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）

% U2 y% k( R% ?2 `' zd [(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?
$ c( Y" O1 O; g$ l/ o0 J) G) A
多谢了！
7 Y5 \+ B' p" m* v% u
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
2 m( V& |( I! Q) C9 c
3 ^$ W& L* |/ f5 ^) v& ^4 D  b(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
  b$ m2 s6 G$ s; d9 ?(a+bx+kx)c=sx
# E% s9 g' }3 a: ^5 w$ ~c=sx/(a+bx+kx)

十年移民路_

Solution:

( L& U" c8 W8 d- r: @; O( k& FFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
* i* L! s( z; rso:
% s1 E7 b6 S9 r# n
% L* Q* L" w7 J% ^. s5 N2 @# IbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
( O. C) Z% V0 i, j8 ?
! l) ^1 N! `. L2 p0 O(a+bx) dC(x)/dx  = -(k+b)C(x) +s
- Z2 P4 X. z* S' T, c! k+ Q; d
% j- h6 Q' o( Z) F8 ~2 X! p
& A! D6 N, d) i0 g& q- J3 Uintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

3 B! C. s! B5 ?/ J+ d3 ~{(a+bx)/K} dY(x)/dx=Y(x)
5 y2 I0 L8 O+ O5 o
from 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)
7 L# q7 X/ r4 T  f4 p
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
( e/ i9 p0 ]2 A# m# u. R* @by using early transform, we can have:

0 ~" f6 Q" D9 \$ x8 M-(k+b)C(x)+s = (a+bx)^(k/b+1)
- k5 ^0 l9 d$ ?( {) N9 \1 B
7 m% r$ P! v  C' ~1 b- ifinally:

) m  h6 s2 S- fC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)