Homework 14 Laplace Transforms6.1 # 1 , 3, 5, 6, 8 , 9, 15 , 11, 12, 15, 16, 21, 22, 26

6.2 # 1 , 2, 3, 5, 6 , 7, 8, 11, 12, 13, 17, 21 , 22, 24, 25 , 28, 29 , 30, 36

6.1.1. Is the function

f HtL =t2 0 § t § 1

2 + t 1 < t § 26 - t 2 < t § 3

continuous, piecewise continuous or neither on the interval @1, 3D.

1 2 3

1

2

3

4

SOLUTION. The function has a jump at t = 1 and so it is not continuous. The function is continuous on 0 § t < 1,1 < t < 2, and 2 < t § 3. It also has finite one sided limits at t = 1 and t = 2. So the function is piecewise continuous 6.1.8. Find the Laplace transform of sinh bt.

SOLUTION. This is given in the table as LHsinh btL HsL = bs2 - b2

. We could also write

LHsinh btL HsL = LJ‰ b t - ‰- b t

2N HsL

= 12ILI‰ b tM HsL - LI‰- b tM HsLM

= 12I1

s- b- 1

s+ bM = b

s2 - b2.

6.1.15. Find the Laplace transform of t ‰a t using integration by parts.

SOLUTION. We have that of s > a that

LHt ‰a tL HsL = Ÿ0¶ t ‰a t ‰- s t „ t

= Ÿ0¶ t ‰Ha- sL t „ t

= ta- s

I‰Ha- sL t 0¶ - Ÿ0

¶ ‰Ha- sL t „ t M

= - 1Ha- sL2

‰Ha- sL t 0¶

= 1Ha- sL2

6.1.26. The Gamma function is denoted by GHpL and is defined by

GHpL = Ÿ0¶ xp- 1 ‰- x „ x.

The integral converges as x Ø ¶ for all p. For p < 1, the integrand is unbounded as x Ø 0 + but the improperintegral converges for all p > 0. Show

a) For p > 0, GHp + 1L = p GHpL;

b) GH1L = 1;

c) GHn + 1L = n ! whenever n is a positive integer. (This means that the Gamma function provides an extensionof the factorial);d) GHp + nL êGHpL = p Hp + 1L … Hp + n - 1L;

e) Show GI 12M = p ; and

f) Find GI 32M and GI 11

2M.

SOLUTION. For 0 < p < 1 and 0 < a < b < 1, we have

0 < Ÿab xp- 1 ‰- x „ x § Ÿa

b xp- 1 „ x = bp - ap

p.

This means that Ÿ01 xp- 1 ‰- x „ x converges for 0 < p < 1. We also have that

Ÿ1b xp- 1 ‰- x „ x § Ÿ1

b ‰- x „ x = e- a - ‰- 1 .

So we get that GHpL exists for 0 < p < 1. For p ¥ 1, the integral Ÿ0¶ xp- 1 ‰- x „ x is improper only at ¶. Using

l'Hôpital's rule, we have that

0 § limxضxp

‰xê2= limuض

2p up

‰u§ limuض

2k uk

‰u= limuض

2k k!‰u

= 0

where k is the ceiling of p, i.e., the smallest integer greater than p. So we get that xp

‰xê2§ 1 whenever x is greater than

some number M . This means that

ŸMb xp ‰- x „ x § Ÿ1

b xp ‰- xê2 ‰- xê2 „ x § ŸMb ‰- xê2 „ x < ¶

So the gamma function converges for p > 0.

SOLUTION a). We have that

Ÿab xp ‰- x „ x = - xp ‰- x a

b + Ÿabp xp- 1 ‰- x „ x.

If p > 0, we let a Ø 0 and b Ø ¶ in the previous equation and get

GHp + 1L = Ÿ0¶ xp ‰- x „ x = lim aØ 0

bضŸab xp ‰- x „ x = lim aØ 0

bض

Jxp ‰- x ab + Ÿa

bp xp- 1 ‰- x „ xN = p GHpL.

SOLUTION b). Evaluating G(1), we get

GH1L = Ÿ0¶ x1- 1 ‰- x „ x = Ÿ0

¶ ‰- x „ x = limbض I‰0 - ‰- bM = 1.

SOLUTION c). We can do a formal induction proof on the hypothesis P(n): GHn + 1L = n ! for all positive integers n.The basis step GH1L = 1 is given by b) and the inductive step GHn + 1L = n GHnL = n Hn - 1L ! = n ! is a).SOLUTION d). This also follows from part a).

