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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" A3 Y4 L1 [5 {- o# kd [(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?

! M+ l  r+ n) ^$ U  X( [* T多谢了！
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1 X  Z, T0 h0 A  f[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

, f6 a( I4 Z/ O2 G" o' ^" g(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
; n' b0 r, k0 b* I- ~8 a: f5 S3 m(a+bx+kx)c=sx
: G" E8 j* t; i6 N: Tc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  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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9 R$ W; T5 j$ p0 cintroduce 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)

7 s5 ^* p% v& x- S1 p* A! jfrom 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)

so that:   ln Y(x) =( K/b) ln(a+bx)

; _) h7 a  O0 H! t& C. Uthis means:  Y(x) = (a+bx)^(K/b)
; y3 \& V% \8 x% n  `by using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)

; X8 i9 c. {" |8 ]' afinally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)