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）
5 _+ y( s1 B9 E% _$ G8 d) s
d [(a+bx)c ] /dx = -kc +s
/ B, R2 i* u# U+ Hwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
% U/ b( |4 c, N! Y
多谢了！

# x4 g1 I9 S0 ?  Y[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
! L( ~" Z/ e5 o1 [* Uac+bxc=-kxc+sx
; B+ G1 i/ ]" F2 w6 B(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
0 E1 C4 r" P4 k, i' t% \5 Q
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

0 ~$ i8 H" q- ?! _* bbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
: {2 o" p* @9 C
9 Y( I9 O6 V- D(a+bx) dC(x)/dx  = -(k+b)C(x) +s


* ^8 C7 w0 [# v7 `introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
6 R2 F; k, S2 ?9 r' C9 Q0 Lwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

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

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)
' r9 f% u; _" {9 o
so that:   ln Y(x) =( K/b) ln(a+bx)

- v6 n+ E+ {: |( V& Jthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
' X2 d- C2 t5 A# o1 v, Q
-(k+b)C(x)+s = (a+bx)^(k/b+1)
* K9 E. w3 S/ S- u# S
finally:
+ }& ]; e( j7 ]* U, R! W8 F4 Q
1 ~9 z+ B$ b3 H5 H3 v7 hC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)