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）

& a! a. g8 q' F2 J1 F- ud [(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?
& j" u. @  E& R$ e# Q
2 U9 h1 b9 e* e+ D多谢了！
& q% o0 O2 @& S/ p) q% X
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

. S- K. w! D% a# t  K(a+bx)c  = (-kc +s)x
# p' c. Z' T! e( p9 y6 Yac+bxc=-kxc+sx
5 z/ m6 O2 E9 c  ?0 D# y(a+bx+kx)c=sx
& D2 v" Z! e7 t4 u. |c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
5 s! P3 u4 Z6 V9 L3 L/ \* f6 Gso:

0 k1 q: S5 w# [5 E% |9 tbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
  h$ i8 @1 }- u. |i.e.
: [7 Z0 I% y9 `0 a
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

% k0 y) T  a6 A( F0 i. ?' D& K/ o
* I1 Q2 E* [2 X" s+ f2 Z1 Y: f  Ointroduce 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:
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2 [  [: v& U, q{(a+bx)/K} dY(x)/dx=Y(x)
/ i8 Q5 m' T1 U. A/ L8 o" g
from here, we can get:
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dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

0 ^' ?0 U/ }  m- W) Y5 n. oso that:   ln Y(x) =( K/b) ln(a+bx)

2 W9 \% p0 O6 b4 `this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
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) l( P8 Q1 Y7 ?" L/ F' r$ |-(k+b)C(x)+s = (a+bx)^(k/b+1)

/ i; C. N# D9 e0 b) B. hfinally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)