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）
' w* n% K5 Y) x
- Y  H$ Q- I9 {$ t4 j9 F3 O& u* C" ed [(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！

+ e; u8 m6 b, g( d% F8 u2 f[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
8 F, k0 x6 K9 Z
; X2 {$ P0 {  [7 X& z  {6 D6 U(a+bx)c  = (-kc +s)x
* H) J0 L$ u; k" r% E" E/ U; _1 hac+bxc=-kxc+sx
! r! }8 B8 n! ](a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
) z7 T4 k4 O4 }3 L8 h
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
/ y. o( h* e, j$ g% ~, Mso:
; M3 o' I' h: v1 Y! h) ?
2 Q6 [" |9 P" p: p+ T1 Y% qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
' z0 J: u& e( W# ^. l" A2 N3 Ni.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

) F; ^- a! G' r: f, N0 ?
& X0 c, P1 p3 j+ n7 B0 {introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
( t& r( `7 G! P( D6 p' Vwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
/ t; A; s0 _, n. G& D  J9 Ctherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

6 Q0 I; o' q2 \8 wfrom here, we can get:

# l3 S* x: ^7 rdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
2 P. f/ H$ l6 s7 \& V
1 c2 D9 V2 B4 k: K& O  Z4 wso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
+ F! Y! F, [4 I/ `: {/ v
finally:

' a. s( t0 B# b! VC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)