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）
4 S, W$ z" u# O! c" {( e+ m
d [(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' d( ~* S  U) p4 ?: b多谢了！
  u( J; u5 R( k' \3 C' E( ^; j5 p& j+ ?
0 u. q! L4 }' c& E3 s3 ~[ 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
: y4 i# o  R5 W* O(a+bx+kx)c=sx
& G# O" w' E3 m0 oc=sx/(a+bx+kx)

十年移民路_

Solution:
2 s5 {8 z. V7 K" B* ]/ {4 o2 c- T
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
8 y# e( `$ F+ M1 a2 u) c0 P
& A( `8 H& X3 m- s6 L) ?bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
1 v% F  ]9 }$ B/ A& S3 V6 a# K2 M
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


. y8 B! Q7 I" P* e# K( l: H4 Ointroduce 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:

3 F# a+ C1 g3 U, O{(a+bx)/K} dY(x)/dx=Y(x)
7 o, D4 a8 T3 G. E" X% ~5 X
8 A/ v) s* g. F( F% l; i9 }from here, we can get:

3 W/ x% O0 ~: `, ~dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
/ F3 s0 Y9 |5 P7 Q- A/ `
6 i- Q+ {/ X4 R% y6 t8 rso that:   ln Y(x) =( K/b) ln(a+bx)

6 Z: g  Y8 y2 [5 x6 Z  }. Bthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

1 l% U1 y! D4 ~* R. _. l( m- r-(k+b)C(x)+s = (a+bx)^(k/b+1)
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finally:
3 n/ w. b* K* P  @
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)