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）
. a6 d: H  e/ g4 r$ S6 R  ~' E2 K% f7 O6 |
( z8 O  z# [9 d( z6 w# Hd [(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?
2 O6 {9 c" E6 u. n4 `2 w
多谢了！
7 `, K) B1 ~! G7 g% F! ?( B1 @) m
# a2 Z+ A. c' g5 N' X[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
5 d$ h2 S) p, J
7 I, j- V1 ^( U# q(a+bx)c  = (-kc +s)x
0 s1 k' r" I6 e9 sac+bxc=-kxc+sx
4 M) k  [# Q+ ~* N( }" W+ g4 M(a+bx+kx)c=sx
4 {, \  U" u4 C, Q7 n1 D1 ic=sx/(a+bx+kx)

十年移民路_

Solution:

4 [( f2 ^+ _  R% MFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
8 ~  O6 h1 }8 P8 S. x7 d( s5 Sso:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
& U( Z3 R" X+ `
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
4 Q. ~6 ]) E; W+ y

  S( d& c# r0 _# q3 q& [9 L3 uintroduce 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:

{(a+bx)/K} dY(x)/dx=Y(x)

! Q; X# _8 a2 B2 dfrom here, we can get:

& D  i+ |0 h: E: W) n% j+ kdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
( G, V/ ?' ~/ P! B
/ u- {7 b/ Q3 q* S7 L. ^3 |so that:   ln Y(x) =( K/b) ln(a+bx)

1 Q# g1 l/ Q5 Q) M  y8 Othis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
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* W. g, f6 R; R% w' e& e-(k+b)C(x)+s = (a+bx)^(k/b+1)
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% E8 k. Z) H$ d, ~finally:
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+ r* x" A- I9 J" M2 b7 l9 wC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)