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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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?
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$ R7 ]% ?4 C/ d* N多谢了！
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% I5 |; F5 ]0 \+ \[ 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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9 j% a: _+ h4 x  Z! _(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
& W  W1 I1 j% H4 R* w(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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% [4 Z  U& L1 J- hFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
8 z3 ^& p! O7 ]' Si.e.

3 f1 h4 l  b  V  ~; Z(a+bx) dC(x)/dx  = -(k+b)C(x) +s

8 J  c# G2 r; U- s
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
; f# {0 [& J( p. \2 `1 Fwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

" r2 Q8 P( E2 Q{(a+bx)/K} dY(x)/dx=Y(x)
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- }5 @  g" d. u7 E6 h  P: _% _3 l  B" Zfrom 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)

  D- U0 E8 g+ V- {so that:   ln Y(x) =( K/b) ln(a+bx)
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this means:  Y(x) = (a+bx)^(K/b)
. H  W, B/ z$ |% K! gby using early transform, we can have:
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. _& b4 b  g0 H+ S-(k+b)C(x)+s = (a+bx)^(k/b+1)
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1 O6 h4 m$ ?! `3 c& C* n$ sfinally:
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4 n. N: y$ g' E+ mC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)