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）

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?
. U, R4 d. T7 I* A8 O3 N8 _" c
多谢了！
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) q0 }4 [: V) M, H3 s# m1 c; n[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
5 d) a- U1 A8 F* o6 v8 _" T2 U! r
2 x8 r7 }4 q1 m4 S, R9 g" t(a+bx)c  = (-kc +s)x
: F6 b7 [" W, p: z1 aac+bxc=-kxc+sx
(a+bx+kx)c=sx
; i: S* K7 X/ I; [# z9 Q" Mc=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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$ P8 P6 Y8 G2 @( K: _bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
* A" ?+ P1 f  ]! N- F* q, Bi.e.
& j! Y& g: V" ?5 T* |; l" a
1 n6 r, U# x0 a% Y3 P(a+bx) dC(x)/dx  = -(k+b)C(x) +s

$ R( K! f% S& v. Z
; l( z# ?  i6 _$ S/ L% `% xintroduce 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
4 l% q* X! a# H- j% atherefore:
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. j1 o7 @$ C- d7 F' D/ ]0 X9 Q{(a+bx)/K} dY(x)/dx=Y(x)

/ R. m2 X5 S. S1 H, bfrom here, we can get:
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+ ^) `! T( R9 f1 ?# YdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
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# C. [  Y; `2 `0 Gso that:   ln Y(x) =( K/b) ln(a+bx)
  o5 p4 r5 D8 V4 W* Z" F: a, G
* X  K. J  v6 j! fthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

* C( ]9 e0 z" w/ u  b+ D* f-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
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( j6 D- \7 N) X2 SC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)