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）
7 o* R' X  }& z# c0 W
d [(a+bx)c ] /dx = -kc +s
) Y, O- N$ W/ I7 m5 swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

. E6 P. G6 z3 ~% |0 }; P多谢了！
- |( c6 t5 q# ^5 a$ i1 y
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

2 U6 _+ V% T' _9 H(a+bx)c  = (-kc +s)x
4 M5 Y2 q* k1 S, d' l/ m7 s& gac+bxc=-kxc+sx
(a+bx+kx)c=sx
0 j. X$ b7 u  {) P) V" r- E3 yc=sx/(a+bx+kx)

十年移民路_

Solution:
! h$ B) _+ }6 ]
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
, W7 w+ W# f) Y6 s& [$ z- cso:
3 e$ C# F' p' `
; ~; c/ y8 n; Y2 vbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
& |% k3 S" G* C% J
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

# P7 e' T9 [" t( K# g' T
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
0 j6 w2 }- j% ?/ |* t' U' N6 Z+ mwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
1 g+ E# {' m- y& `; w" ktherefore:
0 C2 J9 B/ P$ i6 X6 A  s2 y1 r
  P  m' h' S) N8 Z{(a+bx)/K} dY(x)/dx=Y(x)
7 M) M2 D. G- y# X# h
from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
3 H5 G( m; V, i- c% O; q8 G
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

! u% ^2 Y! `  c( U7 |* q( @-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 {0 w* a- r3 F. W* o$ [finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)