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）

" N  t0 l8 @' L8 ld [(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?

$ M# J% q4 q7 W- d多谢了！
0 f+ N/ E+ U; g  u: e8 N
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
8 T2 |. X( `4 T1 k% u
3 x3 {" W, E" F! S* ~; ~" X(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
) G/ H( \# M6 z3 P- Ac=sx/(a+bx+kx)

十年移民路_

Solution:
$ g6 x" K0 T0 ], e( f4 `
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

$ k6 }+ S7 ]2 w; lbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 d/ G3 O3 k+ u) \" _i.e.

% ^8 Y7 \3 R5 k1 Q$ L6 I9 w# V(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 U" A8 j: A! m% R( f, M8 f

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
: O- b% [8 k4 W: a% S" m7 ^which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
$ Y& g, u0 ?9 ^+ S7 q% D+ btherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
" A, b( C8 i  y" r
from here, we can get:
% [' H5 k/ A; F
  c; ^! p& S! w" wdY(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)

this means:  Y(x) = (a+bx)^(K/b)
2 s9 Q0 o+ T7 ?by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
" j) u5 l' l$ P8 l
, w4 G8 w) i5 C( Bfinally:
% t* v% q# G& w& @% w' F
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)