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）
# ^0 f+ P( H. Z+ P7 p
d [(a+bx)c ] /dx = -kc +s
9 }% ~( D9 F# Lwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

& T: i1 X" \$ n' m( u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
( [* K, K/ M% U7 s2 Lac+bxc=-kxc+sx
2 B* p  f9 u* L! o  V" Q% O(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
  n: V2 d7 F& `) _4 a( @0 `! _3 C
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
6 a* }* I$ [! d3 z, I! yso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
" T6 T* o/ J, G! L
- m/ m+ g$ _: H1 A/ E0 h  f2 l1 u(a+bx) dC(x)/dx  = -(k+b)C(x) +s
% m% i2 `' q1 N/ E" R) o* p

6 ?! m( H& H1 h' X) \introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
9 q) o1 q7 p0 W+ Hwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
2 y' z. x, K3 C$ R" n! k* D, Otherefore:
( Q) o0 l3 \9 M0 a- m6 t
{(a+bx)/K} dY(x)/dx=Y(x)

: {; k# J6 a) |6 R8 ~& w+ o& h& Rfrom here, we can get:
3 a$ l# |8 U8 [, p& N
3 c1 C0 O6 {' M$ [5 {0 ^/ SdY(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)
3 h3 }* s* _1 M! L
8 |# D; C! ?4 d+ D2 m, ]7 e) {7 kthis means:  Y(x) = (a+bx)^(K/b)
+ t. J% i$ F1 N# R' Hby using early transform, we can have:

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

finally:
* T$ q4 f* g2 K9 r# S, Q4 @
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)