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）

; ?6 B6 D$ m. G. zd [(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?

1 w/ H' e% o) L) v1 T8 t多谢了！
8 w& W+ e5 y# {; ~
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
% y/ a& K3 F6 \  f) y+ O
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
* R1 [" {4 T) Y' e0 W" ^i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
; g; S0 a, W7 b/ ?2 B4 ?

" O/ @" v$ t1 \0 Rintroduce 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
( O1 g4 C: k; H' Y  l$ }therefore:
2 [: ^( ~: d4 a
{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
7 ~& N; Y6 z7 X+ j8 a6 Z1 X
) y0 w: M0 x5 _! vso that:   ln Y(x) =( K/b) ln(a+bx)
1 u" u5 T* O' O; y
8 \  E3 O# c7 D. |) A+ |! Sthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
, |, k5 Q; ^) S5 ~9 H
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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