t) =

�e2t(2e−t)

W (t)dt = −2t

3.


9e−t − 2

3te−t. As −2e−t/9 is a solution of the

homogeneous equation, we can omit it and conclude that yp(t) = −2

3te−t is a particular

solution of the inhomogeneous problem.

12. The solution of the homogeneous equation is yh(t) = c1e−t + c2te−t. The functionsy1(t) = e−t and y2(t) = te−t form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = e−2t. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

te−t(3e−t)

W (t)dt = −3

2t2

u2(t) =

�e−t(3e−t)

W (t)dt = 3t.

Therefore, a particular solution is Y (t) = −32t

2e−t + 3t2e−t = 32t

2e−t.

13. The solution of the homogeneous equation is yh(t) = c1et/2 + c2tet/2. The functionsy1(t) = et/2 and y2(t) = tet/2 form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = et. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

tet/2(4et/2)

W (t)dt = −2t2

u2(t) =

�et/2(4et/2)

W (t)dt = 4t.

Therefore, a particular solution is Y (t) = −2t2et/2 + 4t2et/2 = 2t2et/2.

14. The solution of the homogeneous equation is yh(t) = c1 cos(t) + c2 sin(t). The functionsy1(t) = cos(t) and y2(t) = sin(t) form a fundamental set of solutions. The Wronskian ofthese functions is W (y1, y2) = 1. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

sin(t) tan(t)

W (t)dt = sin(t)− ln(sec(t) + tan(t))


u2(t) =

�cos(t) tan(t)

W (t)dt = − cos(t).

Therefore, a particular solution is Y (t) = (sin(t)− ln(sec(t)+ tan(t))) cos(t)− cos(t) sin(t) =− cos(t) ln(sec(t) + tan(t)). Therefore, the general solution is y(t) = c1 cos(t) + c2 sin(t) −cos(t) ln(sec(t) + tan(t)).

15. The solution of the homogeneous equation is yh(t) = c1 cos(3t)+c2 sin(3t). The functionsy1(t) = cos(3t) and y2(t) = sin(3t) form a fundamental set of solutions. The Wronskian ofthese functions is W (y1, y2) = 3. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

sin(3t)(9 sec2(3t))

W (t)dt = − csc(3t)

u2(t) =

�cos(3t)(9 sec2(3t))

W (t)dt = ln(sec(3t) + tan(3t)).

Therefore, a particular solution is Y (t) = −1 + sin(3t) ln(sec(3t) + tan(3t)). Therefore, thegeneral solution is y(t) = c1 cos(3t) + c2 sin(3t) + sin(3t) ln(sec(3t) + tan(3t))− 1.

16. The solution of the homogeneous equation is yh(t) = c1e−2t + c2te−2t. The functionsy1(t) = e−2t and y2(t) = te−2t form a fundamental set of solutions. The Wronskian of thesefunctions is W (y1, y2) = e−4t. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

te−2t(t−2e−2t)

W (t)dt = − ln(t)

u2(t) =

�e−2t(t−2e−2t)

W (t)dt = −1

t.

Therefore, a particular solution is Y (t) = −e−2t ln(t)− e−2t. Therefore, the general solutionis y(t) = c1e−2t + c2te−2t − e−2t ln(t).

17. The solution of the homogeneous equation is yh(t) = c1 cos(2t)+c2 sin(2t). The functionsy1(t) = cos(2t) and y2(t) = sin(2t) form a fundamental set of solutions. The Wronskian ofthese functions is W (y1, y2) = 2. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

sin(2t)(3 csc(2t))

W (t)dt = −3

2t

u2(t) =

�cos(2t)(3 csc(2t))

W (t)dt =

3

4ln | sin(2t)|.


2t cos(2t) +

3

4(sin(2t)) ln | sin(2t)|. Therefore,

the general solution is y(t) = c1 cos(2t) + c2 sin(2t)− 32t cos(2t) +

34 sin(2t) ln | sin(2t)|.

18. The solution of the homogeneous equation is yh(t) = c1 cos(t/2) + c2 sin(t/2). Thefunctions y1(t) = cos(t/2) and y2(t) = sin(t/2) form a fundamental set of solutions. The


Wronskian of these functions is W (y1, y2) = 1/2. Writing the ODE in standard form, we see

that g(t) = sec(t/2)/2. Using the method of variation of parameters, a particular solution

is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

cos(t/2)(sec(t/2))

2W (t)dt = 2 ln[cos(t/2)]

u2(t) =

�sin(t/2)(sec(t/2))

2W (t)dt = t.

Therefore, a particular solution is Y (t) = 2 cos(t/2) ln[cos(t/2)] + t sin(t/2). Therefore, the

general solution is y(t) = c1 cos(t/2) + c2 sin(t/2) + 2 cos(t/2) ln[cos(t/2)] + t sin(t/2).

19. The solution of the homogeneous equation is yh(t) = c1et+c2tet. The functions y1(t) = et

and y2(t) = tet form a fundamental set of solutions. The Wronskian of these functions is

W (y1, y2) = e2t. Using the method of variation of parameters, a particular solution is given

by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

tet(et)

W (t)(1 + t2)dt = −1

2ln(1 + t2)

u2(t) =

�et(et)

W (t)(1 + t2)dt = arctan(t).


2et ln(1 + t2) + tet arctan(t). Therefore, the

general solution is y(t) = c1et + c2tet − 12e

t ln(1 + t2) + tet arctan(t).

