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）
. ?* W9 D, R, ]* n& V
2 P  D! m4 h  p, A, Vd [(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?

多谢了！

- b( N( K5 v# K( ?) X, N4 L[ 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
; I" ^5 [/ i5 k; _) A& Rac+bxc=-kxc+sx
* d' j  ~% P- S(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
. ~* ]/ ]  z3 m' O) o- P( }" i3 [+ E
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 ^+ H9 K; a, ?( i  s# P6 mso:
- m+ @% z2 Z8 F- o
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

5 |* u% I$ Q4 d/ m(a+bx) dC(x)/dx  = -(k+b)C(x) +s
! B1 k0 o: ^% _  Y4 {3 t
; d% ^4 K, R( L/ H5 j8 \% J
! m0 X& t  c( D) ?. j' p( Dintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
1 x  o  E1 x* ^1 p; Dwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
1 @8 U* `0 Z/ x7 x' o* b+ t
  H1 s+ p- j* @0 f{(a+bx)/K} dY(x)/dx=Y(x)

" T: g! c9 ]9 |, N! Pfrom here, we can get:
" q! @7 q9 ?# t( I
; l: ?) F& N: _: `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)

this means:  Y(x) = (a+bx)^(K/b)
& w+ e& w+ _% M7 c6 ?: v* U) Wby using early transform, we can have:
! Y$ N2 n' c# n- v* O
0 v+ v# G0 ~1 B  T1 t. |3 u-(k+b)C(x)+s = (a+bx)^(k/b+1)

+ w! [9 S; [  ~' O8 _3 xfinally:

" d  L: l: F, a* q+ S3 @* XC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)