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）

, L$ g. g3 J+ _d [(a+bx)c ] /dx = -kc +s
. c2 B8 w/ p; f) s4 m5 owhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
; X; G0 T; u. [1 w* Z
( ^  U6 G! g" ^4 L4 E4 A' z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
6 N* d3 Q. V6 C  o. t8 j( eac+bxc=-kxc+sx
; \; [7 Y% A" |) V(a+bx+kx)c=sx
* @5 E, i* S4 b# {3 Qc=sx/(a+bx+kx)

十年移民路_

Solution:

: A( n& \- H7 sFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
9 l4 X3 u! J5 C( Uso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
  S# Y7 W# a6 v2 x( R7 N/ F
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

" L) `5 [0 w5 T6 S5 D2 a
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
6 l$ C/ Z5 r2 Q( ?! B3 a4 ~+ e5 @  hwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

- f" |5 ?0 f$ R2 Q0 F: G5 ^/ [1 f" W{(a+bx)/K} dY(x)/dx=Y(x)
( y7 m6 c2 I/ ^$ C. _( ?5 c
" a: E' ^  }: X. Wfrom here, we can get:

7 ?4 I5 g. t( x( V6 y0 N- U& KdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

so that:   ln Y(x) =( K/b) ln(a+bx)
6 D! ^) Z! |' A3 C. x- p
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
) m% f) W1 S  }& s2 |, B# a
" b! ?, e9 ^- Y% n, V-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
" G- `$ Q" |4 s: L  T% ^0 e
) }% B% R9 X/ Q+ k8 LC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)