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请教大家一道微分题,多谢!(原题贴错了,现纠正)

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发表于 2005-4-3 19:44 | 显示全部楼层 |阅读模式
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请教大家一道微分题,多谢!(原题贴错了,现纠正)
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1 u9 s# y. R& k4 y8 Ed [(a+bx)c ] /dx = -kc +s ; I2 k: v9 A* p. C  E; f
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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+ Y9 c2 K  Q+ K9 ?' h多谢了!8 n2 C( Q5 ]% {# r# k  r( j( I  E) _

* @' i  c, d& w+ h[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]
鲜花(0) 鸡蛋(0)
发表于 2005-4-5 02:47 | 显示全部楼层

供参考

d [(a+bx)c ] /dx = -kc +s
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(a+bx)c  = (-kc +s)x% z5 F$ `' L- }; n: q+ s
ac+bxc=-kxc+sx
: c/ k* p% m4 S' U/ ], u: e(a+bx+kx)c=sx
' R8 i* O$ _* n" x; Mc=sx/(a+bx+kx)
鲜花(19) 鸡蛋(0)
发表于 2005-4-5 22:11 | 显示全部楼层
Solution:
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9 G; \! h, d! ~1 QFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
, V4 A/ C3 K$ dso:6 P) B3 }; p# t1 V5 }4 b

( c0 g( O4 ?8 ybC(x) + (a+bx) dC(x)/dx = -kC(x) +s
' Y( L, d; T% ^! W; Wi.e.
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% q& h' C$ g' _* Z: K; i  O" O(a+bx) dC(x)/dx  = -(k+b)C(x) +s
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) x$ ^$ [0 _' d+ h
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b) / ]8 T6 w+ h# x
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
8 G) c: y) |5 Z% s1 h- O6 atherefore:% e# `3 J; J' J0 ~1 X! ?* x1 g

# o+ x8 f4 H/ D& w7 m0 c9 O" u{(a+bx)/K} dY(x)/dx=Y(x)
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( _  O  c- Y& Y1 lfrom here, we can get:3 }2 j) C8 ?& `4 O
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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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0 C6 l6 L7 O3 m: a, lso that:   ln Y(x) =( K/b) ln(a+bx)
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this means:  Y(x) = (a+bx)^(K/b)& i, h. E. F; |" j' L
by using early transform, we can have:) I  R4 g3 m6 s$ K9 j
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-(k+b)C(x)+s = (a+bx)^(k/b+1)0 ]7 d, k3 o4 e5 u

3 f; W' e  g2 Sfinally:4 t8 M0 Q1 Z% K% D& \2 v* q3 h& q! D
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)
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