解 a)特征方程det(?E?A)???21?1??2?3?0，得?1,2??3是特征??2值．对应的特征向量分别为u1????1??1????，，??0,??0为任意常u??2????2?3??2?3????． 3t?(2?3)e?e3t?3t?e数．所以方程组x??Ax的一个基解矩阵为?(t)???(2?3)e??3t?3t?e?1expAt??(t)?(0)???(2?3)e???3?(2?3)e? ??6?e?3t3t3t??11???? ?3t??(2?3)e??2?32?3?e3t3t?1?(2?3)e?e3te(2?3)e3t3t?e???． ?3t??(2?3)e?3tb)由第3题a)立即得到方程组x??Ax的一个基解矩阵为
?e?t?(t)????e?t??e?t?1expAt??(t)?(0)????e?t?e5t??． 5t?2e??1e5t??11???? 5t??2e???12??e5t?e?t?1?e5t?2e?t?． ??5t?t5t?t??3?2(e?e)2e?e?c)由第3题b)立即得到方程组x??Ax的一个基解矩阵为
?0??2t?(t)??e?e?2t?e?te?t0e2t??2te?． e2t???1?0??2t?1expAt??(t)?(0)??e?e?2t??e2t?2t?2t ??e?e?e2t?e?2t?e?te?t0e2t??011????2te??111?
??e2t???101?e2t?e?t??2t?te?e?． e2t???31?(??3)(?2?4??3)?0，特征值??1e?t?e2te?2t?e?t?e2te?2t?e2t??1d)特征方程det(?E?A)??8?50??1?1为?1??3，?2,33??3????????2?7．对应的特征向量分别为u1???7?，u2????5?47?，
?1?7??4???????3???u3???5?47?，?,?,?均为不等于零的任意常数．故方程组x??Ax的一个基解矩阵
??1?7???为
??3e?3t??(t)??7e?3t??4e?3t??13e(2?7)t(47?5)e(2?7)t(7?1)e(2?7)t??3e(2?7)t?(2?7)t?． (47?5)e?(7?1)e(2?7)t??由expAt??(t)?(0)立即可得expAt???1(t)?2(t)?3(t)?，其中列向量函数
?42e?3t?3(7?37)e(2?7)t?3(7?37)e(2?7)t??1??3t(2?7)t(2?7)t?， ?1(t)???98e?(137?49)e?(49?137)e84???56e?3t?2(14?57)e(2?7)t?2(14?57)e(2?7)t?????42e?3t?3(?7?57)e(2?7)t?3(7?57)e(2?7)t??1??3t(2?7)t(2?7)t??98e?(175?537)e?， ?1(t)??(175?537)e252???56e?3t?2(14?7)e(2?7)t?2(14?7)e(2?7)t??????84e?3t?3(14?7)e(2?7)t?3(14?7)e(2?7)t??1??3t(2?7)t(2?7)t?196e?(617?98)e?． ?1(t)??(617?98)e126??112e?3t?(7?137)e(2?7)t?(7?137)e(2?7)t????（该题计算量太大，作为该法的习题不是太好！）
5．试求方程组x??Ax的一个基解矩阵，并求满足初始条件?(0)??的解?(t)：
?12??3???，a)A?????43??3??； ?????103??0?????b)A??81?1?，????2?；
?51?1???7??????121??1?????c)A??1?11?，???0?．
?0??201?????1?e?t解 a)由上题b)知expAt???t3???ee5t??2?1????，所以所求解为 ?5t??2e??11??e?t?2e5t???(t)?(expAt)?????e?t?4e5t?．
??b)由上题d)知expAt??(t)??1(0)，其中
??3e?3t??(t)??7e?3t??4e?3t?所以所求解为
3e(2?7)t(47?5)e(2?7)t(7?1)e(2?7)t??3e(2?7)t?(2?7)t?． (47?5)e?(7?1)e(2?7)t???1432??42??0?????1?(t)?(expAt)???(t)??3(37?7)57?72(14?7)???2?
252?????3(37?7)57?72(14?7)???7??3t(2?7)t(2?7)t?546e?3(27?91)e?9(35?47)e1???1274?e?3t?(511?3747)e(2?7)t?3(289?1607)e(2?126???728e?3t?(77?897)e(2?7)t?3(7?317)e(2?7)t???7)t?． ???c)由第3题c)知，矩阵A的特征值为?1,2??1，?3?3．对应于特征值?3?3的特
?2??846??u1???????2征向量v???1? （??0的任意常数）．又由(?1E?A)u??423??u2??0，得到
?2??846??u??????3?3?3????1??2???????????11u??3?3?，由???0????1???? （?,?是任意常数）?解出
3?3??0??2???(4??2?)???(4??2?)?????????,??,???．依公式（5.52），得满足初始条件?(0)??的解为
?2??2t?12t??13t1?t?3t?t?(t)?eEv?e[E?t(A?E)]u?e?1??e?t14??4?0?2??2t??2????t???1?
