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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( w8 v/ f5 s/ F9 M: L1 Z$ |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

- u, F- p# u4 z/ ?/ F) p+ X- V(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
) ?# U- @; R8 [, m# l(a+bx+kx)c=sx
9 S3 s, R' h$ |- rc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
: G6 M: j$ N% _6 [1 iso:
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  w: v' `) a' X$ B" H6 G! n2 ubC(x) + (a+bx) dC(x)/dx = -kC(x) +s
  c8 J. C& P/ D) @$ ri.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


3 e, T! ^: J( ~; @introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
0 D2 |4 i6 ]8 M- z0 t* ?( e4 `which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
% `! D# _% }9 K3 W& |9 \, p7 btherefore:
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{(a+bx)/K} dY(x)/dx=Y(x)

0 F$ w$ F1 p' h4 T; E4 {! M3 ?" Ofrom here, we can get:

, ]* W6 e! B! r$ n$ a* AdY(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)
6 ~& X  K$ I" L+ f2 d
2 I- O  @& i8 [0 `- ]this means:  Y(x) = (a+bx)^(K/b)
" M8 O7 d5 y  W2 f% q& N' cby using early transform, we can have:

; J3 `& J# Q, C3 l  Q6 E-(k+b)C(x)+s = (a+bx)^(k/b+1)
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finally:
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7 u- \6 J. i* RC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)