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）
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3 O$ {3 X" \% Ud [(a+bx)c ] /dx = -kc +s
+ p, @# i9 A0 Q& S* twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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' s% z9 u* G; U1 \- T1 T7 H5 J多谢了！
  W, ]. E5 S! l
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

* t0 k4 g! `( C; \+ s) I(a+bx)c  = (-kc +s)x
$ X  }! G( Y& i1 Cac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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, w) |+ w. H. E5 CFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

5 k7 J% g& H0 Y0 B( YbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
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(a+bx) dC(x)/dx  = -(k+b)C(x) +s
0 L6 i9 l4 _1 s+ N2 x* `0 P

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
. \7 c- E8 U7 j, f& r# ztherefore:

+ U& [; {# f0 D$ P3 H9 q8 n" ~{(a+bx)/K} dY(x)/dx=Y(x)

7 N  @* Y) f# v8 E5 r% P8 ^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)
# Z1 ~+ X% j# h* K8 Y# ^: F0 F
so that:   ln Y(x) =( K/b) ln(a+bx)
/ C% R: q0 m: P+ |* x
this means:  Y(x) = (a+bx)^(K/b)
1 b* Y3 A' H& p. l1 i- p# uby using early transform, we can have:
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7 p" h" G8 o" ]8 z$ N9 e-(k+b)C(x)+s = (a+bx)^(k/b+1)
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finally:
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8 |' ?" b2 I5 WC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)