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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- Q" I# ?6 l2 \# pd [(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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4 E5 f1 [# u% s1 |! w$ S9 _1 W- Q多谢了！
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7 [& V8 I: n+ h[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

7 H( c0 V. J9 e(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
- j  K+ A. o- ~4 P1 D  W(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

$ S5 @( a4 X0 a' W; N, bFrom:  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

4 a( \2 m- O1 @* b) d
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
+ e9 n' @# U# `: e) W7 {therefore:
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{(a+bx)/K} dY(x)/dx=Y(x)
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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)
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# S$ T. G; t7 N) P# j8 p* c8 R+ s: }this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)

* b0 ^" m1 s8 X! _+ j) H) s+ i. wfinally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)