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; i% s: ^2 M
# F8 p1 N9 `/ N4 M+ f! Kd [(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?

2 Q8 E- d2 u: t# g多谢了！
1 n  t/ Q; g* K& z) X. u& G) v
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
& L# s) r0 `# l( t! ]% u
(a+bx)c  = (-kc +s)x
+ ~+ p( t0 r# t9 ~# {, ^ac+bxc=-kxc+sx
(a+bx+kx)c=sx
! W4 |7 o) C% N' C# Oc=sx/(a+bx+kx)

十年移民路_

Solution:

" a+ n- {4 V0 v4 V. s; D  ~From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

3 n0 u& W$ l3 ^9 wbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
  f: j. m0 w( D6 \2 R9 |
( t% R" A# p. M$ `
$ K& |' _6 |6 G2 Zintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
+ N) O4 E  N2 j* a: z) ?) x& Owhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
" Q& b  T( i# |5 Btherefore:
! }; L2 h* u# r/ j# O6 [7 ]
/ N( }9 e* r" d1 i/ j2 ?{(a+bx)/K} dY(x)/dx=Y(x)

7 }( |' z1 O/ \/ j+ m+ o- C: Dfrom here, we can get:

$ R1 O1 l% t1 s( j0 X# \dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
$ B% u( E- _, u/ |, r6 G4 m
* P  z; \4 p  t8 h, Hso that:   ln Y(x) =( K/b) ln(a+bx)

1 o1 O/ _* E1 k8 |4 n$ M+ Q( a& Qthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)