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）

- x6 i& ?+ y: Y$ Ud [(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?
; N& V3 J- h3 h: B6 v1 L
多谢了！
; P* b* n  G' Z: q$ ?/ e6 X# S# K$ \
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
4 E* s& U7 o4 t* h) n. u
(a+bx)c  = (-kc +s)x
. R  E: A" l6 j6 H* cac+bxc=-kxc+sx
(a+bx+kx)c=sx
, ~  a( ?2 S; `  Dc=sx/(a+bx+kx)

十年移民路_

Solution:
( j' ^) ~0 g& [% B% [
# p+ X" ^. v, H8 MFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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6 m% v. ?9 `* K' \- F- qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


+ _+ M) d8 n* s9 c# ~9 s3 y& Nintroduce 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
4 ^- }; a$ G4 ^+ Y% |therefore:
6 O. d+ ]" s; p# z) J$ |+ U4 o
{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

5 y; X& S2 \' M/ hdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
" B  J; ^5 `7 `9 k6 p( \
8 E* z+ T# E( \so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
; Y- _, ^0 w/ _+ l0 }& ?by using early transform, we can have:

/ {+ ^, E4 \4 K/ C- d' M9 @-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)