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 20:41:5 ~; w4 y" t% ?7 x
请教大家一道微分题，多谢！
% ^! c# T; Y1 N% q3 h- h; i+ @2 `9 b- Z* O0 Z, \$ j8 W7 L8 e
d [(a+bx) ] /dx = -kc +s 8 @; S& S1 s9 |6 x/ f( Q
where: only x and c are unknown, others are all known, requiire c = ?
6 A) N0 | B5 W) k
9 T* J" n# V& g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: _- ` t# V+ J, }
it is too easy:
4 Z- \/ e' w: _3 T4 K; U7 a4 @+ P4 a7 x3 T5 y- [
d(a+bx)/dx = b0 P0 h- m# Q7 `+ H
! C5 e+ K4 B/ |+ T" X2 N
so* v: p0 g$ b" \5 a

( i0 w+ J) c5 eb=-kc+s1 g' k. U! n8 i
3 H4 z6 R! P. T% w2 @2 R! C
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:/ T- {: H$ h- S1 m  d& |
请教大家一道微分题，多谢！; Z' Y9 X- n( B

9 @. W+ g$ @& Q8 V: d* W$ ud [(a+bx) ] /dx = -kc +s * N& d5 C, i: K  m
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
: n4 [: f1 L  M2 B; f# ~, J5 x# {! Q! B, z# m. r
多谢了！
' h: K/ z6 I' ~
- x/ @$ {6 z4 K# v# ^$ D[ 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:
" @4 U: c9 U! C9 S9 rsorry, did you post another question? your statement is still not clear.3 C( O% I; ^9 r

& [( W5 H* o. v* Y& B9 {a, b, s, k are constants? c means c(x) ?
/ ^1 E; _6 i2 \! L$ g2 ]+ Y$ q# X" |# g( x: y
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:
; w' ~# |# F# [" `- Z( d' e对不起，写错了，忘了变量c,应该是7 Y1 w/ B& z8 P# t
d [(a+bx) c] /dx = -kc +s
$ r- Y/ Z" k2 R6 J; O3 h" Y6 M# T# V7 w1 \# @5 Q$ U
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:) R! v$ O* r! ~; v8 ?
do you mean it is:
) M$ L7 _$ j# ]5 j7 F/ z3 G4 Q9 X0 q% ]. _; ?2 V5 R" n
d { (a+bx) C(x)}/dx = -k C(x) + s
9 X1 M# M4 e; T; W# G3 B$ P! T" F, g9 x/ m
where a, b, s are constants, you want to know what is C(x) ?/ o, Y7 O, }1 e0 s" I
! o7 X9 ~$ P* A8 [
if 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:# v! v6 k. P E; t
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:( `3 l$ T* f; `
9楼的答案有点问题吧。
& [0 `: H& _# r& ]0 c  O& O2 `您说：
0 e0 q: N/ f# s) b( `  w/ T( E3 O0 G3 }3 h' r$ Y
d{(a+bx)*C(x)}/dx =-k C(x) + s
- L7 F- l/ ?9 u$ P% t/ l4 |so:( q( u2 N; E/ y& |0 @
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s! V4 _8 v3 F1 s# A" Z
4 A7 l8 O# P/ f; P
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
3 U3 J* O" D3 F( x/ r' l但这是不正确的
8 w! m( P7 j, |- G: cd{(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' Z! g) F6 O- i6 Q( x4 [! h, i- f
....$ R" ]8 h! ^' p* ^% [) }: p. X
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:+ @" e. M' |! _7 D5 l( V
这样算混淆了微分和导数的概念。
/ Z F% F7 X# X G" w& O
, C8 U# O5 s- G9 bd(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$ y/ I: @( j: [2 I& ?* W

) l. w8 h X0 X" q+ G: W4 FIf 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:
! s2 f; f7 p1 F. ^0 o. M4 U9 h这样算混淆了微分和导数的概念。
3 P& ]' u# _9 X2 ?' ?
, K7 M; T; L1 }; S, }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) I+ y: b1 v/ e2 ~# z# b1 a

1 K( T& O' R$ ?Ah, you made a mistake: it should be:/ w$ O; ]# b" I, ]6 w! i

/ J. g+ ?( v T# H) k/ R6 U0 v( `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:( r4 a4 a+ Q9 u: V, d
Yes, there should not be " ' " after "f" and "g", it is a typo. :D; r! X9 r$ d! D$ v- ^) R7 M
[8 f9 S( L+ R# B/ m
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:( _2 f5 X/ E6 a0 ?# k

2 m7 u/ V5 ?% ] }- y, S/ x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。* l( C% [/ H; A' V% ^! @* 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:% m7 b8 U4 P2 o/ ?3 L0 z1 J, D
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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