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?
! b, v, \: u: U% ^, C
2 Y+ N2 q% v( o- p. q+ d( }多谢了！
, n8 R% [8 J& e# e" q9 U, U* p
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

" H& V! R2 h) m* \* a0 F(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

) u8 e8 y& m; x0 k: bFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
, S, q( \* k" I  v: ]9 ~1 B  _
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 h( ]9 E& b/ k& C% Ji.e.
9 [; H) x  n2 D  s/ `5 n
$ p4 W( n& R  p7 L  D(a+bx) dC(x)/dx  = -(k+b)C(x) +s
( a' @, w' L$ _$ k5 \, D/ T  C2 j4 _

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
% Z7 g0 P5 l1 |: xwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
0 @+ E% O8 w: T- J4 m
{(a+bx)/K} dY(x)/dx=Y(x)

* k- W4 B5 j( g" Zfrom here, we can get:

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)
$ y1 ]# ^* o' C+ C( e
- b1 x! x+ p/ `2 @+ }this means:  Y(x) = (a+bx)^(K/b)
& J5 z" m- S1 M7 e8 Kby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
* i, `$ L& \+ _" r% O( [) A6 O
8 G, ?6 T* I2 v, M% x, Ufinally:
: Q9 a% y2 r- J; X# F% u4 ~
/ F4 K7 O- i8 QC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)