SOLUTION e). We have

GI 12M = Ÿ0

¶ x1ê2- 1 ‰- x „ x = Ÿ0¶ 1

u‰- u

2H2 u „uL = 2 Ÿ0

¶‰-u2„u

using the change of coordinates x = u2. Then a polar change of coordinates is used to give

IŸ0¶‰-u

2„uM2 = Ÿ0

¶‰-u2„u Ÿ0

¶‰-v2„v

= Ÿ0¶Ÿ0¶‰-u

2 - v2 „u „v

2 Hw14aplace.nb

= Ÿ0pê2Ÿ0¶r‰-r2 „ r „q = p

2J- ‰- r

2

2N 0

¶ = p

4,

and finally,

GI 12M = 2 Ÿ0

¶‰-u2„u = p .

SOLUTION f). We have that

GI 32M = 1

2GI 1

2M = p

2 and

GI 112M = 9

2GI 9

2M = 9

2ÿ 72GI 7

2M = 9

2ÿ 72ÿ 52GI 5

2M = 9

2ÿ 72ÿ 52ÿ 32GI 3

2M = 9

2ÿ 72ÿ 52ÿ 32ÿ p

2= 945

2p .

6.2.1. Find the inverse Laplace transform of 3s2 + 4

.

SOLUTION. We have that

L- 1 93

s2 + 4= = 3

2L 9

2s2 + 4

= = 32

sin 2 t.

6.2.5. Find the inverse Laplace transform of 2 s+ 2s2 + 2 s+ 5

.

SOLUTION. We have that

L- 1 92 s+ 2

s2 + 2 s+ 5= = L- 1 :

2 Hs+ 1LHs+ 1L2 + 4

> = 2 L- 1 9l- 1*s

s2 + 4= = 2 ‰- t 2 L- 1 9

ss2 + 4

= = 2 ‰- t cos 2 t.

6.2.6. Find the inverse Laplace transform of 2 s- 3s2 - 4

.

SOLUTION. Writing the function in terms of partial fractions gives2 s- 3s2 - 4

= As+ 2

+ Bs- 2

= HA+ BL s+ 2 A- 2 Bs2 - 4

.

So we need

A + B = 2 and 2 A - 2 B = - 3

and A = 1 ê4 and B = 7 ê4. So we get

L- 1 92 s- 3s2 - 4

= = 14

L- 1 91

s+ 2= + 7

4L-1 9

1s- 2

= = 14‰- 2 t + 7

4‰ 2 t.

6.2.12. Solve y²″ + 3 y£ + 2 y = 0, yH0L = 1, y£H0L = 0 using Laplace transforms

SOLUTION. The characteristc polynomial is pLHsL = s2 + 3 s + 2 = Hs + 1L Hs + 2L. Applying the Laplace transform tothe equation gives

L 8y²″ + 3 y£ + 2 y < HsL 0 = pLHsL L 8y< HsL - pL@1DHsL yH0L - pL

@2DHsL y£H0L

= pLHsL L 8y< HsL - Hs + 3L = 0.

So we get

L 8y< = s+ 3s2+3 s+2

= 2s+1

- 1s+2

and

y = 2 ‰- t - ‰- 2 t.6.2.13. Solve y²″ - 2 y£ + 2 y = 0, yH0L = 0, y£H0L = 1 using Laplace transforms

SOLUTION. The characteristic polynomial is pLHsL = s2 - 2 s + 2. Taking the Laplace transforms of both sides weget

pLHsL L 8y< - pl@1DHsL yH0L - pL

@2DHsL y£H0L = L 80< = 0 or

Hw14aplace.nb 3

Is2 - 2 s + 2M L 8y< - Hs - 2L H0L - 1 = 0

and

L 8y< = 1s2 - 2 s+ 2

= 1s2 - 2 s+ 1+ 1

= 1Hs- 1L2 + 1

and

y = L-1 :s

Hs- 1L2 + 1> HtL = L-1 :

1Hs- 1L2 + 1

> = ‰t sin t

6.2.17. Solve yH4L - 4 y£££ + 6 y²″ - 4 y£ + y = 0, yH0L = 0, y£H0L = 1, y²″H0L = 0, y£££H0L = 1 using Laplacetransforms.SOLUTION. Taking the Laplace transform of both sides of the differential equation gives

