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?

多谢了！

4 f& V2 F" g9 n' G0 j3 x7 j[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
- Y( x  @5 e3 X+ O6 D" n& j
: v# r; }# k% z(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
! f& {3 f$ w. R7 X/ F: gc=sx/(a+bx+kx)

十年移民路_

Solution:
, V3 j0 O% i+ Q, p5 w9 F
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
+ Y. M5 U3 r0 W% ~) q1 h$ E
' ~- S8 u/ B8 u( Q. d0 M+ {3 MbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
9 p5 n! {% C7 G4 m; }i.e.

9 o. @/ z7 B' r' N- V* a: L8 W" t6 K9 a- K(a+bx) dC(x)/dx  = -(k+b)C(x) +s
/ K  s9 o6 z4 v

; R# q+ C8 |; Uintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
2 C2 n  [6 w9 j8 J* F; Gwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
- i/ |, ?1 Q$ G# r1 Xtherefore:
/ d  T' f& H7 {! O
{(a+bx)/K} dY(x)/dx=Y(x)

7 h8 E, G1 V, v) ~% A( A6 G% Ofrom here, we can get:

  j: \$ O$ R5 P* edY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
. M; r" I# v4 K+ I/ e4 t- R; W
9 G3 h; y; T4 x6 B5 b& tso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
( a  ~9 o1 h2 R
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

8 D/ s: ^% n% z: O3 Q) [: q2 WC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)