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:; v; }% u, V( h8 b9 U n+ R2 }
请教大家一道微分题，多谢！
1 V" b! Z8 g+ N3 D- T
% P! A- C+ y$ D. ^! x3 @' V5 ]+ Od [(a+bx) ] /dx = -kc +s
; C: K2 D% U" m' Ywhere: only x and c are unknown, others are all known, requiire c = ?
2 A& L, T3 ]3 C
8 m+ `3 @' \* {9 q& M9 q0 s& p, 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:
( y& Z8 n- O8 Q$ Uit is too easy:
& i! Y9 f: }- z  S2 y; [; Q8 @9 u& D
d(a+bx)/dx = b
5 r+ |8 i5 M( u- X9 {1 k5 N) U( k! B' M( e  R. g1 O% ~
so
" X% z/ G4 M% N5 @1 n! z1 }
, y3 P. ~8 W1 [' ~b=-kc+s' U2 j9 l$ ?* }2 s+ B+ r% |$ O

+ W: b+ N3 b2 s9 ~) X$ nfinally:  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:: _: B) _) g1 e2 w
请教大家一道微分题，多谢！! y6 c. W" D! h- S' \+ R

* M+ Q* w. J" r6 D, c6 Od [(a+bx) ] /dx = -kc +s 6 i9 J: h& ~- V$ K
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?4 @9 h P& q3 b
6 Y1 D- J2 u5 o& p2 ~+ X
多谢了！
- z3 \, D' |8 s2 y' l" c3 d3 u$ X5 k
[ 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:/ F/ H" R0 A# v, K% |; n5 a1 {
sorry, did you post another question? your statement is still not clear.
; A* `1 l1 w4 k" u& P
( O: g z: ?% O" ma, b, s, k are constants? c means c(x) ?
. X' l8 B% E' |- @# [
4 B8 I! R; ]( v+ t `3 _" E, @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:
K! R2 T/ Q, _对不起，写错了，忘了变量c,应该是
4 \$ L! f5 V5 x% i6 P4 o) ~2 yd [(a+bx) c] /dx = -kc +s
% @- Z/ E& u) K
: _/ e1 y, `; z# x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:8 D( K m% p. x
do you mean it is:
. }: d+ y+ M& [$ h$ f- k* ^( N. S
d { (a+bx) C(x)}/dx = -k C(x) + s
+ K6 C8 m" a! z
7 \& i6 {6 m$ g) K, gwhere a, b, s are constants, you want to know what is C(x) ?0 K/ Q, S" V3 B, h8 ], \9 S
% K8 i: I! D" |# v! s% s! P2 k
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:
$ G6 b* Y5 o; G6 e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:
" S. J8 J, C3 M# W2 k. }9楼的答案有点问题吧。! r3 S! ]$ c# F) F
您说：! f9 V  O) N2 Z4 p4 k+ V

" i- p9 A; }! t/ l% z0 |6 Cd{(a+bx)*C(x)}/dx =-k C(x) + s
4 @' f8 [6 t" N4 [: [2 ]so:' y% `% b! |5 ?- U4 J5 @
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s! q& O2 F# [% H' j" [' B# T+ I
1 b( |( g' v5 ?
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx( m+ K5 A5 I$ }
但这是不正确的9 [/ Y; X2 I5 W  ?) l# p1 ]6 w# _' Y
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
" ^% @7 d# \% u....
6 I* u$ H9 B: B. Cd{(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:
# y4 _; t4 c9 @5 e, v8 s0 s这样算混淆了微分和导数的概念。
* U) o$ A$ r `2 M4 Z8 T! P5 ~: B7 u0 ^. i9 r2 F+ Q$ Q
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
% |1 ]! e8 r* n: z% b% ^3 o. f
If 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:
& t% U7 v5 ]& ?; v这样算混淆了微分和导数的概念。- M, R: g+ K) F

7 J/ m% Q- \' { [( a' R( x: V# ^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
+ ~2 W4 W/ K' k& M! G+ C8 Z" I( m) l" `
Ah, you made a mistake: it should be:1 S8 z8 m* d' p2 g: X- R# A
5 B( z; N2 \) @ X
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:' N4 M/ V& r/ \& b7 }
Yes, there should not be " ' " after "f" and "g", it is a typo. :D$ x- }& ?3 V9 Y7 L2 O/ X u8 y

1 t( ~% d Z9 S5 R% ^) K' SBut 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:% E( `' a- N2 r6 r( S7 |

) m2 ~* E4 b, C6 i- i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。
% E6 n; Q5 W# [8 S7 y% L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:
% [3 Z1 o2 g% L: s2 D6 l$ K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。

账号		自动登录	找回密码
密码			注册

数学高手请进，多谢！

浏览过的版块