復(fù)變函數(shù)與積分變換2011 20122習(xí)題解答八

f(t)cos

tt L[f(t)]23estdtestcostdt 1e2e

2f(t)cost(t)sintu(t

s sL[f(t)]cost(t)estdtsintu(t)estdtcostest 1s21s21f(t)t21L[f(t)]L[t2]L[1]2!121 f(t)1tet

s2L[f(t)]L[1]L[tet]1s

(sf(t)tcostf(t)tcostteiteit2 1 s2L[f(t)] L[teit] L[teit] 2(s (si)2 (s2s令g(t)cost,則G(s)L[g(t)] ss2

G'(s)

1(s21)2L[tcostL[tg(tGs2即Lf(t(s21)2f(t)e2tsin6t解：因為L[sin6t] ,s2

(s2)236f(t)cos2t解：f(t) 2 s2Lf(tL[]2

2L[cos2t]

2s

2s24

4)f(t)u(1et)1,1et t解：因為u(1et) 0,1etLf(t1estdt1 f(t)(t1)2etf(t(t1)2ett2et2tetet

即：u(1et) t s24sL[f(t)]L[t2et]2L[tet]L[et] 2 (s (s

s1

(s 令g(t)et 則G(s)L[g(t)]1

(s1)3 ,(s

L[et]12

s24sLf(t

e]2L[te]L[e]

(s f(t)tneat 因為L[tn] ,

L[f(t)]L[tneat]

(sa)n1g(teat，則G(sL[g(t

故Gn)

(sa)n1Lf(t

(sa)n1f(tte3tsin2t,Lf(tf(tte3tsin2t

1te3t(e2tie2ti)

1[te(2i3)tte(2i3)t],L[f(t)]1L[te(2i3)t]1L[te(2i3)t]

2

4(s23)2

(s2i

(s2i

[(s3)?

f(t)

tte3tsint

Lf(t0g(te3tsin2t,G(s)Lg(t)].g(s)L[e3te2tie2ti]1

2 2is2i s2i3 (s3) L[0g(t)dtsG(ss[(s3)24

t ' sin2tdt]L[t0g(t)dt]s[(s3)24]'所以Lf(t

2(3s212ss2[(s3)24]2F(s) ;(s2設(shè)F1(s) , F(s)

F(sF1(sF2s2 s2而f(t)L1[F(s)]2sin f(t)L1[F(s)]cos (t)2tsinicosi(t)dtsinitf(tf(t

i F(sRe(s12s11,s21F(ssF(s0.8.3 f(t)Res[F(s)est,1]Res[F(s)est,1] F(s) (s2設(shè)F1(s) , F(s)

F(sF1(sF2s2 s2而f(tf(tL1[F(s1sin f(t)

f(t)f(t)

tsinisini(t)d

[cositcos(2iitt 2ttcositisinittchtsht F(sRe(s12s11,s21F(ssF(s0.8.3

tet tete tchtf(t)Res[F(s)est,1]Res[F(s)est,1] 由(1)題知L[tsht(s21)2,故L[0tshtdt(s21)2即 ]1ttshtdt1tt(etet)dttchtsht0(s2 2 4 0f(t)sinktt解：設(shè)Lf(tF由拉氏變換的微分性質(zhì)得：F'(s)L[tf(t)]L[tsinkt] s2kdsds所以Lf(tF(ss2kte3tsin

kf(t)0

dtt

解：在(1)k2得t

]arccos 2 te3tsin set

dt] (1)

dtt

L1

t

L[1]

t 故G(s)lns

即 lnst所以L1etln(s 所以

ete2tdt

1

tdtL1et lnsln(s ln 0

te2t0

1解：te2tdt 11 ;s2解：因L[sinkt] s2k故 ]1sins2 s (s1)(s

取k2得L[sin2t] .s2 (s1)(s2) s1 s2而L1 2

11 et2e2ts s21

11 所以L(s1)(s2) e ;(s

(s

F(s0.8.3 L1 Res(s1)4 (s

因為este(s1)tetet1(s1)t

t

t 故(s (s1)4(s1)3 3!(s1) 1 et t所以L

et(s1)42s3s2

2s解：設(shè)F(s) ,則F(s)在Re(s)0內(nèi)具有兩個單極點s s3i.除此s2F(ssF(s

2i L12s3Res[F(s)est,3i]Res[F(s)est,3i]11e3it1 2i ?

2cos3tsins (sa)(sF(s

s(sa)(sb)2s1

s2bF(sF(s8.3L1[F(s)]Res[F(s)est,a]Res[F(s)est,b] ca2eatcbt ;(s21)(s4

(a 解：設(shè)F(s) (s21)(s44)

F(s) s2

F(s) s42顯然L1[F1(s)]sint 下面用留數(shù)法來求L1[F2容易求得F2

2(cos2kisin2k)

k0,1,2, s21 s31 s41用留數(shù)法求F2s)

(1i (1i L[F2(s)]

Res[F2(s)e,sn]8 ]2sintshttL1[F(s)]L1[F(s)F(s)]L1[F(s)]L1[F(s)] t 21t[costcos(2t)]shd1(1cht)cost1sintsht4 2s ;s24s 2s ,則F(s)2(s2)1 2(s s24s (s2)2 (s2)2 (s2)2因為L[sin3t] ,s2

L[cos3t] .s2所以L[e2tsin3t] L[e2tcos3t] s (s2)2 (s2)2故L1[F(s2e2tcos3t1e2tsin3?

