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）

, b( y0 C2 G9 W9 [d [(a+bx)c ] /dx = -kc +s
  z" r5 {* B& V9 Y% pwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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3 d8 O2 d. R# R+ @" \# T( f多谢了！
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9 T, Z: N5 J( Z  ^- r7 q[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
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8 N0 S5 r: S5 D0 o3 [  R: u; j(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
% s/ D9 d/ V! W' s. X( m7 Z4 Rc=sx/(a+bx+kx)

十年移民路_

Solution:
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1 e# U8 [( t- c4 nFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

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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
  b4 I. l: p% U; R1 L. ]2 [therefore:
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{(a+bx)/K} dY(x)/dx=Y(x)
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+ S6 m+ C" l5 Rfrom 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)
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9 s% Z% q1 ]2 r! G. lso that:   ln Y(x) =( K/b) ln(a+bx)
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this means:  Y(x) = (a+bx)^(K/b)
% s: @0 `2 V0 w1 U2 d5 tby using early transform, we can have:
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* A* y4 N7 p! Z. }0 _-(k+b)C(x)+s = (a+bx)^(k/b+1)

. `  L# `2 r, `( ~0 C7 @finally:
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8 ^( J& f0 @$ k2 x( X- A* lC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)