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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

* V0 q, ]5 l/ [/ g  g8 |9 u0 m多谢了！
4 ~( C) Q# d0 |0 o$ o. E# E
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
5 Y7 {. \6 o* K0 m. ?2 r5 T# `
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
) s7 `- h2 T0 N' h/ w; f. e(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
  G, E2 T) `/ a+ b3 O8 ~
2 e  k* s$ @' \7 o+ E$ bFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
' \6 Y5 G* x0 ?1 Q$ T1 E4 W5 S6 Q
. B$ T( U3 H$ b9 ~" b% c* YbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
! O1 ?- _( M) I- v" D5 Bi.e.

0 V$ a+ r8 ~& m1 Q2 H(a+bx) dC(x)/dx  = -(k+b)C(x) +s
' W+ L6 D5 R+ z: N; Q; D
9 ^$ N* `2 B) X
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
7 D$ x$ L8 s7 d& c6 z! ^% owhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
8 {+ P9 m% r! b
3 v1 T( }% g) h% a) K/ ?from here, we can get:
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* y. p& I; b) A0 u' }9 E6 F9 a; I+ k2 vdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

3 X  j1 N! D' G! v; h# nso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
- l' @% }$ Q4 X8 f0 T& I8 I& [by using early transform, we can have:

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

finally:
$ q; X: z$ k5 |: y% k
& V. `. t+ Q- T# l2 I1 C1 ~C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)