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）
) `( N% G: T! C8 _  A" O3 D
d [(a+bx)c ] /dx = -kc +s
9 b3 \, }& v% |) H$ h7 x0 v  ^where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
- `  K, O, S1 {0 R1 q
' j4 ~- o, N, K. U1 F1 _多谢了！
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[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
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(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
$ e& m* s1 C7 H, y, m% F(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

, ^8 W$ v: p% C0 uFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

: U& f$ u- l/ VbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
; |4 k3 Z% o) _! l! D# b, R, C6 ui.e.
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$ y6 i  L9 w" }& L$ f6 a9 K1 c" ~, m(a+bx) dC(x)/dx  = -(k+b)C(x) +s

" C8 i# c0 t( b: B, L# E- r
3 r" Y8 t( j4 ]  Y3 B# h" Vintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
8 D+ B9 w! v/ R' i# s- H+ Rwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

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

from here, we can get:
2 p0 y2 f' j# }% j4 v  F
+ a6 N& e+ L2 tdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

/ v2 D: G2 @3 p5 sso that:   ln Y(x) =( K/b) ln(a+bx)
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% r7 I2 k% N* G  F/ x8 Bthis means:  Y(x) = (a+bx)^(K/b)
, ?1 `/ |6 {- h2 n6 ?by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
3 T5 J3 }. ]9 {: t2 f3 t7 ^8 B' Z5 J
2 v. L, D& x1 V* Lfinally:
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# G+ J; a1 O9 n9 l# tC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)