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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' V$ Y' G. L. q. Sd [(a+bx)c ] /dx = -kc +s
7 R/ e8 e6 N2 G& A) v* a% F: N- Wwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

* q: i# U, b* b3 P) M' I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
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(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
/ @6 X5 `$ t$ \3 k1 H- H  |(a+bx+kx)c=sx
" _0 S% ^1 C! z/ V7 x5 Xc=sx/(a+bx+kx)

十年移民路_

Solution:
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- j6 B! u& H% c- T& ?" vFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

% f4 r$ Q* s% U- U! Z2 r$ z0 J* @5 \+ @bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

2 K( k& n% q0 r1 m' d3 D  m) j
1 m" a, e+ n6 V  ]0 r/ mintroduce 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:
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{(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)
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so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
' W: ^& a% R, Z8 O. i% U& h1 dby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
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! E& y0 G7 P+ f; @1 Y7 d" {8 }2 h% `finally:
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! T5 d" ?; k; {. V5 ~3 N1 NC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)