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Math 201 Homework 6 with Solutions

Warning. Looking at the solutions before seriously trying the problems willseverely damage the value of this document.

Problem 1. (7.4 6) Determine the inverse Laplace transform of

3

(2 s +5)3. (1)

(Ans.3

16t2 e−5t/2)

Problem 2. (7.4 18) Determine the partial fraction expansion for

3 s2 + 5 s +3

s4 + s3. (2)

(Ans.1

s+

2

s2+

3

s3−

1

s + 1)

Problem 3. (7.4 24) Determine L−1{F } for

F (s)=7 s2− 41 s + 84

(s− 1) (s2− 4 s + 13). (3)

(Ans. 5 et + 2 e2t cos 3 t− 5 e2t sin 3 t)

Problem 4. (7.4 34) Compute L−1{F } for

F (s)= ln

(

s− 4

s− 3

)

. (4)

(Ans.e3t

− e4t

t)

Problem 5. (7.5 10) Use Laplace transform to solve

y ′′− 4 y = 4 t− 8 e−2t, y(0) =0, y ′(0) =5 (5)

(Ans. − t− e−2t + 2 e−2t t + e2t)


y ′′′+ y ′′+ 3 y ′− 5 y = 16 e−t, y(0) =0, y ′(0) =2, y ′′(0)=− 4. (6)

1

(Ans. − 2 e−t + et + e−t cos 2 t)


t y ′′− t y ′+ y = 2; y(0) =2, y ′(0) =− 1. (7)

(Ans. 2− t)

Problem 8. (7.6 14) Determine an inverse Laplace transform of

e−3s

s2 + 9. (8)

(Ans.1

3u(t− 3) sin 3 (t− 3))

Problem 9. (7.6 26) Determine L{f } with the “sawtooth function” f(t)given by

f(t)

1

a 2 a 3 a t

(Ans.1− e

−as− a s e

−as

a s2(

1− e−as) )


y ′′+ 3 y ′ +2 y = g(t), y(0)= 2, y ′(0)=− 1 (9)

where

g(t) =

{

e−t 06 t < 31 3< t

. (10)

(Ans. 2 e−t + e−t t + u(t− 3)[

1

2+

(

4− t− e3)

e−t +e6− 2 e

3

2e−2t

]

)

2

Solution 1. (7.4 6) First write

3

(2 s + 5)3=

3

8

1

(s +5/2)3. (11)

Now it’s clear that we should use

L{

eat f(t)}

=F (s− a)� L−1{F (s− a)}= eat f(t). (12)

We have a =− 5/2 and F (s)=3

8

1

s3. Now recall

L{

t2}

=2

s3� L−1

{

2

s3

}

= t2. (13)

We have

f(t) =L−1

{

3

8

1

s3

}

=3

16L−1

{

2

s3

}

=3

16t2. (14)

Therefore the answer is3

16t2 e−5t/2. (15)

Solution 2. (7.4 18)

1. First check that the degree of the numerator is less than that of thedenominator.

2. Factorize the denominator:

s4 + s3 = s3 (s + 1). (16)

3. Write down the partial fraction expansion:

3 s2 + 5 s + 3

s4 + s3=

A

s+

B

s2+

C

s3+

D

s +1. (17)

4. Try to determine the coefficients.As

A

s+

B

s2+

C

s3+

D

s + 1=

A s2(s + 1)+B s (s + 1) +C (s + 1) +D s3

s3 (s + 1), (18)

We need

3 s2 + 5 s + 3= As2 (s + 1) +B s (s +1)+ C (s +1)+ D s3. (19)

− Setting s =0: C = 3.− Setting s =− 1: −D = 1� D =− 1.− Comparing coefficients for s3: 0= A+ D� A = 1.− Comparing coefficients for s2: 3= A+ B� B = 2.

Summarizing, we have

3 s2 + 5 s +3

s4 + s3=

1

s+

2

s2+

3

s3−

1

s + 1. (20)

Math 201 Homework 6 with Solutions 3


1. Check degree of numerator < degree of denominator.

2. Factorize denominator. Already done as

s2− 4 s + 13= (s− 2)2 + 32 (21)

cannot be factorized any further.

3. Write down the form of the expansion:

7 s2− 41 s + 84

(s− 1) (s2− 4 s + 13)=

A

s− 1+

B (s− 2)+ 3C

(s− 2)2 + 32. (22)

4. Determine the coefficients.

First write

A

s− 1+

B (s− 2) +3 C

(s− 2)2 + 32=

A[

(s− 2)2+ 9

]

+ [B (s− 2) + 3 C] (s− 1)

(s− 1)[

(s− 2)2 +32] . (23)

Therefore

7 s2− 41 s+ 84= A[

(s− 2)2 + 9

]

+ [B (s− 2)+ 3C] (s− 1). (24)

− Setting s =1: 50= 10A� A= 5.

