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）
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- s7 Y1 X' C) W  ]1 M9 ^d [(a+bx)c ] /dx = -kc +s
* ^1 |+ |' V3 I1 J( u5 c( I, `where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

" c) ~. X, G2 L7 N多谢了！
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[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

& y( C9 ?% k5 c7 R7 a(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
# w5 H! g  L* X0 B(a+bx+kx)c=sx
/ F& y# T2 E" E- Gc=sx/(a+bx+kx)

十年移民路_

Solution:
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From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
2 j4 L1 A$ M  V- fso:
" }) q  h/ a' U6 r# ^9 h9 b( N" m$ t
* M$ J2 T8 z' s; m5 R( |' j+ YbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# f1 Q0 q- D* {8 }2 g3 bi.e.

4 V4 U; \! h5 l3 w4 e3 a(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
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{(a+bx)/K} dY(x)/dx=Y(x)

% J  g) p0 R" ofrom here, we can get:
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7 \- n  O( @  I$ kdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

1 ^' H0 l4 Y& ?0 ~, oso that:   ln Y(x) =( K/b) ln(a+bx)
* E3 u0 X. n, ?- C
! _0 a' _1 N* fthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

; b6 f- z* n" w. F+ `-(k+b)C(x)+s = (a+bx)^(k/b+1)
, W6 h1 y/ x! k7 o' l# M# ?( U; P3 e" l
finally:
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2 z1 }0 S4 V1 S7 S% E2 d7 tC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)