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
3 A$ P4 J. G8 r, o1 ywhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

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
2 O+ _: D* i2 S8 H7 u) k. J
(a+bx)c  = (-kc +s)x
( w  D! r1 C* J2 F6 ]' _. Rac+bxc=-kxc+sx
(a+bx+kx)c=sx
6 ~$ \" d& C  r- W* @$ v$ ~9 Bc=sx/(a+bx+kx)

十年移民路_

Solution:
# b$ @2 J; ~( m
6 Z/ o& g/ Y+ C, L5 S7 ~From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

- V- ?) K$ U. S: qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
0 O7 Q7 e# l) m6 _' I1 S9 _
9 z3 o. x  a' g5 i& W(a+bx) dC(x)/dx  = -(k+b)C(x) +s

: L: q: \7 @$ r
$ u! v) s& Y2 j+ n1 n- }9 z3 l' Ointroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
; `+ i. h2 ?. E  u0 N. I$ F& ^which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

+ T3 y5 _( p6 K+ p{(a+bx)/K} dY(x)/dx=Y(x)
9 h1 q3 P! K1 q
from here, we can get:
, {7 r7 D3 g$ V$ P) D+ Y' P' T
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
: D$ f0 r2 E$ m2 V/ B# o" O
so that:   ln Y(x) =( K/b) ln(a+bx)
, S0 N/ |+ }$ {8 }
0 _$ u! p! V) K3 ythis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

. ~$ ]- r0 a" c' E: kfinally:
7 W2 ^8 ]: I. p: S
  j* ^3 ^6 N( E/ J; Z) t$ P$ o: `C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)