L 8L@yD< HsL = pLHsL L 8y< - pL@2DHsL y£H0L - pL

@4DHsL y£££H0L = 0

Since pLHsL = s4 - 4 s3 + 6 s2 - 4 s + 1 = Hs - 1L4, pL@2D = s2 - 4 s + 6 , and pL

@4DHsL = 1, we get

L 8y< = s2 - 4 s+ 7Hs- 1L4

= 1Hs- 1L2

- 2Hs- 1L3

+ 4Hs- 1L4

and

y = L- 1 :1

Hs- 1L2> - 2 L- 1 :

1Hs- 1L3

> + 4 L- 1 :1

Hs- 1L4> = ‰tJ t

GH2L- 2 t2

GH3L+ 4 t3

GH4LN = = ‰tJt - t2 + 2 t3

3N .

since L 8tn ‰a t< HsL = la*L 8tn < HsL = la*GHn+ 1Lsn+ 1

= GHn+ 1LHs- aLn+ 1

where a = 1 in the present problem.

6.2.21. Solve y²″ - 2 y£ + 2 y = cos t, yH0L = 1, y£H0L = 0 using Laplace transforms.

SOLUTION. We have that

Is2 - 2 s + 2M L 8y< - HyH0L Hs - 2L + y£H0L sL = L 8y²″ - 2 y£ + 2 y< HsL = L 8cos t< HsL

or

L 8y< HsL = 1s2 - 2 s+ 2

Js

s2 + 1+ s - 2N = 4- s

5 Is2 - 2 s+ 2M+ 1

5Js- 2s2 + 1

N + s- 2s2 - 2 s+ 2

= 15J

4 s- 6Hs- 1L2 + 1

+ Js- 2s2 + 1

N N

where we have used partial fractions

1s2 - 2 s+ 2

Js

s2 + 1N = A s+ B

s2 - 2 s+ 2+ C s+D

s2 + 1=

HA s+ BL Is2 + 1M+ Is2 - 2 s+ 2M HCs+DL

Is2 - 2 s+DM Is2 + 1M

with

HA s + BL Is2 + 1M + Is2 - 2 s + 2M HCs + DL = s

The like powers of s give the equations

A + C ‡ 0A + 2C - 2D - 1 ‡ 0B - 2C + D ‡ 0B + 2D ‡ 0

which have the solution 9- 15

, 45

, 15

, - 25=, Now we can take the inverse Laplace transform to get

L-1 :15J

4 s- 6Hs- 1L2 + 1

+ Js- 2s2 + 1

N N> = 15

L- 1 :4 Js- 1

Hs- 1L2 + 1N - 2 J

1Hs- 1L2 + 1

N + ss2+ 1

- 2 1s2 + 1

>

= 15H4 ‰t cos t - 2 et sin t + cos t - 2 sin tL

4 Hw14aplace.nb

which is the solution for the initial problem.

6.2.25. Find the Laplace transform Y HsL = L 8y< HsL of the solution of the initial value problem

y²″ + y =t 0 § t < 10 1 § t <¶

, yH0L = 0, y£H0L = 0.

SOLUTION. The forcing factor can be written as t - t H1HtL where H is the Heaviside function. We can also use

pLHsL L 8y< = pLHsL L 8y< - pL@1DHsL yH0L - pL

@2DHsL y£ H0L = Ÿ01t ‰- s t „ t

= t ‰- s t

- s 01 + Ÿ0

1 ‰- s t

s„ t = -‰- s

s+ 1- ‰- s

s2= 1- ‰- sHs+ 1L

s2

and

L 8y< = 1s2 + 1

J1- ‰- sHs+ 1L

s2N.

and

L 8y< = J1

s2 + 1N J

1s2

- ‰- s

sN.

We could use the tables to evaluate L 8 f < HsL. We have that

L 8 t - t H1HtL< HsL = L 8t< HsL - L 8Ht - 1L HH1HtLL< HsL - L 8H1HtL< HsL

= 1s2

- L 8l1* t H1HtL< HsL - L 8l1*1 H1HtL< HsL

= 1s2

- ‰- s L 8t< HsL - ‰- s L 81< HsL

= 1s2

- ‰- s

s2- ‰- s

s= 1-‰- sHs+ 1L

s2.

6.2.29. Find the Laplace transform of t ‰a t.

SOLUTION. We can work this out directly. However, here we use a tranform formula. We have

L 8 t ‰a t< HsL = la*L 8t<@sL = la*1s2

= 1Hs- aL2

,

Hw14aplace.nb 5