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）
/ a" d' e% U& Z6 _& [
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?

多谢了！
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[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
) F2 i' d* u7 ^% O) {' A5 V
+ {+ X6 I8 s4 P; f1 u(a+bx)c  = (-kc +s)x
2 W, F- z: s9 R9 tac+bxc=-kxc+sx
6 }: _# p) `! t(a+bx+kx)c=sx
- `7 b4 A# J- T8 m; jc=sx/(a+bx+kx)

十年移民路_

Solution:

  ^0 ^4 Z' G6 V" l4 o; G7 C0 ?From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
3 S/ D! Y7 h( T- \  \, b2 D
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

' G# `  q, V- r% @" d: T
2 K; S$ v3 g0 l7 s" K8 `( `introduce 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:
: K0 N0 m: [& C2 j) }% }4 m/ m5 {3 r
{(a+bx)/K} dY(x)/dx=Y(x)
9 F* m& O9 f+ M, d& i
from here, we can get:

+ }& E1 c* Q; J3 C. @! wdY(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)

this means:  Y(x) = (a+bx)^(K/b)
/ r6 |& J! ?6 Q4 ^3 fby using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)
- {2 H" x% }  D$ a
finally:
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, W! e9 y- o0 ~- }/ T" ZC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)