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）

# U( \& z( I5 M- Kd [(a+bx)c ] /dx = -kc +s
3 k- T/ ~. F+ hwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
3 E1 w8 x# v/ l: ]+ m+ B9 Q8 w
多谢了！
( z/ d5 L0 A) |2 S3 ~
3 f& S8 H4 n) s0 }+ ]; X4 Z6 ]4 u% M[ 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
7 S$ p& u/ X6 q8 |3 Q8 kac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
( ]* D. V2 A0 \
" g  a6 J, D- y  X( C5 P8 EbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

# i) w, g. ?& [( I% l, [
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
" i& r) l7 Q. P) |( j% Y4 qwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
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% y' s* R6 Q% D$ a. A' O{(a+bx)/K} dY(x)/dx=Y(x)
+ M: l7 ^" c7 X$ _
from here, we can get:
9 L  k9 K3 j: `. v; g) F( m
& E, O8 H+ H/ J! y' Y/ ]dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

5 a/ @2 V9 c5 R) d- Fso that:   ln Y(x) =( K/b) ln(a+bx)

- P/ i$ ]$ w1 A: H6 }7 ythis means:  Y(x) = (a+bx)^(K/b)
  B! ^! {0 S( Yby using early transform, we can have:
1 n  c+ r& X9 F( e
) ^2 s, R8 }& K  m# g, ]3 W-(k+b)C(x)+s = (a+bx)^(k/b+1)
8 G4 ], K: X) U! T: H* J
finally:
/ k5 X/ V, Z& n
+ s- w/ u+ j. B$ w: U9 lC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)