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）

& [/ z( `1 @0 z5 `: @; qd [(a+bx)c ] /dx = -kc +s
: y. g; `. M, A: ~( R+ H0 w, l8 vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
3 J! h. V% k! A4 W( r
  Y9 v; P" |' E! M[ 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
& q. q' Z, Q4 t5 h/ @ac+bxc=-kxc+sx
* u( }9 y3 W2 F' i$ x/ ~( J(a+bx+kx)c=sx
+ A, F. z) F- H: M7 Jc=sx/(a+bx+kx)

十年移民路_

Solution:
  Q4 H) |5 v4 ^& E- C1 o, ?# ]: q
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
- @- h/ o2 ]- v
( \8 P% e3 j! m3 z* ubC(x) + (a+bx) dC(x)/dx = -kC(x) +s
( C( i4 e5 [2 _i.e.
2 r2 F- r; n5 b8 N
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
" Z/ b6 H/ b$ |! C9 c& f9 g- ]6 {9 lwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
/ a' ~6 m, y7 Z! d$ L- a
4 V, P) L! e( {* u2 L4 f{(a+bx)/K} dY(x)/dx=Y(x)

% A+ E2 U; [$ y- E3 L3 M% k/ [from here, we can get:
" H. Y8 m5 m: c" r) e) R9 Z! J
) Z  ]" k! W/ q: u2 a1 U/ b6 YdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

0 x: _/ D  `6 t; Bso that:   ln Y(x) =( K/b) ln(a+bx)

# K  n. h) N( \% A9 _9 wthis means:  Y(x) = (a+bx)^(K/b)
( Y/ p; w7 \3 F. I5 M. r8 mby using early transform, we can have:
! Z7 K! O0 {! t6 I
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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