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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d [(a+bx)c ] /dx = -kc +s
6 e  y1 T! U+ o0 E8 D& a( _where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

! m9 X3 v! z; _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

(a+bx)c  = (-kc +s)x
- u7 P" Z9 P; Q- lac+bxc=-kxc+sx
(a+bx+kx)c=sx
& w+ ?2 p4 s" Jc=sx/(a+bx+kx)

十年移民路_

Solution:

+ _3 y  Q4 n- y& R3 r5 vFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 F& F2 V0 m  B4 C
3 C$ U9 q3 c4 C; ?
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
1 \8 r3 W. Q# R# e1 s( xwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
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3 ~+ q2 O' l$ p/ ]; V{(a+bx)/K} dY(x)/dx=Y(x)
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# A: n2 y% [( N' v4 @' A9 o0 bfrom here, we can get:
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1 a/ l3 Y+ r/ [0 E0 Z: A: o2 ]dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
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so that:   ln Y(x) =( K/b) ln(a+bx)
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this means:  Y(x) = (a+bx)^(K/b)
9 M+ M8 P' E% r% Vby using early transform, we can have:
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2 b: ?$ V8 e% U/ L4 `% @' i$ C-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
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5 `/ i) Y% K$ U; a" n+ fC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)