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）

- f# T6 `: C0 {# g/ {/ c5 u+ 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?
0 w2 j9 V1 }! m' K
多谢了！

' V* I( z- ?  ~' v6 I. ~[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

0 d& g, N2 H; J* K; b5 v* T) L& c(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
6 k& j1 }1 M/ A$ m0 Oc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

3 p8 Q7 L$ x' Y: B; S+ Q+ N3 X9 ebC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

6 w5 J; p' @& T/ T! A3 ~(a+bx) dC(x)/dx  = -(k+b)C(x) +s

: q2 _! [* g) S9 _. U6 w# T1 j
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
) }0 z, d& j0 P  S6 s1 Bwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
( I8 G; s% ]+ s% T% {/ y' ~
' R( s( F7 \6 z  f0 Y4 I& Zfrom here, we can get:
4 u# {/ ]" d( }
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)

1 [  ?. G) x* a5 K! @this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

: T: u6 g6 b  A: j# B* M9 B-(k+b)C(x)+s = (a+bx)^(k/b+1)

# w8 _6 w% q* M! hfinally:

3 w# `# b: h/ Q6 i" ]! W- I$ _' c: GC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)