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）

d [(a+bx)c ] /dx = -kc +s
& r& P8 w# `2 _3 w1 S8 q4 nwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
+ D) g3 o7 q; B3 [& h/ ?) {6 [
/ V7 j9 A, R1 ]: ~9 V/ _多谢了！

8 ~4 l. y, v% A: L! ]+ O[ 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
: m( |1 t8 n  P; W  ?# }) Jac+bxc=-kxc+sx
(a+bx+kx)c=sx
7 W( M" [9 ]. i+ x3 G( {  tc=sx/(a+bx+kx)

十年移民路_

Solution:
, u% @/ t( C# a0 S" p; F! @4 ]
+ ?- Y1 O. o3 s# G* J& S. NFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
) @4 z) b* D  f2 m: b5 oso:
7 u7 B* G. r0 n1 b! E
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
: N  w: a+ |8 Ki.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
* I8 {* E8 g/ N% c! M4 y) b# j, h# M+ X9 t
: k4 ~7 I# O+ [" ?9 o0 \  N& K
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
5 N' w7 l8 a0 B: X9 w1 {+ X( v8 }0 Etherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
2 ?4 V) O& v$ p( A. s' V
. l; Z, i1 D2 N) h7 `from here, we can get:

  X: U* i; [' g+ a; \/ {- FdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

- Y( g! n3 q4 M* ^' Y. n2 S3 `4 L5 [7 c0 Xso that:   ln Y(x) =( K/b) ln(a+bx)
1 z$ n# J5 @' O
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

/ s1 \+ V  \5 l5 [( V1 Z' T-(k+b)C(x)+s = (a+bx)^(k/b+1)
* |6 }# S$ D: q
7 Z2 l5 S4 r* Wfinally:

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