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）

4 f; s; l" ^$ n* z1 Td [(a+bx)c ] /dx = -kc +s
$ ~* F( m- _' f6 p/ F' G2 E) y. J& Swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

. i( \+ Q' d6 t+ w9 _多谢了！

, U% [/ V- r. C$ _( u. Y  w% z[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
2 i7 F0 _, a) f' v+ T3 E
* l! h0 ^' w5 O0 l* f(a+bx)c  = (-kc +s)x
7 Y' i  M6 O" L2 E- z+ ?( sac+bxc=-kxc+sx
- x1 S0 U) s0 L3 ?. q(a+bx+kx)c=sx
% K7 g# D4 J& F; ?9 R2 `c=sx/(a+bx+kx)

十年移民路_

Solution:
: M6 ^5 ?5 ^3 ]- O0 i# j, k
7 k1 K. Q* _$ vFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
8 L4 g1 X  U/ {1 \; mso:
7 L8 _+ X7 j. l( V+ [1 r$ z( A
! C9 s6 m: N+ q* v' g' ebC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
' H  z- B6 S2 e9 P0 |9 G
7 L0 }8 V- L/ A$ K9 M(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
  c5 A/ B" u# W+ ~7 |; m: Kwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
( s- u2 a7 [/ r& Z- c4 E& p. Gtherefore:
' I, K2 |0 c" K% q" d& J
( _7 J" x) p5 I* X2 A{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

0 ^( }% W& q5 `! ?: X- KdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
' L1 @* E5 b9 I3 b
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
! |0 K& D+ B, S0 q+ k! A& E5 z7 vby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
9 j* e" I/ t, f
' I8 O9 H/ o+ B5 ]2 _6 nfinally:

0 ^* y0 Y- G$ d4 D. r5 H! y; T: OC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)