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# h' I7 g* G/ D- E( O. od [(a+bx)c ] /dx = -kc +s
) R5 {( i- r9 T/ L  Z+ Jwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

1 L# t' V7 L9 i7 P4 v. n3 h多谢了！
. ]  I1 j3 [; d' v8 n  X( y
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

9 G! N; w$ ~% v; U* \! ~(a+bx)c  = (-kc +s)x
9 C1 t% O& w* t* b. Qac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
; j0 }5 k/ v" ?  ?so:
# p! [# g( p# F4 K& r) U; T% K
4 Q# U. v0 M8 }; vbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
  x# C/ A  w( S$ G) ~i.e.
2 L0 Q1 C4 r3 U: D% @( k
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
5 u; F+ Y! o, S% r* m" O" m

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
6 H  n: M: z1 ^- f% T$ Q" v' k+ h  Ztherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
" g# v8 v( b& l
$ b8 f7 ]1 C" I# hfrom here, we can get:
% r; @5 w) U) l5 k# {) R; L
( h. R- x: ~& E+ b- ?5 C; PdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

" i' j2 C. m8 {; \* z7 R4 Q0 Lso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
& S: n5 Z- w5 y+ \% Sby using early transform, we can have:
. F4 E, B2 P4 X. M$ T6 L% I
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

! F1 v: V  o3 FC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)