− Setting s =2: 30= 9A +3 C� C =− 5.

− Comparing coefficients for s2: 7= A+ B� B = 2.

So the partial fraction expansion reads

7 s2− 41 s + 84

(s− 1) (s2− 4 s + 13)=

5

s− 1+

2 (s− 2)− 5 · 3

(s− 2)2 + 32. (25)

5. Computing the inverse transform.

We have

L−1{F } = L−1

{

5

s− 1+

2 (s− 2)− 5 · 3

(s− 2)2 +32

}

= L−1

{

5

s− 1

}

+ 2L−1

{

s− 2

(s− 2)2 + 32

}

− 5L−1

{

3

(s− 2)2 +32

}

= 5 etL−1

{

1

s

}

+ 2 e2tL−1

{

s

s2 + 32

}

− 5 e2tL−1

{

3

s2 +32

}

= 5 et +2 e2t cos 3 t− 5 e2t sin 3 t. (26)

4

Solution 4. (7.4 34) Note that ln(

s − 4

s − 3

)

is not similar to anything in the

transformation table. However, if we taked

dswe get

F ′(s)= [ln(s− 4)− ln(s− 3)]′ =

1

s− 4−

1

s− 3(27)

which can be easily inverted.Now recall

L{tn f(t)}= (− 1)nF (n)(s)� L−1

{

F (n)(s)}

= (− t)n

f(t). (28)

We have

L−1{F ′(s)}= e4t − e3t =(− t) f(t) � f(t) =e3t − e4t

t. (29)


1. Transform the LHS (left hand side):

L{y ′′− 4 y} = L{y ′′}− 4L{y}

= s2 Y − s y(0)− y ′(0)− 4 Y

=(

s2− 4)

Y − 5. (30)

2. Transform the RHS:

L{

4 t− 8 e−2t}

= 4L{t}− 8L{

e−2t}

= 41

s2− 8

1

s +2. (31)

3. The transformed equation reads

(

s2− 4)

Y − 5=4

s2−

8

s + 2(32)

which gives

Y (s)=5

s2− 4+

4

s2 (s2− 4)−

8

(s2− 4) (s +2). (33)

4. Compute the inverse transform.First write

Y (s)=5 s2 (s + 2)+ 4 (s+ 2)− 8 s2

s2 (s2− 4) (s + 2). (34)

Factorize the denominator:

s2(

s2− 4)

(s + 2) = s2 (s + 2)2(s− 2). (35)

So the expansion takes the form

Y (s)=A

s+

B

s2+

C

s+ 2+

D

(s + 2)2+

E

s− 2(36)


which gives

5 s2 (s + 2) + 4 (s + 2) − 8 s2 = (A s + B) (s + 2)2

(s − 2) +

[C (s + 2) +D] s2 (s− 2)+ E s2 (s + 2)2. (37)

− Setting s =0: 8=− 8B� B =− 1.− Setting s =2: 64= 64E� E =1.− Setting s =− 2: − 32=− 16D� D =2.− Comparing coefficients for s4: 0= A+ C + E� A+ C =− 1.− Comparing coefficients for s:1 4=− 8 A− 4B� A= 0.− Thus A+ C =− 1� C =− 1.Summarizing, we have

Y (s)=−1

s2−

1

s + 2+

2

(s + 2)2+

1

s− 2. (38)

This gives

L−1{Y }=− t− e−2t + 2 e−2t t + e2t. (39)


1. Transform the LHS:

L{y ′′′+ y ′′+ 3 y ′− 5 y} = L{y ′′′}+L{y ′′}+ 3L{y ′}− 5L{y}

= s3 Y − s2 y(0)− s y ′(0)− y ′′(0)

+ s2 Y − s y(0)− y ′(0)

+3 s Y − 3 y(0)+5 Y

=(

s3 + s2 + 3 s− 5)

Y

− 2 s +4− 2

=(

s3 + s2 + 3 s− 5)

Y − 2 s +2. (40)


L{

16 e−t}

= 16L{

e−t}

= 161

s +1. (41)

3. The transformed equation reads

(

s3 + s2 + 3 s− 5)

Y − 2 s+ 2 =16

s +1(42)

which gives

Y =2 s− 2

(s3 + s2 + 3 s− 5)+

16

(s3 + s2 +3 s− 5) (s + 1). (43)

4. Computing the inverse transform.We have

Y =2 (s− 1) (s+ 1)+ 16

(s3 + s2 + 3 s− 5) (s + 1). (44)

1. The reason is that, by observation we see most of the terms on both sidescontains a factor s2 so they won’t contribute to the linear term s.

