请教大家一道微分题，多谢！(原题贴错了，现纠正）

醉酒当歌

请教大家一道微分题，多谢！(原题贴错了，现纠正）

- b% {/ h. m0 xd [(a+bx)c ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
( d7 t3 K2 w7 t+ }( n- K
: c1 t, Q" Q- B多谢了！
; Y2 [/ f) }4 l# M( m
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
4 S/ ~) z7 q- y. V2 M& W, M: e) f: t
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
" k+ J& a& _) X3 N) z# p
! C( d# u/ X5 t/ b* j7 yFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
' i  L( c- y0 X( c+ b& ]4 Y; w. P
$ o) L: Y5 A9 y& K+ r1 `- hbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 F! d1 J# N) m4 O/ xi.e.

7 u  L* b3 M( U$ z/ {! M(a+bx) dC(x)/dx  = -(k+b)C(x) +s

! W& r. k! a9 G( C* s- L+ n
- w. ?" s' i; c5 m3 r: `introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
& s) S" h, o+ M% P& {6 Awhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
) W  x6 O. f# A! K1 J$ J' q
{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

" s0 I3 W8 p$ ]: e9 E! UdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

. d% Y" E% q" x1 Bso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
" Y1 I) E4 O7 i2 k7 rby using early transform, we can have:
/ f+ @" @; v9 O$ ?6 q7 I: r
( R4 s* }+ m) K  q6 x7 Z-(k+b)C(x)+s = (a+bx)^(k/b+1)
) u+ J: W6 o% Z- U# p% P' Z: i" o
+ F8 F/ J/ i( l7 P% }finally:
1 g2 I; x' c+ @1 w# j4 o
) n- @- o7 {: J5 Z+ TC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)