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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

; Z- y. d. D5 u: H  q- R0 h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

; A8 l, m1 z+ C, t: ^( R(a+bx)c  = (-kc +s)x
! F( F: U6 s' Wac+bxc=-kxc+sx
(a+bx+kx)c=sx
7 o0 q3 C3 Z" Ic=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
9 \6 o1 ~  K; j! s  E; Oso:
! q* K) Z/ G* f; p: S. X4 [
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
% M( O+ X: O+ @7 R1 Wi.e.
% X/ H! W* a/ N  K7 u% x" n6 ?
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
" {1 n' B8 t, ]1 r* _, D" L

1 r1 d7 K& L, p4 p! ~* rintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
2 B5 ?% y# t5 L/ Dwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
  x4 Z" e0 M5 S( [- u. @therefore:

2 r7 n7 R; J2 W- h% S' s# @# |{(a+bx)/K} dY(x)/dx=Y(x)
, [" m( l' f  C; e
" K7 s3 b2 {1 s. y( a# m$ Z5 ?$ |: V1 x* Tfrom 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)

so that:   ln Y(x) =( K/b) ln(a+bx)

- U- l$ {& M2 n: G6 K  \this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
+ r0 h. ?3 x. a/ f# g
9 f$ W, @8 J* T% P-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
2 v7 D6 S1 Q1 o$ k9 y" @$ e
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)