12-11

12-11习题12111试用幂级数求下列各微分方程的解(1)yxyx1解设方程的解为10nnnxaay代入方程得111011xxaxaxnannnnnn即0])2[()12()1(112021nnnnxaanxaaa可见a1102a2a010(n2)an2an0(n12)于是11a2102aa!!313a!!4104aa!)!12(112kak!)!2(102kaak所以]!)!2(1!)!12(1[120120kkkxkaxkay1201120)2(!1)1(!)!12(1kkkkxkaxka11220!)!12(1)1(12kkxxkea即原方程的通解为1122!)!12(112kkxxkCey(2)yxyy0解设方程的解为0nnnxay代入方程得0)1(01122nnnnnnnnnxaxnaxxann即0])1()1)(2[(21220nnnnxanannaa于是0221aa1331aa1112!)!12()1(akakk02!)!2()1(akakk所以]!)!12()1(!)!2()1([12112010kkkkkxkaxkaxaay11211020!)!12()1()2(!!1kkkkkxkaxka1121120!)!12()1(2kkkxxkaea即原方程的通解为1121221!)!12()1(2kkkxxkCeCy(3)xy(xm)ymy0(m为自然数)解设方程的解为0nnnxay代入方程得0)()1(01122nnnnnnnnnxamxnamxxannx即0])())(1[()(1110nnnnxamnamnnaam可见(a0a1)m0(nm)[(n1)an1an]0(nm)于是a0a