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?

2 }; Q  z. @7 o: R% C( k多谢了！
; x7 E. U; _0 Z; P8 |" o
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
& ]+ ], c4 Q$ C$ ?/ l4 ?5 D
, Y. y0 F+ b9 e: [8 |(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
+ J: Q" P6 a" y$ [( d* r- H(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
0 r8 N( q8 P2 L, K1 Dso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
' w' g7 O2 o1 f+ {+ Ni.e.
! _( [+ h+ T9 |* e2 S1 U
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


2 k" h; f5 F; `3 vintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
- N, z) k: V6 z5 `which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
1 ^% P/ Y7 g- k* \3 Itherefore:
5 N( P( f; h  I
% z5 Z! u# x9 I5 w{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
6 e* Q2 m' @0 n
; o' f" H& E- ndY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

; H9 l8 f  D" A- H, b" Dso that:   ln Y(x) =( K/b) ln(a+bx)
  K( R" j$ @7 V# ]
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

5 e+ c% _; r7 Z& P! S& z; w-(k+b)C(x)+s = (a+bx)^(k/b+1)
+ a' S6 ]9 N1 P2 o& z: q& R
( ]* }* V. _* T$ X1 dfinally:

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