1e2 e2 s2s2因為L11 s211e2s所以

t(t2)u(t 1e2s 2t t2 0t22s s(s1)(s2s

1

s(s1)(s 2s因為L11 L11et L11e2t 1 2s 3 所以?

s(s1)(s2)2 2 s45s2解 解 s45s2 3s2 s24 1 sin 1 1sin而 s2 所以L1 1(sint1sin2t)1sint1sin2t?

s2 (s24s5)2

s s 1 解：設(shè)F(s)

(s24s

[(s2)2

g(t)

(s21)2則L1[F(se2tg(t設(shè)F(s) , F(s) .則F(s)F(s)F s2 s2 f(t)L1[F(s)]cost f(t)L1[F(s)]sint g(t)f(t)f(t)

tcossin(t)d1

t[sintsin(2t)]d1tsin 2 所以L1[F(s1e2ttsin2s24s (s24ss24s (s 解：設(shè)F(s(s24s13)2[(s2)29]2則L1[F(se2tg(t

g(t)L1 (s29)2設(shè)F(s) F(s) .則f(t)f(t)L1[F(s)]L1[F(s)]cos3t. s2 s2 g(t)

f(t)f(t)

tcos3cos3(t)d

t 2t1tcos3t1sin 所以L1[F(s)]1tcos3t ?

s s33s26ss s3

6s4

3(s (s1)2311 1 因為L e

L(s1)23 t 1 t (t 0故由卷積定理知：L(s1)33(s1) 01 3 又因為L(s1)233

sin3 d3(1

t)e33所以L1 s 1et(22 t 3 t33 s2 s2 解：設(shè)F(s)

ln(s1)ln(s1)ln

L[f(t)]FL[tf(t)]F'(s) 故tf(tetet

1

2s e 也即f(t1u(t

[F(s)] u(t)

t

t 所以01u(t)d0dtmtn（mn為正整數(shù)

ndm

tmn (t)d0(t)m1m1(t tm1(t)n1dm1

00m1d(t00tet00

tmn1(t)n(n1)(m (mn t(t)dm(m (mn1)(m tmn1 m!n! tmn1.(m1) (mn1) (mn1)!解：tetd0sintcost

t0

et

tetdt0

tett解：tsincos(t)d1t[sintsin(2t)]d1tsint1cos(2t)t1tsin 2 u(ta)f(t a解：因為u(a) a 當(dāng)0at0u(af(t)daf(t)d t故當(dāng)at0u(af(t)d 0a即u(taf(t(ta)f(t)

a 解：設(shè)au,則0(af(t)da(uf(tu 所以當(dāng)ta0即at當(dāng)ta a0即0at時，由函數(shù)的篩選性質(zhì)得 a(u)f(tua)du(u)f(tua)duf(tFL[tf(t)dt] 證明：因為

[F(s)]f(t

L1F(st1f()dtf(t t也即t

f(t)dt]

F

L1

tsin (a (s2a2)2 證明：因為L1 cos L1 1sin L1

cosasina(t)d [sinatsina(2tt tt(s2a2)2t

a 2atsinat

1cosa(2t)t

y''4y'3yet y(0)y'(0)解：設(shè)LyY(s).s2Y(s)sy(0)y'(0)4sY(s)4y(0)3Y(s)s26s

1代入y(0)y'(0)1,得：Y(s(s1)2(s3 s26s s26s yL1[Y(s)]Res(s1)2(s1[(72t)et3e3t4

est,1Res (s1)2(s

est,y'''3y''3y'y y''(0)y'(0)1,y(0)解：設(shè)L[y]Y s3Y(s)s2y(0)sy'(0)y''(0)3[s2Y(s)sy(0)y'(0)]3[sY(s)y(0)]Y(s)1sY(s2s1s(syL1[Y(s)]Res[Y(s)est,0]Res[Y(s)est,1]1ety''2y'2y2etcost y(0)y'(0)解：設(shè)L[y]Y s2Y(s)sy(0)y'(0)2[sY(s)y(0)]2Y(s) s (s1)2s2Y(s)2sY(s)2Y(s)2 s ,(s1)2即：Y(s)2 [(s1)21 e

e因為L(s1)2

(s1)2

sint ttyL1[Y(s)]

20

cosetsin(t)dtetsinty''3y'2yu(t1) y(0)0,y'(0)1解：設(shè)L[y]Y s2Y(s)sy(0)y'(0)3[sY(s)y(0)]2Y(s)L[u(t Y(s)L[u(t1)] .因 s23s

s23s s

s23s所 L1 ete2t L1L[u(t1)] s3s2

tu(1)[e(t)e2(t)]d0

t1u()[e(t1)e2(t1) t1[e(t1)e2(t1)]d11e2(t1)e(t1)u(t yL1[Y(s)]ete2t11e2(t1)e(t1)u(t1) y(4)y'''cost y(0)y'(0)y'''(0)0,y''(0)2解：設(shè)L[y]Y s4Y(s)s3y(0)s2y'(0)sy''(0)y'''(0)s3Y(s)s2y(0)sy'(0)y''(0)

,s22s32s23s Y(s)s3(s1)(s21)s2(s1)(s21)s3由例8.16知：L

tsint,又因為L11et.s2(s2(s21) t(sin)e(t)d

d

s2(s1)(s21)

因為L12t2

t1etcostsin. .y

e costsin

L[Y(s)]

t1 ?

x'xyety'3x2y2et x(0)y(0)