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）
4 ?; Q2 B) z/ D, `6 I1 Y
8 e& }3 g$ N- D: s+ Z; Md [(a+bx)c ] /dx = -kc +s
9 M3 F* T: ?0 C% b. u6 {9 T! ]" w" ~where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
) z6 k1 {' V7 {/ y, w% g! ~8 ~& q
0 c: T$ P& x( ^多谢了！

2 W# ]2 j8 g8 x3 r$ `9 `[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
  u1 c, h# k( R
# o+ H7 J$ a( K  F(a+bx)c  = (-kc +s)x
8 C. N8 L+ ?( }3 R4 p, S; S! Mac+bxc=-kxc+sx
7 X% C% e" M4 e  s( Y5 e1 ?(a+bx+kx)c=sx
2 R! d* e' c3 k  X# Uc=sx/(a+bx+kx)

十年移民路_

Solution:
' b) ]4 J$ X2 j! f5 ]  J1 W' O
- A8 Y& [# C+ B& F' M/ M8 [From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
9 [1 R6 L2 E' h" T4 ]3 H0 H
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
% W, ?1 N3 \3 B/ B- O8 v* s
& r* C- C$ ^5 M6 K$ E$ j4 Q1 H(a+bx) dC(x)/dx  = -(k+b)C(x) +s
7 B  J/ h5 J8 h" r9 U! r4 G
# ^+ ^9 X: E% F/ |2 Q
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
4 v4 U  T$ C( Z, F& |' Bwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
% d' d0 x3 D5 F# r  P( o% Q# |therefore:
& n% P, a: x- _- `
6 Q* `4 v" Z1 I- ?; g{(a+bx)/K} dY(x)/dx=Y(x)

' ?4 T2 v/ {% afrom here, we can get:
; |1 f( I( F% n$ T
dY(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)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
- ~( p$ A; l. C* _$ S$ Z4 c
# v1 r1 @! G* @+ |+ P, Z# B7 B-(k+b)C(x)+s = (a+bx)^(k/b+1)

* L' e6 }( M- _' ffinally:

% E* g6 _  D; \: j3 GC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)