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）
: W2 X8 D$ O0 U4 b
9 _- e7 h- d2 k  e! v5 cd [(a+bx)c ] /dx = -kc +s
+ E: N! \5 G* i) swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
2 o& b: [9 _4 C& C( r, w
多谢了！

; U8 D. l' p5 G7 s' x) O[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
+ Q+ Z% q6 S( |' \: R4 y% y8 [4 y
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
. }$ G( ~7 f! M" n! n* \c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
5 I6 P5 J) g2 u6 T3 B6 D6 Wso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
2 A; z4 ^: _0 D) @: P2 G9 Y
* r& c. H; e! s' G1 Y(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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
therefore:

1 ?# Y6 a! |3 q6 p{(a+bx)/K} dY(x)/dx=Y(x)
! R& ~& w/ C& m- Q2 ~: q
from here, we can get:
2 F0 o2 F  l5 U, X
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

so that:   ln Y(x) =( K/b) ln(a+bx)
% B4 g! }0 K& h2 {  b" A
7 f( m1 _" y5 W6 j  {. Cthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
% t! \# p9 U, A9 Y& g1 |
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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