北科大数理方程 第6章习题答案(2)

t 0

f ( )d ) t

1 s

L ( f ( t ))

可得t

h

12

1

h sx

s s h

e

a

) h f ( )d 02

2h

e0

h

x 2a

e

2

d d

d ) d e h ( x e 0 2a

t

2

2

[ e

h

x 2a 1 2a 1 2a t t

e

2

d ] |

0t -( hx2

t

2

t

e0

-h

x

2 a

e

4a

d(

x 2a

)]

2

=

2

( e

ht

e

2

d

e0

4a

2

)

d 此 令 = 2

x 2a

=

( e

ht

e

2

d

x 2a t

-(

hx2

2 2

e

4a

)

d )