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
1 M7 c- }, W$ g3 q5 Twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
% d# K  R! v8 W: D$ i( l. p: R
多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
" Y; }! K' l! j2 l
(a+bx)c  = (-kc +s)x
4 S$ l0 y7 x9 }" |ac+bxc=-kxc+sx
- R) t3 V) c5 M& c2 Y1 D: K(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
3 \* G- }' m9 O
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
: R8 N% I/ s% \2 v* O$ I4 O- @
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
- J0 B5 N( M+ xi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
) L) D5 A6 M. `: o0 j0 \& }which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
& e! O* H$ x; ^& X! Ztherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

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)
6 H+ \( D7 r" y' C
- n. K5 T9 b0 oso that:   ln Y(x) =( K/b) ln(a+bx)

1 I$ _  t' b8 t# i4 n# W* fthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
0 l- ]1 `1 x( {  b
-(k+b)C(x)+s = (a+bx)^(k/b+1)
: K( N8 W& \8 J/ d, n
% e5 y% |( a1 R1 I- t8 o  Vfinally:
1 e$ E0 |7 a7 D' G
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)