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 `8 P, d2 x5 v+ |! Y4 J
d [(a+bx)c ] /dx = -kc +s
  R  [; d/ i; R! ^" B5 gwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
1 b/ [8 w( Y/ ]4 P5 X- c. u2 I
+ b/ r7 P* |8 Z+ M$ w$ y- B多谢了！

$ k8 J/ r4 w& r  ~( g5 e+ h% |) e[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

: {. b! `* u& O6 i' i(a+bx)c  = (-kc +s)x
! V, q# H  A' Mac+bxc=-kxc+sx
" _& `* o* \9 N6 w(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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7 Z6 A0 |/ A. g2 a5 t" EbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
8 r3 J8 b% S& v  }: o/ E

7 `4 A! D1 V0 Z/ A5 m  Dintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
4 G  l4 b5 K% x$ ^which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
. D4 a- u) D5 u& I
{(a+bx)/K} dY(x)/dx=Y(x)

' y* q+ G1 I" q) Qfrom 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)

so that:   ln Y(x) =( K/b) ln(a+bx)
# {% y+ N4 f# Y& ?$ `- S
" e4 a, Q, I0 \% V( o8 A9 D3 `, Sthis means:  Y(x) = (a+bx)^(K/b)
; ?1 x: t( W9 Y; R& @+ h6 \7 hby using early transform, we can have:

4 E( r' j5 \0 e% X+ m. l-(k+b)C(x)+s = (a+bx)^(k/b+1)
! @0 h6 t9 f4 k( `9 z$ x" x
finally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)