数学高手请进，多谢！

醉酒当歌 · 发表于 2005-4-3 19:41

请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?

多谢了！

[ Last edited by 醉酒当歌 on 2005-4-3 at 07:56 PM ]

十年移民路_ · 发表于 2005-4-3 19:52

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
$ S M, Q8 {7 W: \0 A+ e请教大家一道微分题，多谢！
$ i; x# S& u0 ~9 c# ?# J' m4 M0 D3 b) T3 ^/ R% V; d. }
d [(a+bx) ] /dx = -kc +s
) T \/ Y* p$ N# a qwhere: only x and c are unknown, others are all known, requiire c = ?
5 N; C; a. I+ Z" h1 h
/ E- S& v# T, A2 L7 i1 [" h多谢了！

it is too easy:

d(a+bx)/dx = b

so

b=-kc+s

finally: c= (s-b)/k

醉酒当歌 · 发表于 2005-4-3 19:57

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:: a/ s8 O; p3 L
it is too easy:5 U/ X) J9 }" z; @

$ [. v, e( H! f2 d0 t) m! |d(a+bx)/dx = b; j( u( C/ S  `+ u& f

4 Z4 }' _9 g# C( m7 Wso0 @4 D0 s% ?) ?' C
: P7 m9 W  D0 u
b=-kc+s
3 z% b* ~- v7 z! V
finally:  c= (s-b)/k

the right side cannnot contain c, but b contains c, so it is still not solved.

十年移民路_ · 发表于 2005-4-3 20:51

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
! K( ?  W0 v7 q& X# @请教大家一道微分题，多谢！# ]% ]# q1 G% N. z
) a0 d$ i6 I( F! j1 N1 x- f+ w6 F
d [(a+bx) ] /dx = -kc +s
) Q. A  x( s! S. N  U) J6 Q3 ~8 ]where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?, g" B' ~. _% I; |& b6 f' `$ ]+ Z) K4 `

$ @2 Q6 C1 z2 U! ^) B多谢了！8 K  G: P" H' f7 H  o3 d

: j3 p6 t/ P9 s  s* b[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.

a, b, s, k are constants? c means c(x) ?

please tell me.

醉酒当歌 · 发表于 2005-4-4 11:30

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
1 o' ~) B2 H! V* Rsorry, did you post another question? your statement is still not clear.
7 W8 g% @; `2 g3 V
1 z7 h3 G/ o' g4 a/ t7 Q/ ya, b, s, k are constants? c means c(x) ?
( q ^. W+ d. {& O8 u# E* @* s" Y$ A4 j" l8 V
please tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

十年移民路_ · 发表于 2005-4-4 18:36

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
) Y0 D, ~# s; f" a# K8 ^ o6 u- M对不起，写错了，忘了变量c,应该是
& ]1 E. [. x* ~& I# V2 Qd [(a+bx) c] /dx = -kc +s
8 V3 J- d- Y5 }1 C& ]8 L; G: O& q3 o
多谢

do you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?

if so, this needs to solve the differential equation. please comfirm it first.

醉酒当歌 · 发表于 2005-4-4 18:43

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:+ ~/ E; S% c' `. N" q+ z) n- ]$ d2 A
do you mean it is:: S& J8 z( b5 v4 S# p( c. S  r& f9 _
7 x- H* U! T0 S' K9 k) F
d { (a+bx) C(x)}/dx = -k C(x) + s  t* ~. j& I) a6 Q/ S
. X! t8 q& `6 e2 e+ D, |, J- v
where a, b, s are constants, you want to know what is C(x) ?7 u2 ~1 I8 X& D' y, p  X4 s0 K% z

& N' {, K# s$ a% N/ N9 K. p& lif so, this needs to solve the differential equation. please comfirm it fi ...

多谢你,你说对了,就是说用x的函数表示c. 也就是 c(x), 多谢你了。请帮忙。

十年移民路_ · 发表于 2005-4-4 19:17

answer:

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

十年移民路_ · 发表于 2005-4-4 22:26

this answer is the good one.

procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(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:

{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

so that: ln Y(x) =( K/b) ln(a+bx)

this 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)

醉酒当歌 · 发表于 2005-4-5 13:33

果然是高手，多谢了，打印出来好好研究一下。高等数学学得实在不好。你一定是工程师吧，而且数学相当好。再谢！

十年移民路_ · 发表于 2005-4-5 20:41

Originally posted by 醉酒当歌 at 2005-4-5 14:33:8 m% I: v U( q @& A4 H/ F
果然是高手，多谢了，打印出来好好研究一下。高等数学学得实在不好。你一定是工程师吧，而且数学相当好。再谢！

请你再仔细检查一遍小地方，但我可以保证方法和过程是没有问题的。
是的，我现在是注册工程师，在上大学时，曾把“基米多维奇”和“樊映川”两本习题集通做一遍。（共大约一万多题吧）。这种题真的不是难题。

