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）
3 ~. s9 M+ f; F- ]. K
- X5 ~8 S/ F) w9 h% @1 }% I  W/ R- Dd [(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?
8 s/ m  h$ X2 {9 C4 o2 N' s2 V5 n. X
# F0 B& v$ i. |多谢了！

8 V7 I. g; W2 I; _[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
8 k/ S. S0 Q1 f
& I; }7 x3 z4 ^$ z! A% \4 q4 T9 \(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
1 f+ w8 U* z, x7 }/ x
+ {1 M1 h$ X; p. E) |From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
0 |( q! s- k5 Kso:
/ `" p. n6 W3 z
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
# R! l; i6 d7 B1 v# t
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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
! i, n0 R8 h# rtherefore:
& p0 I. u' d. v9 R7 a0 Y* {- p
{(a+bx)/K} dY(x)/dx=Y(x)

7 m" f4 x4 V1 P/ C/ w7 Bfrom here, we can get:

) H7 t6 j0 P) g  g; W" x$ bdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

4 _' y3 S- ?5 K  d. ^$ ?8 [* Zso that:   ln Y(x) =( K/b) ln(a+bx)
9 P! ]& w& E- x' V
+ ~5 Y+ Z0 _0 @5 w: O# l6 }this means:  Y(x) = (a+bx)^(K/b)
1 c" @( B* C' ?1 Y% nby using early transform, we can have:

' r& A0 `9 i; w$ |3 C-(k+b)C(x)+s = (a+bx)^(k/b+1)
% O4 p! o( L* c; V( @' @4 q
1 J( U. G; \  Y$ s; D% Mfinally:
6 l* B0 Y+ \0 C6 k' l7 q
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)