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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