6

First we need to factorize the denominator. To do this, we need to guessa root for s3 + s2 + 3 s− 5. Try s =1 and we are lucky. So we factorize

s3 + s2 +3 s− 5 = (s− 1)(

s2 + 2 s +5)

= (s− 1)[

(s +1)2 + 22

]

. (45)

Thus we write

Y =A

s +1+

B

s− 1+

C (s + 1)+ 2D

(s + 1)2 +22. (46)

This gives

2 s2 + 14 = A (s− 1)[

(s + 1)2 +4

]

+B (s + 1)[

(s +1)2 +4

]

+ [C (s + 1)+ 2D] (s + 1) (s− 1). (47)

− Setting s =1: 16= 16B� B = 1;− Setting s =− 1: 16=− 8 A� A=− 2;− Comparing s3: 0= A+ B + C� C = 1;− Setting s =0: 14=− 5 A+ 5 B − [C + 2D]� D = 0.So

Y =− 2

s +1+

1

s− 1+

s + 1

(s + 1)2 +22. (48)

Taking inverse transform:

y =− 2 e−t + et + e−t cos 2 t. (49)



L{t y ′′− t y ′+ y} = L{t y ′′}−L{t y ′}+L{y}

= −d

dsL{y ′′}+

d

dsL{y ′}+L{y}

= −d

ds

[

s2 Y − s y(0)− y ′(0)]

+d

ds[s Y − y(0)] + Y

= − 2 s Y − s2 Y ′+2 + Y + s Y ′+ Y

= −(

s2− s)

Y ′+ (2− 2 s)Y + 2. (50)


L{2}=2

s. (51)

3. The transformed equation now reads

−(

s2− s)

Y ′ +2 (1− s)Y +2 =2

s. (52)


This is a linear equation which can be re-written into the standard form:

Y ′+2

sY =

2

s2. (53)

The integrating factor is

e∫ 2

s = s2. (54)

Apply this integrating factor we obtain

[

s2 Y]′

= 2� Y =1

s2 [2 s +C] =2

s+

C

s2. (55)

4. Taking inverse transform. We get

y = 2 +C t. (56)

5. Now using y ′(0) =− 1 again, we get C =− 1. So the solution is

y = 2− t. (57)

Solution 8. (7.6 14) Spotting e−3s we know that we need to use

L{f(t)u(t− a)}= e−asL{f(t + a)} (58)

which gives

L−1{

e−as F (s)}

=u(t− a) f(t− a). (59)

In this problem F (s)=1

s2 + 9and a = 3. We compute

f(t) =L−1

{

1

s2 + 9

}

=1

3sin 3 t (60)

Therefore

L−1

{

e−3s

s2 + 9

}

=1

3u(t− 3) sin 3 (t− 3). (61)

Solution 9. (7.6 26) It is clear that the function is periodic with perioda. To compute its Laplace transform we need to first write down its formulain one period [0, a]. Inside this interval the function is linear (because itsgraph is a straight line) so

f(t) =A t +B for 0 <t < a. (62)

Now as f(0) =0, f(a) =1, we have f(t)= t/a.Now recall the formula for transforming a function with period T :

L{f }(s)=

∫

0

Te−st f(t) dt

1− e−sT. (63)

We have here T = a and f(t)= t/a from 0 to a. Thus

L{f }(s)=

∫

0

ae−st t/adt

1− e−sa=

1− e−as − a s e−as

a s2 (1− e−as). (64)

8

Solution 10. (7.6 40)


L{y ′′+3 y ′+ 2 y} = L{y ′′}+ 3L{y ′}+2L{y}

= s2 Y − s y(0)− y ′(0)

+ 3 [s Y − y(0)] + 2Y

=(

s2 +3 s + 2)

Y − 2 s+ 1− 6

=(

s2 +3 s + 2)

Y − 2 s− 5. (65)


First write

g(t)= e−t + u(t− 3)(

1− e−t)

. (66)

Now compute

L{g} = L{

e−t}

+L{

u(t− 3)(

1− e−t)}

=1

s + 1+ e−3sL

{

1− e−(t+3)}

=1

s + 1+ e−3s

[

1

s−

e−3

s + 1

]

. (67)

3. The transformed equation now reads

(

s2 +3 s + 2)

