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
8 I$ X3 ^% O" K0 ]8 f0 [where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

* m& s& ~% S2 b6 M8 [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
' G9 s; ?5 t$ ?: b4 W$ c6 U  |+ ]
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
* b" }# L: o- G9 ^c=sx/(a+bx+kx)

十年移民路_

Solution:
; }1 q- w& b$ f" n4 V
! h$ c& Y8 n+ e5 \# I1 a$ JFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
! b6 r6 O8 w. I* B+ Xso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

: }1 u5 `- ^3 \  }* N(a+bx) dC(x)/dx  = -(k+b)C(x) +s


, B' L1 m" [, t( w- k$ T" M+ ~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
therefore:
! C4 {" c" {. k4 }# M& M
0 X8 ^  D0 |  v7 @{(a+bx)/K} dY(x)/dx=Y(x)

" j6 O2 d5 B. m* d( D0 jfrom here, we can get:
+ d, k: j' f/ X, ]4 V* V
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

% W/ d5 g. Q4 Xso that:   ln Y(x) =( K/b) ln(a+bx)
  d+ a- ?8 Q: A  Q
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
& d+ t- R! c& g. `( }0 t2 R% L
% ^' {) S* t' X7 M1 W-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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