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）

5 y1 S5 l! f6 Jd [(a+bx)c ] /dx = -kc +s
) \' N" k" `' K2 ~where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
) d8 A5 e6 \8 \4 w
多谢了！

% d0 Y) v8 W, v. R[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

7 \$ e' I7 _; p/ C0 z1 _' s: @(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
5 v" f1 h, W# Q* L- `4 K+ g5 o! ^c=sx/(a+bx+kx)

十年移民路_

Solution:
* s6 B7 `5 {8 G4 z
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
$ }( Q0 v- u( ~7 O$ q( pso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

: u: M+ x8 n) f$ N0 G. i(a+bx) dC(x)/dx  = -(k+b)C(x) +s

. s' \  |" _' c+ p9 _9 r
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
$ C5 V/ H: G" ~3 `8 U/ Btherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
2 o; d% W+ h; D1 v7 F" d/ `3 o
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
) r* g$ A& t. e" d8 P
5 M, G3 [  W7 R% Q4 vso that:   ln Y(x) =( K/b) ln(a+bx)
( L8 t7 ~" ?& d' T0 k
0 d* q5 P) ?, B% j) d* `this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

9 j* W1 g! b4 E6 U2 J. w8 A-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

6 T; s! ^# _# G# o9 C, W* A) SC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)