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 d* ~+ J6 ]- Ed [(a+bx)c ] /dx = -kc +s
( Q. S3 T8 \) k9 u5 B' g, uwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

2 _1 B4 ]+ }1 _2 \+ N" h多谢了！
3 t% C- |/ v3 v9 y6 X4 Y
[ 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/ Z' d" R2 P7 G3 z4 r& Q' Lac+bxc=-kxc+sx
( u4 s+ _' q8 m( m(a+bx+kx)c=sx
8 ~% n# @% C+ C; l: f1 I  B5 ?c=sx/(a+bx+kx)

十年移民路_

Solution:
( E" v5 z+ W& ^# h
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
- w8 v% r0 [4 [3 V% h% d1 `so:
3 e4 D9 V2 G; T) k8 \* n
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
( Y" f. o& o: I9 y: o3 {6 `0 e) I
. }* i7 D( u! S$ u1 a(a+bx) dC(x)/dx  = -(k+b)C(x) +s
% z, R* w5 }. Q  r& C7 B

2 c! f; Q6 a4 p9 S. _+ Vintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
+ W# f2 A: n5 U, X5 t( c& Dwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

+ r3 \2 x4 B, V/ x) a. [3 ]{(a+bx)/K} dY(x)/dx=Y(x)
) L* W+ f* v7 G3 ~$ x& c( R) w
/ Y$ B- J+ D1 ]% X( J  dfrom here, we can get:

, f% R5 K5 u9 M* \dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
7 l7 L& h  U4 n$ r% d- _+ K1 [' r
+ \' |) R' A: w, j( gso that:   ln Y(x) =( K/b) ln(a+bx)
* O3 X) I) c4 m& C, l- ^5 y
  x# l* ^) w3 M! B4 L- Ithis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

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

6 J5 j/ F+ N  \! Y1 V% {0 j' hfinally:
. I# i" R1 z7 X- g  M
; E1 X* M% y6 Y: ^$ @C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)