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）

. J) g! f' K+ n5 H* i- Kd [(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！
8 M# o- a+ v8 H' |( s6 K
[ 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
0 d* h# Z! Y0 w# [( P, ^ac+bxc=-kxc+sx
& a$ [3 W  Z5 }(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

% f! N7 O$ W( z% J' JFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
2 V4 I" d: B% M; F2 k/ Aso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
/ i7 b$ @+ u# F! h+ _6 @i.e.
6 q9 q; X6 {0 s; W
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
% W% R" D& T8 \3 }/ f9 J- t7 J

5 b& z- }. T) ]2 h# Eintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
* S' w2 T. \3 ^+ X! ^which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
1 {% B* l) {  n) |& d0 \) _therefore:

+ i8 O1 f; t& p{(a+bx)/K} dY(x)/dx=Y(x)

, E# Q+ e  i$ d, i) g2 |! O+ h+ ~: {from here, we can get:

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)

( s9 Q% L: h6 V# Fthis means:  Y(x) = (a+bx)^(K/b)
) P3 L6 `3 ~9 u) v/ zby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)