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）
% J+ v/ M% x* `6 Y& F2 D8 S1 P+ y
6 s7 |3 y' m- _, i6 U$ \9 o4 A$ ]d [(a+bx)c ] /dx = -kc +s
( I2 w) S. Y6 g) V% lwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
8 u& c# h3 O: H4 ?8 k( J) D* j
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
* {% N; K# ]  K( Q(a+bx+kx)c=sx
& N# Y* A6 U' O- \7 @2 _, n/ s8 mc=sx/(a+bx+kx)

十年移民路_

Solution:
1 p& Z, i$ n7 T; l
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
: K( ~5 w9 m. p1 g0 ]- dso:

4 `2 Y4 f3 S0 J7 U8 S7 v1 nbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
. K6 K8 k  v- v1 l6 z9 X
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
% ^* |+ O( J5 V$ ?
) @* \% Q3 J  G' g4 Q6 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
5 X7 o, J/ D1 Z( Y4 Ztherefore:
' w- G4 Y( X+ \0 E) K9 [% Y
/ ^9 B' M' m/ K' i. F{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
4 p$ ]" w! H, G. X: C+ X
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
5 h1 |/ V& m/ O# ^- h, k- z4 V
so that:   ln Y(x) =( K/b) ln(a+bx)
0 |' w- `2 A. l: J: E4 e! Q
9 ]! D; v) G  G$ v4 Q3 H* F+ R$ Ithis means:  Y(x) = (a+bx)^(K/b)
( m/ T% U, e. [% |  j2 B' ~7 sby using early transform, we can have:
- W7 C1 j- Z1 a/ M: x: D
-(k+b)C(x)+s = (a+bx)^(k/b+1)
+ x3 M* y. P5 ^3 I  j+ K
+ y) h$ _7 |  @; R: m9 [) Jfinally:

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