复变函数与积分变换（修订版）课后答案（复旦大学出版社）
L[?1s(s?a)222]?L(?1ss?a22?as?a22?1a
?1?)F(s)?1(s?4)2222??2??22216(s?4)8(s?4)s?422212(s?4)21s?42
又因为
L(cosat)?ss?a22,L(sinat)?as?a22
?116s?4?18(s?4)?故
L(F(s))?116L(?1所以，根据卷积定理
L(??t?12s?42)?18L(?1s?4(s?4)22)?2116sin2t?18t?cos2t
ss?a22?11?)?cosat?sinat22s?aaaa(2)：
F(s)??1s?5s?413s?11313?111cosa???sin(at?a?)d???0aat2a?sinat?t012[sinat?sin(2a??at)]d?42?13s?1222(12?1s?42)(12?1
?12s?21)?16)2s?222
15.利用卷积定理证明
L[?1L(F(s))?L(1s(s?1)]?2πet?t0e?y2s?1162L(?1)dy
(3)
?sint?sin2t)证明：
L[?11s(s?1)1s(s?1)]?L[?11ss?1 et?1]F(s)?s?2(s?4s?5)?122?1211??()?222[(s?2)?1]2(s?2)?1
t?e?2ts?2L[?1]?2π?t0e?y2dy故L(F(s))?
(4)
F(s)??sint
因为 L(?12s?3s?3(s?1)(s?3)14,B??14221s)?1πt?12,L(?11s?1?As?132?Bs?3?C(s?3)2?D(s?3)3)?e
?A?t,C?,D?3
所以，根据卷积定理有
L[?2π?11s(s?1)et]?1π?t?12?e?2t2?πett0yet?12(t?y)dy?21π2πetet?t0y2edy?y2?1?y故
134?4?2?23s?1s?3(s?3)(s?3)?13?t0edy??????y令y?uπ?0e?u2du??t0edyF(s)?
16. 求下列函数的拉普拉斯逆变换.
(1)F(s)?1(s?4)22
且
1s?5s?442(2)F(s)?
(1s?3)???1(s?3)2,(1s?3)???2?1(s?3)3(3)F(s)?s?2(s?4s?5) 2s?3s?32(s?1)(s?3)
222
所以
L(F(s))??114(4)F(s)?解:(1)
e?t?14e?3t?32t?e?3t?3t?e2?3t
17.求下列微分方程的解
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复变函数与积分变换（修订版）课后答案（复旦大学出版社）
(1)y???2y??3y?e?t,y(0)?0,y?(0)?1
(2)y???y??4sint?5cos2t,y(0)??1,y?(0)??2
(3)y???2y??2y?2et?cos2t,y(0)?y?(0)?0
(4)y????y??e2t,y(0)?y?(0)?y??(0)?0 (5)y(4)?2y???y?0,y(0)?y?(0)?y???(0)?0,y??(0)?1
解: (1)设
L[y(t)]?Y(s),L[(y?(t)]?sY(s)?y(0)?sY(s),L[(y??(t)]?sY(s)?sy(0)?y?(0)?sY(s)?122s?Y(s)?2s?Y(s)?2Y(s)?2?2s?1(s?1)?12
(s?2s?2)Y(s)?Y(s)?2(s?1)[(s?1)?1]222(s?1)(s?1)?122??[1(s?1)?12]?
因为由拉氏变换的微分性质知，若L[f(t)]=F(s),则
L[(?t)?f(t)]?F?(s)
即
L[F?(s)]?(?t)?f(t)?(?t)?L[F(s)]?1?1方程两边取拉氏变换,得
s?Y(s)?1?2s?Y(s)?3Y(s)?(s?2s?3)Y(s)?221s?1
因为L?1[所以
L{?11s?1?1?s?2s?1
1(s?1)?12]?e?sintt
Y(s)?s?2(s?1)(s?2s?3)2?s?2(s?1)(s?1)(s?3)2(s?1)[(s?1)?1]?12}??L[(21?11(s?1)?1t2)?]s1??1,s2?1,s3??3为Y(s)的三个一级极点,
??(?t)L[则
3(s?1)?12]?t?e?sint
y(t)?L[Y(s)]??Res[?1?Res[Y(s)?ek?1stst;sk](s?2)?est故有y?t??t?et?sint
;1](s?2)?e(s?1)(s?1)(s?3)(s?2)?e3818st;?1]?Res[(s?1)(s?1)(s?3)
(4)方程两边取拉氏变换,设L[y(t)]=Y(s),得
s?Y(s)?s?y(0)?s?y?(0)?y??(0)?s?Y(s)?y(0)?s?Y(s)?s?Y(s)?332?Res[??14e(s?1)(s?1)(s?3)?t;?3]1s?2?e?te?3t1s?2?1s(s?2)(s?1)2(2) 方程两边同时取拉氏变换,得
