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）
  C. r. J1 D  j3 D' \8 o+ L
' P8 }4 s& D4 Ld [(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?

) n* I( ]+ v. n2 ?9 a, A多谢了！

4 c. G  b  s& E[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
2 h0 j, U% l0 q0 k
* V* e, Y' {, g(a+bx)c  = (-kc +s)x
: F2 u- @7 ?8 u; o. bac+bxc=-kxc+sx
% S7 F3 ]0 x' C; B# O. W(a+bx+kx)c=sx
8 D& x8 S: x, f8 rc=sx/(a+bx+kx)

十年移民路_

Solution:
/ q# l9 Q0 m8 {7 F, u8 Z6 D2 |1 u, F
/ k) ?- c: ?/ \$ O/ m$ c! k: ]From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
/ Y; i# V, Z6 F/ F8 ^
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
$ x% Y' N7 s% {7 c. d/ _6 li.e.
7 ]# ]$ N& ~) z; ]6 R
+ w6 k. T- |7 B(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
. Z$ t0 u1 Q1 z2 }which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
. c- R- ^6 D) Z( K
{(a+bx)/K} dY(x)/dx=Y(x)
& b: e% w8 U1 k7 `2 E; R* E( {" E
: [! x# ]1 Q7 S6 w; N4 r( efrom here, we can get:

* E$ \5 X2 ~) R* Y# ~; L% YdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
; R- P) X( h! \
so that:   ln Y(x) =( K/b) ln(a+bx)

6 E' L% j- x) l2 l$ c" Mthis means:  Y(x) = (a+bx)^(K/b)
8 B1 ], U# u/ A) \5 G* I2 Yby using early transform, we can have:

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

9 g) F) D2 I8 w/ ]finally:

' i( m8 o2 A& h0 O. r5 ~C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)