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）
7 s' C1 ~( L1 |
d [(a+bx)c ] /dx = -kc +s
' C$ _" ~7 D2 b  r& `# J+ Q( bwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
* ]. d5 \8 L- H. j1 y
多谢了！
8 O  D( g2 Q2 X% _2 p
$ ?! w1 @7 j1 g  E# P[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
. R$ O! o0 ^5 h. V9 }7 g, X
(a+bx)c  = (-kc +s)x
, B* Q( f+ i8 }% y, Q( g. F4 T( \ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

% L4 C. ]; a% U( p( S% @4 v- P4 _1 wFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
* U3 x: D0 z6 e! \! X' pso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
: J3 G& D: O7 ^/ ~9 r6 @
7 n1 P9 N- B. S! U1 [- I: d  y) I
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
7 E: w3 i5 [' s! J# Ytherefore:

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

0 k$ _) {3 Y) ~) z. lfrom here, we can get:
4 O" L' q* y' c
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
/ `5 y3 ?' I: d# z$ m
so that:   ln Y(x) =( K/b) ln(a+bx)

+ G9 W1 I1 p. h# u% A$ `- }this means:  Y(x) = (a+bx)^(K/b)
1 s1 M" f* j; q; e# l: wby using early transform, we can have:

* @# @$ |3 }, X: d- D: G-(k+b)C(x)+s = (a+bx)^(k/b+1)
9 w5 b( g' |& Q0 R$ r
finally:

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