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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
; i( |" R5 q# y% R
4 U4 i2 c; Y8 K1 p' Y/ L$ P多谢了！
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[ 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
ac+bxc=-kxc+sx
: K) A' c. z7 b8 h; z6 l' R5 n& L# z(a+bx+kx)c=sx
- H! J5 s9 Q) \- ec=sx/(a+bx+kx)

十年移民路_

Solution:
" x: x, V7 R; d7 M
! P. }. @, p3 k" C. ]From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
+ H% k8 K6 T8 u' ^, d# si.e.

* y( W% R2 S' _# S5 E4 R7 \(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
1 p* @, G+ T( N4 J# j5 i- V+ Fwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

7 o+ k2 k. D) H" Y{(a+bx)/K} dY(x)/dx=Y(x)
/ p; J* F; s# @
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)

3 @3 I. I" i% U+ I& Sthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
. v4 d3 P- d8 A. X6 h5 k) F/ Q% i
3 Q9 q3 r5 p9 T1 A* q$ E. Y' [! Rfinally:
6 `% Z- v0 k9 i% b
( |# q4 @: o& B1 \5 O9 m: mC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)