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）
* a. ^2 v# r; Q  I, z  ]: Z& v0 q
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?
4 J3 `0 j/ h, j0 f7 }% x% p9 F
2 {; A! C8 J- |' J+ M& `1 d2 O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

: S$ V+ q4 ?$ M3 K(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
  S% Z9 ]7 ^: P5 I7 Y! N. g- q(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
& k! T! }! X, M4 o% _: _
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
+ U& W" g: u0 v$ D2 z; o; k+ g, C' G
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
. f$ l8 r- {) _' M2 x" di.e.
; T( s3 W: m7 P% |- I4 @! F) k3 l2 y
) H/ t: y  `4 U(a+bx) dC(x)/dx  = -(k+b)C(x) +s
! j! |5 k: U7 l3 [9 ~; @

1 R4 ~" O/ g, yintroduce 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
therefore:

3 J9 s# X. d8 o3 `{(a+bx)/K} dY(x)/dx=Y(x)
& o) V/ ~& i/ [0 t
: F7 D( K/ q! D. j" H* H' bfrom here, we can get:

* x8 _/ R9 m& p9 k: {8 i' r& ~! SdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

1 y% ]& E" ]1 Y9 X: k+ bso that:   ln Y(x) =( K/b) ln(a+bx)

. z6 h6 f, E  P/ O/ z4 A& k# H  v. Jthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
  B* Q4 @7 X  b+ b
finally:

6 k; H) D. @6 C% W; X- n* G: mC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)