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 [(a+bx)c ] /dx = -kc +s
* F6 E8 }) }2 l. ?: y9 W' b$ Lwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
/ s8 F, ^8 n# k) M" e0 K
多谢了！
) F- E) E: U4 v0 J, s
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

( |& j( u  N6 N% x. y(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
' U5 u5 m2 T$ `' L% d; v; j2 D(a+bx+kx)c=sx
2 i/ V6 H  Y8 Ec=sx/(a+bx+kx)

十年移民路_

Solution:
4 L, j9 |! D0 E: f% k& w) ?- w. {
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
3 l: r. s6 d1 Fso:
+ Q; i3 J, d) \( [; q% m5 E
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
. G, \8 g: y+ Q0 q. c; n0 Q0 L+ p2 pi.e.
, S# |: L$ C& h
6 I: i" O" J3 a  ]- l7 n(a+bx) dC(x)/dx  = -(k+b)C(x) +s

+ W2 F, t. E1 l  Z
' c7 F5 U, Y; E( L" {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
therefore:
0 u3 c- s' g# C  K& H3 h
{(a+bx)/K} dY(x)/dx=Y(x)
1 }7 s$ r) d5 T. P1 a0 n4 E3 }  B
from here, we can get:
# @$ I% S6 D* m5 t
# F+ J( L( O2 V# Z# SdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

; B$ B+ g. ?8 l7 I$ G; N- i  hso that:   ln Y(x) =( K/b) ln(a+bx)

& w, q) I- n6 q2 cthis means:  Y(x) = (a+bx)^(K/b)
- @! r2 X' ?9 c) X9 Zby using early transform, we can have:

1 Q- U* w% h! @7 m-(k+b)C(x)+s = (a+bx)^(k/b+1)
+ ?' x* C7 K4 R; g
% Z% e+ P! c5 G* J& bfinally:
& ~4 x" J8 {! ]; t0 P/ x
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)