高等数学第十二章答案 同济五版1211.docxVIP

习题12?11

1?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y?a

0

???a

n

n?1

xn?代入方程得

??na

n

n?1

xn?1?a

0

x???a

n

n?1

xn?1?x?1?

即 (a

1

?1)?(2a?a

2 0

?1)x???

[(n?2)a

n?2

?a]xn?1?0?

n

可见 a1

?1?0?2a?a

n?1

0?2?1?0?(n?2)

0

?2

n

?a?0(n?1?2????)?

n

1?a 1 1?a

2于是 a

2

1

?1? a?

2

0? a

2 3

? ? a?

3! 4

4!0?????

1 1?a

a

2k?1

? ? a

(2k?1)!! 2k

?(2k

0

)!!

?????

?? 1 1?a

所以 y?a

0

? [(2k?1)!!x2k?1?(2k

k?1

0x2k]

)!!

?a???

1 x2k?1?(1?a

)??

1 x2

1 x2

0

k?1

(2k?1)!! 0

x2 ?? 1

k?1

k! 2

??1?(1?a

0

)e2?

k?1

(2k?1)!!

x2k?1?

?x2

?

即原方程的通解为y Ce2

(2)y???xy??y?0?

?1?

??

k?1

1 x2k?1?(2k?1)!!

解设方程的解为y???a

n

n?0

xn?代入方程得

??

n?2

n(n?1)a

n

xn?2?x??na

n

n?1

xn?1???a

n

n?0

xn?0?

即 a?2a

0 2

???

[(n?2)(n?1)a

n?2

?(n?1)a

n

]xn?0?

于是 a

??1a

n?1

?a

??1a

?????a

?(?1)k?1

a?a ?(?1)k

a?????

2 2 0 3 31

2k?1

(2k?1)!!1 2k (2k)!!0

?? (?1)ka

(?1)ka

所以 y?a

0

?ax?

1

[

k?1

(2k

0x2k???1

)!! (2k?1)!!

x2k?1]

?a?? 1(?x2)k?a??

(?1)k?1

x2k?1

0 k! 2 1 (2k?1)!!

k?0 k?1

e?a

e

?a

2

?a??

(?1)k?1

x2k?1?

0

即原方程的通解为

1

k?1

(2k?1)!!

2y?Ce?x2

2

1

?C??

2

k?1

(?1)k?1(2k?1)!!

x2k?1?

(3)xy???(x?m)y??my?0(m为自然数)?

解设方程的解为y???a

n

n?0

xn?代入方程得

x??

n?2

n(n?1)a

n

xn?2?(x?m)??na

n

n?1

xn?1?m??a

n

n?0

xn?0?

即 m(a

0

?a)???

1

[(n?1)(n?m)a

n?1

?(n?m)a

n

]xn?0?

n?1

0 1 ?1可见 (a?a)m?0?(n?m)[(n?1)a ?a]?0(n?m

0 1 ?1

n n

a于是 a?a?a???m?1 (n?m?2)?a

a

?1a

(n?m)?

0 1n n(n?1)???(m?

0 1

n n!1

所以 y?a

??m

a0xn?a

a

xm?1? ??

a

m?1 xn

0n?1n!

0

m?1

n?m?2

n(n?1)???(m?2)

?a?mxn?a

xn?1?(m?1)!a

?? xn

0n?0n!

0

m?1

m?1

n?m?2n!

?a?mxn?(m?1)!a

?? xn

0 n!

m?1 n!

n?0 n?m?1

?a?mxn?(m?1)!a

(ex??mxn)

0n?0n!

0

n?0n!

m?1

?(m?1)!a

ex?[a

?(m?1)!a

]?m

xn?

m?1 0

即原方程的通解为

m?1

n?0n!

n!y?Cex?