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）
& X# }1 ^; ~8 W: j  X
- Y7 e  L3 T! |d [(a+bx)c ] /dx = -kc +s
# L; T% n1 j- ^) w8 n7 K) z; `where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

% J  D0 m4 H% ]! ~0 w8 o! Q* Y多谢了！
6 m: n1 K5 d3 ]4 q7 ^5 v) @
: a% Z) S! F7 v( v[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
8 C  R3 X* f! g  v* {& p- k
(a+bx)c  = (-kc +s)x
9 T0 y3 m# o* A; W( Kac+bxc=-kxc+sx
+ g. O' S& }1 ?  X+ C5 ^! S7 l(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

, r" r" {( A* X, c/ _5 LFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
5 b) Z4 o, B2 D1 x9 `so:
4 u& {0 ]0 e) y  n0 y
: @& Q' h6 i/ _0 O0 N0 w' ebC(x) + (a+bx) dC(x)/dx = -kC(x) +s
) C1 i$ b) r! A/ a/ B; Qi.e.
  _8 ?4 Y* |0 M/ T" w. \
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


9 b1 l. f" z7 ^  vintroduce 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
" {8 m& U: g. h* h/ d0 ?! Wtherefore:

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

7 {1 n* S, Q. w5 ^& Y& Gfrom here, we can get:

+ p; d  C' m; L' y9 VdY(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)
/ Y& Q! p" V. i) b2 m! L
3 d; R" p& D3 \5 c% v% \this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
2 o$ `' ?) w) \! M8 O
2 o6 |5 a/ V8 P( r# x  g) X! g-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 i. W3 X1 J& D6 |finally:

5 t/ v/ L; G: t) jC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)