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）
% a3 }3 `9 d6 g
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?

" ^3 |$ |8 C$ H3 S5 d: N6 g多谢了！

% {; Z1 F+ d. p# G; B& V- r[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

1 t3 N/ y+ D4 {" v3 g) M(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
$ |8 o. c4 O' t: J6 o: N' L(a+bx+kx)c=sx
4 x8 X& r- c8 @; p6 N; y, rc=sx/(a+bx+kx)

十年移民路_

Solution:

+ f" j% c. a) z& @& i4 s. hFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
2 w6 X/ J# e1 f% o3 `so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


: ~; d/ g8 r0 s+ V4 o4 J2 t3 p! `* xintroduce 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:
5 K& C3 K$ _+ N
{(a+bx)/K} dY(x)/dx=Y(x)

2 I$ b8 Z9 F. I3 J+ w' G$ \1 w( [8 Ffrom here, we can get:
# m4 U; ?+ F) Y8 e! h% ], N
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
7 E0 h/ k: r. i% C" j  G7 D
so that:   ln Y(x) =( K/b) ln(a+bx)
  i* S' c: |# v& [/ j8 }
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
0 o( h4 o# ~8 s% y# [% m
3 O- u! b7 M5 |" i' n-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

- e2 [$ x" d! X" i1 i# t. BC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)