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）
& p' ]+ f' j2 ^+ h
8 [8 {3 h( K) y4 Q- e/ ]d [(a+bx)c ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

# C7 u3 [8 N. i0 d) [/ H多谢了！
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4 d0 d% ]7 c  a0 v# h& i  B  d[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
- d- O  `1 i1 H8 X( hac+bxc=-kxc+sx
+ R0 Q$ C0 @  ~; k/ Y2 J. p2 z(a+bx+kx)c=sx
! P' F) d" X7 d; vc=sx/(a+bx+kx)

十年移民路_

Solution:
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2 I4 j/ D4 g% g, RFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

  p; u9 k0 I) i: A7 G2 u0 obC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
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' {6 I, p7 P7 _(a+bx) dC(x)/dx  = -(k+b)C(x) +s

& v, t7 M! b8 [! j9 H, k& v
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
+ O1 _) C& Z# T. n2 _9 X+ ytherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

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)

$ p% N/ E' C# s- p' \7 g8 Z: M1 b0 qso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

, g9 t- i# p, B8 H. y2 e% V( U-(k+b)C(x)+s = (a+bx)^(k/b+1)
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finally:
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% H$ G6 M. s) \7 D) p, U5 |C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)