20. The solution of the homogeneous equation is yh(t) = c1e3t + c2e2t. The functions

y1(t) = e3t and y2(t) = e2t form a fundamental set of solutions. The Wronskian of these

functions is W (y1, y2) = −e5t. Using the method of variation of parameters, a particular

solution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

e2s(g(s))

W (s)ds =

�e−3sg(s)ds

u2(t) =

�e3s(g(s))

W (s)ds = −

�e−2sg(s)ds.

Therefore, a particular solution is Y (t) =�e3(t−s)g(s)ds −

�e2(t−s)g(s)ds. Therefore, the

general solution is y(t) = c1e3t + c2e2t +�e3(t−s)g(s)ds−

�e2(t−s)g(s)ds.

21. The solution of the homogeneous equation is yh(t) = c1 cos(2t)+c2 sin(2t). The functionsy1(t) = cos(2t) and y2(t) = sin(2t) form a fundamental set of solutions. The Wronskian of

these functions is W (y1, y2) = 2. Using the method of variation of parameters, a particular

solution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

sin(2s)(g(s))

W (s)ds = −1

2

�sin(2s)g(s) ds

u2(t) =

�cos(2s)(g(s))

W (s)ds =

1

2

�cos(2s)g(s) ds.


Therefore, a particular solution is

Y (t) = −1

2

�cos(2t) sin(2s)g(s) ds+

1

2

�sin(2t) cos(2s)g(s) ds.

Using the fact that sin(2t) cos(2s) − cos(2t) sin(2s) = sin(2t − 2s), we see that the generalsolution is

y(t) = c1 cos(2t) + c2 sin(2t) +1

2

�sin(2t− 2s)g(s) ds.

22. By direct substitution, it can be verified that y1(t) = t and y2(t) = tet are solutions of thehomogeneous equations. The Wronskian of these functions is W (y1, y2) = t2et. Rewritingthe equation in standard form, we have

y�� − t(t+ 2)

t2y� +

t+ 2

t2y = 2t.

Therefore, g(t) = 2t. Using the method of variation of parameters, a particular solution isgiven by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

tet(2t)

W (t)dt = −2t

u2(t) =

�t(2t)

W (t)dt = −2e−t.

Therefore, a particular solution is Y (t) = −2t2 − 2t2 = −4t2.

23. By direct substitution, it can be verified that y1(t) = 1+ t and y2(t) = et are solutions ofthe homogeneous equations. The Wronskian of these functions is W (y1, y2) = tet. Rewritingthe equation in standard form, we have

y�� − 1 + t

ty� +

1

ty = te2t.

Therefore, g(t) = te2t. Using the method of variation of parameters, a particular solution isgiven by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

et(te2t)

W (t)dt = −1

2e2t

u2(t) =

�(1 + t)(te2t)

W (t)dt = tet.


2(1 + t)e2t + te2t =

1

2(t− 1)e2t.

24. By direct substitution, it can be verified that y1(t) = et and y2(t) = t are solutionsof the homogeneous equations. The Wronskian of these functions is W (y1, y2) = (1 − t)et.Rewriting the equation in standard form, we have

y�� − t

1− ty� − 1

1− ty = 2(1− t)e−t.


Therefore, g(t) = 2(1 − t)e−t. Using the method of variation of parameters, a particularsolution is given by Y (t) = u1(t)y1(t) + u2(t)y2(t) where

u1(t) = −�

t(2(1− t)e−t)

W (t)dt = te−2t +

1

2e−2t

u2(t) =

�2(1− t)

W (t)dt = −2e−t.

Therefore, a particular solution is Y (t) = te−t + 12e

−t − 2te−t =�12 − t

�e−t.

25. By direct substitution, it can be verified that y1(x) = x−1/2 sin x and y2(x) = x−1/2 cos xare solutions of the homogeneous equations. The Wronskian of these functions isW (y1, y2) =−1/x. Rewriting the equation in standard form, we have

y�� +1

xy� +

x2 − 0.25

x2y = 3

sin x

x1/2.

Therefore, g(x) = 3 sin(x)x−1/2. Using the method of variation of parameters, a particularsolution is given by Y (x) = u1(x)y1(x) + u2(x)y2(x) where

u1(x) = −�

x−1/2 cos(x)(3 sin(x)x−1/2)

W (x)dx = −3

2cos2(x)

u2(x) =

�x−1/2 sin(x)(3 sin(x)x−1/2)

W (x)dx =

3

2cos(x) sin(x)− 3x

2.

Therefore, a particular solution is

Y (x) = −3

2cos2(x) sin(x)x−1/2 +

�3

2cos(x) sin(x)− 3x

2

�cos(x)x−1/2

= −3

2x1/2 cos(x).

26. By direct substitution, it can be verified that y1(x) = ex and y2(x) = x are solutionsof the homogeneous equations. The Wronskian of these functions is W (y1, y2) = (1 − x)ex.Rewriting the equation in standard form, we have

y�� +x

1− xy� − 1

1− xy =

g(x)

1− x.

Therefore, the inhomogeneous term is g(x)/(1 − x). Using the method of variation of pa-rameters, a particular solution is given by Y (x) = u1(x)y1(x) + u2(x)y2(x) where

u1(x) = −�

sg(s)

(1− s)W (s)ds

u2(x) =

�esg(s)

(1− s)W (s)ds.

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