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）

4 d0 Z' Z9 f; i4 _" ]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?
) f( u+ |; h# h+ S7 i1 a
多谢了！
9 I, W! `: }/ ]
- U8 {3 F2 i( P( R) D4 D% d[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
, H6 E' ^. j/ E) g' M! I% J4 w+ ~5 Tc=sx/(a+bx+kx)

十年移民路_

Solution:
5 E: }/ r4 A  G+ U0 ^& a
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
) U" D8 m2 h. U* ?- M( Eso:
% C* w* e# H! y4 K
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
* R' R; ~/ E5 G: k& Vi.e.

7 C$ d' w- w4 E5 _(a+bx) dC(x)/dx  = -(k+b)C(x) +s

- c  L5 o9 `! X7 Y$ F. m: N
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
1 z4 Q) K# V1 {) l1 n* U2 q* P" Q* jtherefore:

4 z- V% ?; \4 \9 C" Y{(a+bx)/K} dY(x)/dx=Y(x)

3 X$ C: s; v" B; m8 H9 g* @from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

, A" I$ t; b( i( |so that:   ln Y(x) =( K/b) ln(a+bx)

. m2 c+ G0 ]6 w0 `4 I8 Kthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
. z& l) t& O' e3 }
. w0 s3 g; q( d- ~-(k+b)C(x)+s = (a+bx)^(k/b+1)

5 z" ?' ?7 y; z( Z+ o$ V; d1 j$ ^8 I  Rfinally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)