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）

9 Z! j0 e/ E# C: }/ gd [(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?

0 Z& e, D  `/ y) l多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
1 X& ^# R( a1 Y! i
(a+bx)c  = (-kc +s)x
# K& B, j; S3 O: H$ }- t$ t7 dac+bxc=-kxc+sx
8 ?3 [2 w( B7 r4 ?5 R(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

: B& v. t+ U* T* c  RFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
+ T3 R7 i5 ~+ Kso:
! R3 ]- m% a# S( X1 b- }
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 j1 a/ H! P) f4 d

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
/ j4 y5 r% r7 @0 k- swhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

- F2 W+ F  [3 Z. b{(a+bx)/K} dY(x)/dx=Y(x)
: O- ^3 v9 K$ t$ B- S, R
from here, we can get:
# F4 Y/ ^) s$ d; V% Y7 T
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
/ V& i& t7 |. O7 E) C, _/ G1 u- Z7 E
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

8 W) t- X: J8 w. s/ a-(k+b)C(x)+s = (a+bx)^(k/b+1)
- |" y& C0 P& n7 w0 `% G" g+ Y
finally:

; [$ u- G4 s: P/ ?9 fC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)