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?

8 e2 {0 o- F+ ~1 ~" Y* g/ x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

(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
! C7 R2 F9 |5 l* i. C& p! M- D(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
& M: x1 ?* d6 h: F0 N
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
0 I# u! J' g" a" S6 c. x7 q/ y
+ [* t0 K* r: Q3 m' R7 p0 y' abC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
& d& E) e6 U- Q+ \/ n$ A) {
; l3 ^$ \# q4 Y0 f/ R( e8 d1 s6 e  C(a+bx) dC(x)/dx  = -(k+b)C(x) +s

2 w* T' P1 N2 v' y
/ I9 E$ ^, j" yintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
0 Y$ @3 T) G8 c( K: fwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
& a1 H! m8 r0 u9 x' Z8 x* ttherefore:
! U6 a+ s) z$ H, `
5 @3 q6 F% z  h2 J{(a+bx)/K} dY(x)/dx=Y(x)
* S7 x& M- K0 E& f  C" t" D) v
from here, we can get:

: Q6 f: X' D/ N0 C- d: Z2 {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)
  i1 a7 F: a1 w$ e  M( e! S
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

. r2 [1 l* |0 l6 x! O-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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