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）
5 L8 [8 k; w. ~: ~
+ S& A3 I/ v( R- r* Q0 |1 kd [(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?

多谢了！

% ^# {' f6 N4 I! g# S' E' U[ 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
2 e5 g. y7 m  W, A; N(a+bx+kx)c=sx
) X* s" Y% i9 F. D  m2 zc=sx/(a+bx+kx)

十年移民路_

Solution:
; I% Y' p/ b3 v
* T# K4 n; p- y/ _! @9 b1 R9 mFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
6 l" y& `# A. W1 d! iso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
, g8 u$ l7 V2 i* M$ {
- d, X4 B( b, A- z1 [) n. b8 _(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
6 H$ Q8 o2 S' K: K: D3 U, _2 L3 ztherefore:
6 i3 j/ |# S1 y2 Y
$ [) I/ ?6 ?6 H$ J( ~8 u{(a+bx)/K} dY(x)/dx=Y(x)
4 Z2 E, O, d: Y) ^
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)

/ L8 X6 @9 }/ i9 Q( K7 s6 hso that:   ln Y(x) =( K/b) ln(a+bx)

2 ^7 C; k" m3 ~5 I' Xthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

4 |0 ~* y& g0 K' J-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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