11s2?Y(s)?1?2s?Y(s)?3Y(s)?s?1 (s2?2s?3)Y(s)?1s?2s?1?1?s?1
Y(s)?s?2s?2(s?1)(s2?2s?3)?(s?1)(s?1)(s?3) s1??1,s2?1,s3??3为Y(s)的三个一级极点,
则
y(t)?L?13[Y(s)]??Res[Y(s)?est;sk]k?1?Res[(s?2)?est(s?1)(s?1)(s?3);?1]?Res[(s?2)?est(s?1)(s?1)(s?3);1] Res[(s?2)?est?(s?1)(s?1)(s?3);?3]??1e?t?3et?1e?3t488(2) 方程两边同时取拉氏变换,得
s2?Y(s)?s?2?Y(s)?4?1s2?1?5?ss2?22 (s2?1)Y(s)?4?1s2?1?5?ss2?22?(s?2)Y(s)?45ss?2(s2?1)(s2?1)?(s2?1)(s2?22)?(s2?1)?2(1111s2
s2?1?s2?1)?s?(s2?1?s2?22)?s2?1?s2?1??2ss2?1?s2?22y(t)?L?1[Y(s)]??2sint?cos2t
(3)方程两边取拉氏变换,得
s2?Y(s)?2s?Y(s)?2Y(s)?2?s?1(s?1)2?1 (s2?2s?2)Y(s)?2(s?1)(s?1)2?1Y(s)?2(s?1)1[(s?1)2?1]2??[(s?1)2?1]? 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?1[F?(s)]?(?t)?f(t)?(?t)?L?1[F(s)]
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因为L?1[(s?1)2?1]?et?sint 所以
L?1{2(s?1)[(s?1)2?1]2}??L?1[(1(s?1)2?1)?]??(?t)L?1[1(s?1)2?1]?t?et?sint
故有y?t??t?et?sint
(4)方程两边取拉氏变换,设L[y(t)]=Y(s),得
s3?Y(s)?s2?y(0)?s?y?(0)?y??(0)?s?Y(s)?y(0)?1s?2s3?Y(s)?s?Y(s)?1s?2Y(s)?1s?2?11s(s2?1)?s(s?2)(s2?1)故
y(t)?L?1[Y(s)]?1e?t?1e?2t?3t?e?3t?3t2?e?3t442(5)设L[y(t)]=Y(s),则
L[(y?(t)]?sY(s)?y(0)?sY(s),L[(y??(t)]?s2?Y(s)?sy(0)?y?(0)?s2Y(s)L[(y???(t)]?s3?Y(s)?s2?y(0)?sy?(0)?y??(0)?s3Y(s)?1L[(y(4)(t)]?s4?Y(s)?s3?y(0)?s2?y?(0)?sy??(0)?y???(0)?s4?Y(s)?s
方程两边取拉氏变换,,得
s4?Y(s)?s?2s2?Y(s)?Y(s)?0(s4?2s2?1)?Y(s)?sY(s)?s12s11(s2?1)2?2?(s2?1)2??2?(s2?1)?故
y(t)?L?1[s?1111(s2?1)2]?L[?2?(s2?1)?]?2t?sint18.求下列微分方程组的解
(1)??x??x?y?et ???y??3x?2y?2?etx(0)?y(0)?1
?x??2y??g(t)(2) ??x??y???y?0x(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,
s?X(s)?2s?[?即：s?G(s)]?G(s)s2?1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?得
2s21?s2s?X(s)?(1?2)G(s)?2?G(s)s?1s?11?s22s??1X(s)?G(s)????G(s)2?s?s2?1??s1?s?所以
t?x(t)?L?1[X(s)]?(1?2cost)?g(t)??(1?2cos?)?g(t??)d?0
s?Y(s)?(s?1)X(s)?...(1)??s?1? ?3X(s)?(s?2)?Y(s)?2?1?s?1...(2)?s?1s?1?(2)代入(1)，得
故
x(t)??(1?2cos?)?g(t??)d?0ty(t)???g(?)?cos(t??)d?0t
19.求下列方程的解
(1)x(t)?(2)y(t)?ss?13X(s)?(s?2)?[(s?1)X(s)?]?s?1s?1s?1s(s?2)s2?s?12(s?s?1)X(s)???s?1s?1s?1
1故X(s)?于是有x(t)?et...(3)s?1(3)代入(1)，得
?t0x(t??)?e?d??2t?3
解：(1)设L[x(t)]=X(s), 方程两边取拉氏变换,得
?(t??)?y(?)d??t0tY(s)?(s?1)?(2)设
1s1???y(t)?et s?1s?1s?1L[(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)L[(x??(t)]?s2?X(s),L[(y??(t)]?s2?Y(s),方程两边取拉氏变换,得
123?2?s?1ss12?3sX(s)[1?]?2 s?1s(2?3s)(s?1)?3s2?5s?2352X(s)??????s3s3ss2s3?x(t)??3?5t?t2X(s)?X(s)?
(2)设L[y(t)]=Y(s), 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?
Y(s)?L(t?y(t))?Y(s)?1s2
(1)?s?(2),得
Y(s)???1s?G(s)...(3)s2?1
t011?Y(s)?s2s21Y(s)?2s?1?y(t)?L?1(Y(s))?L?1(
?y(t)?L[Y(s)]??g(t)*cost???g?cos?t???d?(3)代入(1)：
1)?sht2s?1 37 / 37