??2t?1????2?t141214?2(e3t?e?t)??1?3t?t ??e?e?
4?3t?t?2(e?e)??6．试求方程组x??Ax?f(t)的解?(t)：
?et???1??12??a)?(0)???1??，A???43??，f(t)???1?；
??????10??0??0?0???????01?，f(t)??0?； b)?(0)??0?，A??0?0???6?11?6??e?t?????????1??4?3??sint?????，，c)?(0)??A?f(t)?????2?1???2cost??． ?2?????1?e?t解 a)由第4题b)知，expAt???t3???ee5t??2?1???，由公式（5.61）得 ??5t??2e??11??(t)?(expAt)???exp[(t?s)A]f(s)ds
0t1?e?t ???t3??e?e5t??2?1???1?t1?e?(t?s)???????????5t???(t?s)2e??11??1?03??e?e5(t?s)??2?1??es?????ds ??5(t?s)????2e??11??1?95t4??3t?t?e?3e?e??205?． ??4??3et?e?t?9e5t?3???105??2b)由第3题d)知A的特征值?1??1，?2??2，?3??3，对应的特征向量分别为
?1??1??1???????u????1?，v????2?，w????3?，其中?,?,?均是不为零的任意常数．x??Ax?1??4??9???????的一个基解矩阵为?(t)?[e1u?te?2tve?3t?1?e?t??tw]???e?e?t?e?2t?2e?2t4e?2te?3t???3t?3e?． 9e?3t??11??1????1(0)???1?2?3??149???由公式（5.61）得
t51??6?1????6?8?2?，而expAt??(t)??1(0)． 2?31??2??(t)?(expAt)???exp[(t?s)A]f(s)ds
0?e?t1??t???e2??t?ee?2t?2e?2t4e?2te?3t??651??0???????3t?3e???6?8?2??0?
???9e?3t?31???0???2e?2(t?s)?2e?2(t?s)4e?2(t?s)e?3(t?s)??651??0???????3e?3(t?s)???6?8?2??0?ds
?2??e?s?9e?3(t?s)?31??????e?(t?s)1t??(t?s) ????e20??(t?s)?e?e?t?2es?2t?e2s?3t??(2t?3)e?t?4e?2t?e?3t???1t??t1?s?2t2s?3t?t?2t?3t????e?4e?3e?ds??(5?2t)e?8e?3e?．
02??t4?s?2t2s?3t??t?2t?3t?e?8e?9e(2t?7)e?16e?9e????c)A的特征方程det(?E?A)???4?23?(??1)(??2)?0，求解得特征值??1?1??1??3??2??1?1，?2?2，对应的特征向量分别是u??????，其中?,?是不为零的任???，v???意常数．所以方程组x??Ax的一个基解矩阵为?(t)?[eu?1te?2t?etv]???et?3e2t??，从而，2t?2e???23?expAt??(t)?(0)??(t)??1?1??．由公式（5.61）得
???1?(t)?(expAt)???exp[(t?s)A]f(s)ds
0t?e?t???et?3e2t??23???1?t?e(t?s)???????????2t??(t?s)2e??1?1???2?0??e3e2(t?s)???23??sins??????ds ???2(t?s)??1?1?2coss2e??????(3?2?2?1)et?3(?1??2)e2t???4et?3e2t?2sint?cost???????(3??2?)et?2(???)e2t????4et?2e2t?2sint?2cost?
?112?2???(3?2?2?1?4)et?3(?1??2)e2t?2sint?cost?????(3??2??4)et?2(???)e2t?2sint?2cost?．
112?2?mt7．假设m不是矩阵A的特征值，试证非齐线性方程组x??Ax?ce有一解形如
?(t)??emt，其中c,?是常数向量．
证明 设方程组有形如?(t)??e的解，代入方程得m?emtmt?A?emt?cemt，由此得
m??A??c，即(mE?A)??c．因为m不是矩阵A的特征值，故det(mE?A)?0，
?1即矩阵mE?A可逆，得到??(mE?A)c唯一确定．
所以方程组有一解?(t)?(mE?A)ce?1mt??emt
???3x1??2x1?x2??x2?0,?x18．给定方程组 ?
??x?2x?x?x?0．122?1a)试证上面方程组等价于方程组u??Au，其中