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）

% L3 [- V$ G6 x3 ?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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$ v2 l; {8 y4 n# v& D/ [: B% L多谢了！
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% L( D# s- @6 D[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
0 h: R, L" T. M  R
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
/ ~) C' x* t. V; f0 ^: d4 W4 O(a+bx+kx)c=sx
' q0 x" ?- D+ O! G0 }) Jc=sx/(a+bx+kx)

十年移民路_

Solution:
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$ i- g2 o3 c5 R- ?9 X! wFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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5 ^4 w, k# J$ z  [) b' VbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
  A! O3 @( L; i2 I' B$ Z5 a8 |6 Y
6 d  a: b+ Q) c; F9 B6 H
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
$ o# v+ N& S' A4 z. gwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

3 Z1 S$ u! \6 O1 c{(a+bx)/K} dY(x)/dx=Y(x)

7 ?) ]( K: `) n7 R9 O0 zfrom here, we can get:

" |/ B4 h- j- c0 q- WdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

7 h  x3 D( n- @) Q5 {" L! y, Nso that:   ln Y(x) =( K/b) ln(a+bx)
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this means:  Y(x) = (a+bx)^(K/b)
! O  l8 f7 p9 }, J. qby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
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- O4 n& G7 g5 ]9 M9 Rfinally:
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. ?, O5 I1 t% hC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)