龙舞九天 · 发表于 2005-4-6 18:34

9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s

也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
但这是不正确的
d{(a+bx)*C(x)}/dx 意思是说对 (a+bx)*C(x)/x 整个代数式求导数
并不是 (a+bx)*C(x) 的导数除以 x 的导数。
比如说 dx/dy = (x'y-xy')/y2
x' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
最右边是x的平方。

所以您解的有问题。

[ Last edited by 龙舞九天 on 2005-4-6 at 07:35 PM ]

十年移民路_ · 发表于 2005-4-6 18:53

Originally posted by 龙舞九天 at 2005-4-6 19:34:6 c" w% B8 |, g$ K! f7 a
9楼的答案有点问题吧。  G; a0 N; E* @9 t
您说：% b1 Q8 F- t( N( X3 D8 v0 X

  V! k/ a0 P/ p6 Y; h/ ld{(a+bx)*C(x)}/dx =-k C(x) + s- H3 N9 v) U/ |: t! ^. L8 O
so:" V2 \7 u7 j& G- c% g2 H+ D
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s" {8 {) [8 l2 U* x9 t; {! I

% j+ q  `7 K8 y7 k5 T  z也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx; i$ T0 o: }2 w
但这是不正确的# V4 x9 O  w. z
d{(a+bx)* ...

Please note here:

d{(a+bx)*C(x)}/dx means: df(x)/dx, where f(x) = (a+bx)*C(x)
it is a multiple, not division. Right ?

十年移民路_ · 发表于 2005-4-6 18:56

但这是不正确的
d{(a+bx)*C(x)}/dx 意思是说对 (a+bx)*C(x)/x 整个代数式求导数
并不是 (a+bx)*C(x) 的导数除以 x 的导数。

no "x" in that position.

gnome-kde · 发表于 2005-4-6 19:30

d{(a+bx)*C(x)}/dx 意思是说对 (a+bx)*C(x)/x 整个代数式求导数! d4 Y+ L5 ]* L3 G ]
....4 @8 g3 Y) P& c0 d
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

这样算混淆了微分和导数的概念。

d(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h'(x) = [xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2 (应该对吧，我也没验算)

[ Last edited by gnome-kde on 2005-4-6 at 08:31 PM ]

十年移民路_ · 发表于 2005-4-6 19:45

Originally posted by gnome-kde at 2005-4-6 20:30:
* Y2 d. s. l7 m这样算混淆了微分和导数的概念。
9 {0 U1 Y9 D4 B. d: }! `* X) [& H$ X+ e( X2 z+ C4 g2 S9 \
d(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算. k$ X8 {% L" Z2 Y

8 {# z3 X8 t3 O, Y2 BIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

这样算混淆了微分和导数的概念。

d(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算

Ah, you made a mistake: it should be:

d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h'(x) = [xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2 (应该对吧，我也没验算)

for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

h(x)=(a+bx)*C(x)

gnome-kde · 发表于 2005-4-6 19:51

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:0 F0 a+ M' ^. V  X; \+ G6 E
这样算混淆了微分和导数的概念。
& i8 {' `0 ~1 _$ T9 P/ I- Y, h0 u4 K, H! I# x
d(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算; j! K9 d  P/ q( p& E5 H
  }2 z, \8 y; }* H- }3 u2 I% T; o
Ah, you made a mistake: it should be:6 e# C8 h' x& v+ I+ o; E5 A" |5 P
5 i; I* W" z& r1 S
d(f(x)g(x)) ...

Yes, there should not be " ' " after "f" and "g", it is a typo. :D

But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

十年移民路_ · 发表于 2005-4-6 20:05

Originally posted by gnome-kde at 2005-4-6 20:51:
2 q ]0 I' }# A; a: _7 z7 f* AYes, there should not be " ' " after "f" and "g", it is a typo. :D" y0 g% _1 `. N, I& p( Y

2 G5 R4 A) g" l* }! O6 _But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

Thanks, I am happy here to discuss math.

醉酒当歌 · 发表于 2005-4-6 22:24

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:
9 q, ?" l u! d w1 ?6 T, T' u7 J' e
请你再仔细检查一遍小地方，但我可以保证方法和过程是没有问题的。
9 j+ i4 u/ n/ z: @是的，我现在是注册工程师，在上大学时，曾把“基米多维奇”和“樊映川”两本习题集通做一遍。（共大约一万多题吧）。这种题真的 ...

佩服！是不是微分方程如果不能transform就不能求解了，有没有现成的微分方程求解公式？

十年移民路_ · 发表于 2005-4-7 20:44

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
C# M. _5 t; a7 S' c佩服！是不是微分方程如果不能transform就不能求解了，有没有现成的微分方程求解公式？

理论上，只有很少的几类”齐次常微分方程“可以写出其”封闭的解析解“。不作变换，微分方程是非齐次的，可能不太容易作到。一般地说，没有什么现成的微分方程求解公式。但对”线性定常齐次微分方程“是有一个较方便的解法。

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