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）
( I8 E1 T0 _% `: y
d [(a+bx)c ] /dx = -kc +s
. S9 d- P3 W5 y# V* n5 N; \4 bwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

  U5 d+ J" b! s2 |6 Y' K6 Q多谢了！
5 |0 w$ h: \4 Y2 f* h" r+ Q
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
1 A7 N, k8 W  L1 Q& n& U
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
+ ~$ U. q8 K2 d* X% Z( i- C0 o7 A
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
  p* y- v; A6 P: Hso:
" b$ J' H& T6 s7 q5 Y0 s
6 D- c+ Q; f- L. ]3 U: c1 A8 tbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
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(a+bx) dC(x)/dx  = -(k+b)C(x) +s
6 ?! T' B2 ?7 ~$ R5 n

+ M, S; J4 V+ g+ \  p5 A3 [& D1 gintroduce 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
2 W/ Q6 }$ a5 B, stherefore:

2 I- y* M6 l' G4 |# [3 O: r{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
& ?% X) s, A4 Y2 H: Y5 `
& C' ^1 U3 o4 H9 w" _- 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)

' D; ~7 j. q- R; Xthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
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2 u/ m; G$ a2 ^& J& t# ]9 y-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

4 ~2 m$ _7 h1 R( I5 g5 N& SC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)