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）

( ^1 l% M% }$ d. 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?

多谢了！
/ i% h* R4 A) M4 L. w, L$ j+ F1 P2 s
& U* X9 u7 _- r4 M. G$ G( W& ?[ 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
! p$ g& _3 s, J8 z% C9 g(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
% T$ z2 W: s5 |- Y2 G/ Y
. l0 j/ C" X% PFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
$ A; A/ g! \% N2 @+ w3 o# g
' K1 ]3 e% k- C/ h4 `bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
) Q, h9 J: G, Q# I: O& ti.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

7 V$ V. B0 T5 w
3 G" z  c6 }5 d' kintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
7 e# j- v) R8 [- ]" ?* k- z" xwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
0 g4 b! {/ m2 ~% w+ Y- C0 j* ]" Ztherefore:

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

' `& `. S  b/ v% lfrom here, we can get:
/ E5 K3 \4 Y8 ~9 M) d3 N4 @
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

3 Z$ J6 F- R. {4 I' W) y. a" ^so that:   ln Y(x) =( K/b) ln(a+bx)
# Y$ i* S" C  ]! A: U0 H8 b0 b6 I
! U4 i) N! k' m7 |0 i9 qthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
6 o: m6 F: P, f9 `
* n9 u5 m# ]" d4 v" X) Y) @) P-(k+b)C(x)+s = (a+bx)^(k/b+1)
" x% m9 S) P4 L) `- f
finally:

2 Z  p1 j0 X( ^& [- FC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)