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）
; x% ]* O: o" Q/ v
1 u9 s# y. R& k4 y8 Ed [(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?
9 \7 c3 s4 s: l2 y" l- O4 A
+ Y9 c2 K  Q+ K9 ?' h多谢了！

* @' i  c, d& w+ h[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
0 U* i1 t' i  Q0 z' i& [
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
: c/ k* p% m4 S' U/ ], u: e(a+bx+kx)c=sx
' R8 i* O$ _* n" x; Mc=sx/(a+bx+kx)

十年移民路_

Solution:
% L, o. ^# {8 g* p
9 G; \! h, d! ~1 QFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
, V4 A/ C3 K$ dso:

( c0 g( O4 ?8 ybC(x) + (a+bx) dC(x)/dx = -kC(x) +s
' Y( L, d; T% ^! W; Wi.e.
- `* y( `$ `& k7 {/ c
% q& h' C$ g' _* Z: K; i  O" O(a+bx) dC(x)/dx  = -(k+b)C(x) +s
1 p- r$ G- P! `5 w  v$ W' E1 E

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
8 G) c: y) |5 Z% s1 h- O6 atherefore:

# o+ x8 f4 H/ D& w7 m0 c9 O" u{(a+bx)/K} dY(x)/dx=Y(x)
. `- H2 M8 b+ x0 ?& q
( _  O  c- Y& Y1 lfrom 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)
) N  i/ G0 Z; w
0 C6 l6 L7 O3 m: a, lso that:   ln Y(x) =( K/b) ln(a+bx)
) z9 n  A8 ?$ Z, ?' |/ r
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

3 f; W' e  g2 Sfinally:

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