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:# ~0 ?+ \' M- K9 I  L
请教大家一道微分题，多谢！
" w1 C5 s; o4 `9 f" P, Y  e' ?) }' k) s% r2 q
d [(a+bx) ] /dx = -kc +s
- D) m6 O) Q# t- J# w8 z, nwhere: only x and c are unknown, others are all known,  requiire c = ?6 e. N  [. _+ e& R/ z) w* s! V
/ y: G6 H! i  ?9 ^- Q1 h' E
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:
+ I" y" Q' d* Z1 {it is too easy:' G% y4 R- m) L% V" j8 K

d(a+bx)/dx = b
4 t% ~, Y' ?9 R* L  C9 W2 ]' O( w3 P$ J# S* d* u+ o( e) [
so
; O  J9 ^. S  S2 o+ i
2 a- I. X- k2 C: ?$ R; ?% [4 gb=-kc+s2 j: ~- S" f9 U6 F" _

( _- w4 R' \+ G2 v8 A/ h: Jfinally:  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:
" e# D; V* P, ?请教大家一道微分题，多谢！
& V: {  D' M; k1 |2 o5 `+ ^( b: I6 v" V; A/ S8 U
d [(a+bx) ] /dx = -kc +s / R8 r" w4 }* J2 g
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
! Z  z* Y7 ?! x" `/ v! i8 `9 _, @
多谢了！/ m' p. {1 y! G3 g6 ?5 p- m. A
1 Y$ h9 Q3 w& V3 p' p+ i0 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:
# X/ L/ F) {+ Z* Asorry, did you post another question? your statement is still not clear.
* a+ ` W4 d( ]/ Z' s& |9 V/ h6 T! M4 I8 k: M. o
a, b, s, k are constants? c means c(x) ?- T; s# j4 ^5 `
2 [/ w- Q* O2 ~, U2 p5 Q
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:" s0 ?9 {1 Z5 K1 c" [5 C  t8 a) C" Z& f
对不起，写错了，忘了变量c,应该是
0 E* o$ z  r4 ?# P9 Nd [(a+bx) c] /dx = -kc +s
9 y  a" i* N2 t( h- Y$ m
" O% h. O) q' v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:
; s" l7 q8 U7 ~! o7 E( e# ado you mean it is:
$ h' l& D s7 t0 u. Q
, _5 P2 h+ }. L2 q" l2 X! Y+ {4 Dd { (a+bx) C(x)}/dx = -k C(x) + s/ z8 z5 N; z( d# z2 c
+ t/ v% D% {5 s" Y! u
where a, b, s are constants, you want to know what is C(x) ?3 I% w# G V3 g, H

# ]+ @8 N" b, mif 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 14:33:7 x1 M- D# Z$ U0 h7 s+ \
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:' z( e2 x2 E' ^- F+ L+ |
9楼的答案有点问题吧。- e; y, D# a" w2 Y
您说：4 @1 h5 _* |! V h7 v+ U+ C2 v
( @$ X0 p( U& s2 d( S5 ~/ m
d{(a+bx)*C(x)}/dx =-k C(x) + s
& E' |" i# J( _1 Y- p2 hso:6 G: Z2 {# q# d# i1 y0 U: F
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s1 I% J5 G; x" I$ x% V

0 Q- }* Z& X$ i1 H也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx, X7 p' G2 }. t: @
但这是不正确的- ?; y7 a2 s* d$ }8 x3 [
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3 y) E7 b$ o: q5 {/ `% \
....3 \& I/ z+ u% N* @/ H% 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:
' a6 a7 S( P; W Z% ]3 U1 V/ G这样算混淆了微分和导数的概念。- M# p+ g4 X, j% P% g

3 |& i5 o$ o3 w8 N7 Hd(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$ S# V- d3 V3 @( g

% x, T+ y3 _' f9 U3 C4 T* I( A3 NIf 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:
9 W; ^3 k0 f9 b$ M这样算混淆了微分和导数的概念。; w8 ^$ S- _! P2 l: Q& ]. w" `% m
) X/ J+ E1 _. R5 u' c* y8 @
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
5 N! |0 h! w& v7 [3 k) y2 k9 a) E! H5 D: n$ g6 j w
Ah, you made a mistake: it should be:
4 Y7 p. B# k( _0 w
& M! ^: H; |8 ud(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:
( ?, B3 {9 c# bYes, there should not be " ' " after "f" and "g", it is a typo. :D
! z" t! b3 q7 t5 Z, p
7 t& b4 \$ a6 k s: b6 U% R4 [; PBut 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:
6 m1 a. [# k; h7 U ^9 @6 n" ]2 G& K' S
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。
# V" P0 E9 ^4 b* z! U9 y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 23:24:0 I% t6 B/ |5 r4 R
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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