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）
' h' U$ S8 \8 b
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?

多谢了！
  c6 F& t2 I! q; J3 w' o) J6 {, V9 f
2 Q! r  {2 J) A7 e3 m7 I2 S: o! K[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

+ p/ R& H9 ?+ W) C(a+bx)c  = (-kc +s)x
4 n# H8 S6 c  r5 ?; C: k1 Cac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

. l! g- W. }2 a+ _, s5 o( ^From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

3 X5 [8 B9 w2 N8 A1 a$ x+ D! d) SbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

4 B0 w8 K# A4 y% b& M(a+bx) dC(x)/dx  = -(k+b)C(x) +s
) a& `4 F5 \4 s6 K+ E

, x1 B3 x6 n6 S, z$ T$ B4 _2 e( k; Fintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
, Y3 Y  Y4 }) M0 Gwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
- u$ ?8 p5 F3 `/ g
: {5 v* v) `# R+ W{(a+bx)/K} dY(x)/dx=Y(x)
; F0 M1 ^# X* J
( G3 O! L2 r  afrom here, we can get:

) z0 z) A: {0 v6 S: z# odY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

6 W3 Q/ Y! f: jso that:   ln Y(x) =( K/b) ln(a+bx)
; _6 ~! I( _1 W8 X/ r
6 N2 r# c( J( V6 zthis means:  Y(x) = (a+bx)^(K/b)
4 E4 Z; G+ a# vby using early transform, we can have:
+ z$ D, M$ W- n: S: J
-(k+b)C(x)+s = (a+bx)^(k/b+1)
" \" T) S! F( c) V
7 {- v4 h9 `! ~2 X7 d; dfinally:

, L' r7 J- l: w" e% cC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)