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）
. E8 y( p8 Q4 k7 u! k0 P
  C6 S$ A, p$ ^" i8 ?" `d [(a+bx)c ] /dx = -kc +s
4 k: [' [& ~! C' Vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 d- H# V8 _# A9 ^0 U7 N
多谢了！
' R2 {8 @* ?+ t: \
. a# T/ l" S+ `0 ]% D1 ^1 v; g[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

7 S5 k. B* ?: h# I(a+bx)c  = (-kc +s)x
. T8 B( U$ H3 Z% d- J( V7 A: qac+bxc=-kxc+sx
3 D2 I+ S/ F: v  b(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
0 b; W, b8 ?$ c, f: {1 W, L5 Y
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
: \4 p% {- o( W- G8 ?+ m
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce 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)
/ X2 H- x- w8 c
8 A! N" x( l. s3 F' Lfrom here, we can get:

1 B* w+ j" y; M, o  IdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
' ?  t) U3 M& a; m" e
! O' I4 j' v" w( I, W5 |5 ]so that:   ln Y(x) =( K/b) ln(a+bx)

) Y0 N; c% d. Q. Vthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
8 y3 k+ M8 w5 a8 N
-(k+b)C(x)+s = (a+bx)^(k/b+1)
: l; k/ ~+ `3 `% N! R  J8 o+ u
0 R# ^: w) a% o7 f+ c- \finally:
! r0 |8 A3 W4 Z+ f6 `0 Y% J
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)