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）
; ]+ I9 ?' t2 T4 `1 ]3 y' \
% v4 [0 S, o- od [(a+bx)c ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

2 @4 y$ f# o9 }/ _+ D  [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

, p+ p- @7 P. ^/ M(a+bx)c  = (-kc +s)x
$ K: }3 K1 D! \6 zac+bxc=-kxc+sx
% @1 Z& [/ |; I! x. \(a+bx+kx)c=sx
, k% F+ j- l" J7 T3 r0 V9 tc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
+ x4 f. s  d8 o9 r$ r; v3 _so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
8 `8 l* S4 X" k# p( U; K! Xi.e.
3 x7 c3 {8 \* f9 u
2 S' ]/ a% ]4 Q$ D9 O9 H(a+bx) dC(x)/dx  = -(k+b)C(x) +s
: b7 {9 W* l0 _5 v

. F9 J& m7 Q( T) b0 v* o8 hintroduce 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
therefore:
' @" r; ~4 [3 p: J
{(a+bx)/K} dY(x)/dx=Y(x)
# x7 x/ L% a8 e: S) U& Z
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)
! @' k7 Y5 q! |$ ~+ Q) H
so that:   ln Y(x) =( K/b) ln(a+bx)

' R' W9 d' G  N- lthis means:  Y(x) = (a+bx)^(K/b)
8 f$ G: @6 C4 `2 @& v. Rby using early transform, we can have:
/ F; F' v  ~+ \0 D0 [
* M0 Z6 m) S1 Z) l-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

/ `3 Q' `9 v& m( Z, C6 k" Q  K! jC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)