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）
2 j4 w* b2 u3 K1 @; \% F- z& @
d [(a+bx)c ] /dx = -kc +s
3 H8 s+ L1 O2 s% s' kwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

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

sindowa

d [(a+bx)c ] /dx = -kc +s
+ a( R. L* @  w, P0 z2 j2 c/ ^; Q. ~
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
/ M" _' [3 ^" j# z(a+bx+kx)c=sx
7 I7 d+ Y1 M# l4 L" w8 N' {' Tc=sx/(a+bx+kx)

十年移民路_

Solution:
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From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
* Q6 u* Z# V" ?1 o: i* K/ F7 }so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
7 C% b2 ~& T2 {/ h3 G- ?! m
+ _- y9 h5 S$ T( Q0 @8 z(a+bx) dC(x)/dx  = -(k+b)C(x) +s


6 f9 r; {, d7 Aintroduce 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:

' N& _) N9 F) O: _# X{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
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dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
8 l( \4 d7 F' ?2 O$ C
1 }9 w0 U( ?$ T, `: sso that:   ln Y(x) =( K/b) ln(a+bx)
  z+ S! b8 M& j- p# O
this means:  Y(x) = (a+bx)^(K/b)
% ?# E$ P$ O* t8 H( `" O1 Pby using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)
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finally:
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9 \5 B: a7 R/ N. F' Z/ v, UC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)