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）
  E8 f* V6 G' ]$ U# ~5 s, {
" w# S/ `4 h* U" E3 Y6 C; ad [(a+bx)c ] /dx = -kc +s
; O+ [; e7 N- j% ~, L  X, Qwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

# T  D5 Y4 z' g6 ~) E多谢了！
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$ X: Z1 ^* j$ _+ h# W7 `' w3 V* R( y[ 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
3 B" \* W; a  h) k  ~8 f8 S2 v0 oac+bxc=-kxc+sx
(a+bx+kx)c=sx
# L9 ]( t0 p- N: s! J# v! \c=sx/(a+bx+kx)

十年移民路_

Solution:
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From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
7 g3 ]4 A" Q. J) P; a1 hso:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
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  c7 \" }! ^/ X9 b! r, @+ }(a+bx) dC(x)/dx  = -(k+b)C(x) +s
- D6 X2 t  y. O

/ e; |% R, P- ?+ w1 Pintroduce 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
5 }6 _  W! ?4 l' Y) A4 Utherefore:

* S- E9 d! K/ S& [7 N9 g{(a+bx)/K} dY(x)/dx=Y(x)

" J+ _9 V) M! q5 X& b+ H5 G$ K9 @3 rfrom here, we can get:
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4 N2 n, j: H( g% i4 T$ ?dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
1 ]2 M3 `0 \' m* g% [- |0 C* t
' C4 T2 b0 y; D3 p7 g+ Tso that:   ln Y(x) =( K/b) ln(a+bx)

& U1 \, A6 X, nthis means:  Y(x) = (a+bx)^(K/b)
0 f, j/ @' F7 H. N. w/ p- Rby using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)
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finally:

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