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 q3 A$ f5 ]4 h. \
d [(a+bx)c ] /dx = -kc +s
2 ^5 p" T% O7 y! L6 ~where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
1 s1 s% S9 V# y) j+ I8 s+ f
多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
1 R. I2 @: i5 rc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

! k# m+ a0 j* j! a) @$ QbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
- j6 I: z" h5 ?) ~! i) F
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
7 u# A" X: o! X! T9 D

/ A3 f7 Z- ?7 d# Pintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
: @( K! G' \3 ~0 H2 Twhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
* _2 c& H# O0 M
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)
8 v; K; t; h2 d2 e: B0 d
6 j) ^) A2 @9 mso that:   ln Y(x) =( K/b) ln(a+bx)

0 ^, T  K0 L3 E  s! @$ zthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

9 p* m: w, _7 S2 U' b, p: S-(k+b)C(x)+s = (a+bx)^(k/b+1)
; R6 B1 ~! I7 x$ e1 D, U; {
finally:
6 k* j3 x: q% K, }; f) P; T
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)