Y − 2 s− 5=1

s + 1+ e−3s

[

1

s−

e−3

s + 1

]

(68)

which gives

Y =2 s +5

s2 +3 s + 2+

1

(s +1) (s2 + 3 s + 2)+ e−3s

1

s (s2 + 3 s +2)−

e−3se−3

(s2 +3 s + 2) (s + 1). (69)

4. Compute the inverse transforms. We need to write the RHS into partialfractions.

−2 s +5

s2 +3 s + 2. We have

2 s + 5

s2 + 3 s + 2=

2 s + 5

(s + 1) (s + 2)=

A

s + 1+

B

s +2. (70)

We need

2 s + 5= A (s +2)+ B (s +1) (71)

which gives A= 3, B =− 1. So

2 s +5

s2 +3 s + 2=

3

s+ 1−

1

s + 2. (72)


−1

(s +1) (s2 +3 s +2). We have

1

(s + 1) (s2 + 3 s + 2)=

A

s + 1+

B

(s + 1)2+

C

s + 2(73)

which leads to

1= A (s +1) (s +2)+ B (s +2)+ C (s +1)2 (74)

Setting s =− 1,− 2 and comparing s2 terms, we get

B = 1; C =1; A +C =0� A =− 1. (75)

So1

(s + 1) (s2 + 3 s + 2)=−

1

s + 1+

1

(s + 1)2+

1

s + 2. (76)

−1

s (s2 +3 s +2). We have

1

s (s2 + 3 s + 2)=

A

s+

B

s + 1+

C

s + 2(77)

which leads to

1 = A (s + 1) (s + 2)+ B s (s + 2) +C s (s + 1) (78)

and therefore

A= 1/2; B =− 1; C = 1/2. (79)

So1

s (s2 + 3 s + 2)=

1/2

s+

− 1

s+ 1+

1/2

s + 2. (80)

−e−3

(s2 +3 s + 2) (s +1). As

e−3

(s2 +3 s + 2) (s + 1)= e−3 1

(s2 + 3 s + 2) (s + 1)(81)

we have

e−3

(s2 + 3 s + 2) (s + 1)= e−3

[

−1

s + 1+

1

(s + 1)2+

1

s + 2

]

. (82)

Summarizing, we have

Y =2

s +1+

1

(s + 1)2+ e−3s

[

1/2

s−

1− e−3

s + 1−

e−3

(s + 1)2+

1/2− e−3

(s +2)

]

. (83)

10

Taking inverse transform, we finally obtain

y = 2 e−t + e−t t + u(t− 3)

[

1

2−

(

1− e−3)

e−(t−3)− e−3 e−(t−3) (t− 3) +(

1

2− e−3

)

e−2(t−3)

]

. (84)

Simplify, we reach

y =2 e−t + e−t t+ u(t− 3)

[

1

2+

(

4− t− e3)

e−t +e6− 2 e3

2e−2t

]

. (85)

Remark. Note that as the calculation involved is quite complicated – thuserror-prone, it’s a good idea to check the validity of the solution.

− Quick Check. For equations with right hand sides that are piecewisecontinuous (that is the RHS is totally good except for a few jump dis-continuities), the solution should be continuous. Thus if y = p(t) +u(t− 3) q(t), q(3) must be 0. We check

1

2+

(

4− 3− e3)

e−3 +e6− 2 e3

2e−6 =

1

2+ e−3− 1+

1

2− e−3 = 0. (86)

Indeed.Note that “quick check” does not fully guarantee the correctness of

the solution. Use this only when you do not have enough time to do thefull check.

− Full Check. To do a full check, we need to write

y(t) =

2 e−t + e−t t t < 3

2 e−t + e−t t +1

2+

(

4− t− e3)

e−t +e6

− 2 e3

2e−2t t > 3

(87)

which simplifies to

y(t) =

2 e−t + e−t t t < 31

2+

(

6− e3)

e−t +e6

− 2 e3

2e−2t t > 3

(88)

Now for t < 3 we substitute y = 2 e−t + e−t t and get2

y ′′+3 y ′+ 2 y = e−t; (89)

For t > 3 we substitute y =1

2+

(

6− e3)

e−t +e6− 2 e

3

2e−2t and get3

y ′′+ 3 y ′+ 2 y = 1. (90)

2. In fact only need to substitute y = e−t t as e−t solves the homogeneousequation y ′′ +3 y ′ + 2 y = 0.

3. Similarly, the trick to speed up the checking here is to recall that e−t ande−2t solves y ′′ + 3 y ′ +2 y = 0, so only need to substitute y =1/2.


Done.

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