s?Y(s)?s?2?Y(s)?4?221s?12?5?ss?222
Y(s)?1s?2s(s?1)?12
(s?1)Y(s)?4?Y(s)??2(??41s?122?5?ss?25s22故
?(s?2)?s?2(s?1)s22y(t)?L[Y(s)]?2?114e?t?14e?2t?32t?e?3t?3t?e2?3t
(s?1)(s?1)1?122?(s?1)(s?2)1222s?1s?122?12)?s?(s?1s?2?12)?2s?1s?1?22
(5)设L[y(t)]=Y(s),则
L[(y?(t)]?sY(s)?y(0)?sY(s),22L[(y??(t)]?s?Y(s)?sy(0)?y?(0)?sY(s)s?1s?2?s22L[(y???(t)]?s?Y(s)?s?y(0)?sy?(0)?y??(0)?sY(s)?1L[(y(t)]?s?Y(s)?s?y(0)?s?y?(0)?sy??(0)?y???(0)?s?Y(s)?s
方程两边取拉氏变换,,得
(4)4324323
y(t)?L[Y(s)]??2sint?cos2t(3)方程两边取拉氏变换,得
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复变函数与积分变换（修订版）课后答案（复旦大学出版社）
s?Y(s)?s?2s?Y(s)?Y(s)?0(s?2s?1)?Y(s)?sY(s)?s(s?1)224242L[(x(t)]?X(s),L[(y(t)]?Y(s),L[(g(t)]?G(s)L[(x?(t)]?s?X(s),L[(y?(t)]?s?Y(s)?12(s?1)?2s22??12?(1s?12
)?L[(x??(t)]?s?X(s),L[(y??(t)]?s?Y(s),22
方程两边取拉氏变换,得
?s?X(s)?2s?Y(s)?G(s)...(1)?22?s?X(s)?s?Y(s)?Y(s)?0...?2?
故
y(t)?L[?1111?1?]?L[??()]?t?sint 222(s?1)2s?12s18.求下列微分方程组的解
t??x??x?y?e(1)? ?t??y?3x?2y?2?e(1)?s?(2),得
Y(s)??ss?12x(0)?y(0)?1?G(s)...(3)
?1
t0?x??2y??g(t)(2) ?x???y???y?0??y(t)?L[Y(s)]??g(t)*cost???g?cos?t???d?
(3)代入(1)： s?X(s)?2s?[?即：ss?122
x(0)?x?(0)?y(0)?y?(0)?0解:(1) 设
L[(x(t)]?X(s),L[(y(t)]?Y(s)L[(x?(t)]?s?X(s)?x(0)?s?X(s)?1L[(y?(t)]?s?Y(s)?y(0)?s?Y(s)?1,?G(s)]?G(s)
s?X(s)?(1?X(s)?1?s2s22s?1)G(s)?1?s22s?1?G(s)???G(s)?
微分方程组两式的两边同时取拉氏变换,得 1?s?X(s)?1?X(s)?Y(s)???s?1? ?s?Y(s)?1?3X(s)?2Y(s)?2?s?1?2s?1G(s)???2s?s2?1??s1?s所以
?x(t)?L[X(s)]?(1?2cost)?g(t)??(1?2cos?)?g(t??)d?0?1t
故
x(t)?得
s?Y(s)?(s?1)X(s)?...(1)??s?1? ?3X(s)?(s?2)?Y(s)?2?1?s?1...(2)?s?1s?1??t0(1?2cos?)?g(t??)d?ty(t)???g(?)?cos(t??)d?0
19.求下列方程的解
(1)x(t)??x(t??)?ed??2t?30t?(2)代入(1)，得
3X(s)?(s?2)?[(s?1)X(s)?(s?s?1)X(s)?故X(s)?1s?12
ss?1?]?2s?1s?1s?1(2)y(t)??(t??)?y(?)d??t0ts?1s?1?s(s?2)s?1ts?s?1
解：(1)设L[x(t)]=X(s), 方程两边取拉氏变换,
得
X(s)?X(s)?X(s)[1?1s?11s?1]??2s22于是有x(t)?e...(3)?3s(3)代入(1)，得
Y(s)?(s?1)?1s?1?ss?1?1s?1?y(t)?e
t2?3ss
??3s?5s?2s32X(s)?(2?3s)(s?1)s32??3s?5s2?2s3(2)设
?x(t)??3?5t?t
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复变函数与积分变换（修订版）课后答案（复旦大学出版社）
(2)设L[y(t)]=Y(s), 方程两边取拉氏变换,得
Y(s)?L(t?y(t))?Y(s)?Y(s)?1s21s22?Y(s)?11s
s?1?1?12?y(t)?L(Y(s))?L(1s?12)?sht
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