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:
3 B; Q( l3 m8 o) C+ Y" j. x请教大家一道微分题，多谢！$ l% Z/ H9 b3 j5 B- J! ~+ n

+ C, i! K2 U5 C& L: C# D. ad [(a+bx) ] /dx = -kc +s
/ ]% Q6 S/ S+ F: j5 kwhere: only x and c are unknown, others are all known, requiire c = ?+ G7 }: N. ?; g
7 `' [# t# c* b1 i: L
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:
; M- Y; U9 r6 a! pit is too easy:
; ?( r t: ]7 A+ h# E
/ N! V" h& Z0 }9 T/ Pd(a+bx)/dx = b
) k8 M' Y7 E# O+ C, Y) I& E6 E: E" m
so
0 u& [# `; }9 k: N% V- u+ ?& e* }! }# s' O- T% l5 m
b=-kc+s7 v$ V, a, G0 @' k

* d# T# b2 a# ]# |! Hfinally: 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:$ A X f8 @) C4 @3 n
请教大家一道微分题，多谢！
; g6 b7 p; V& o6 E0 P6 {, }
0 \+ f( l, _/ c* |0 Dd [(a+bx) ] /dx = -kc +s
' v1 ^! s R9 Q7 Qwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?- N. d$ r T) m- W- K, [

& Q3 ?6 M9 n4 E2 ^. d* c多谢了！
% W) [; s* a) S
# q- N1 m) F9 q: G1 \[ 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:. M; N/ _. t9 @& h" u$ p- g
sorry, did you post another question? your statement is still not clear.2 K* c/ `8 l+ {+ D8 G9 S) N

4 ^+ I" {. Z$ {7 \0 K+ n3 k" L& Xa, b, s, k are constants? c means c(x) ?
! u9 S' n, u' ?' Z
4 p% F; P' Y9 s- dplease 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:
b9 n `( j7 L3 _4 b对不起，写错了，忘了变量c,应该是' V" S, x) @8 ~# {! A& \; ?! t( i
d [(a+bx) c] /dx = -kc +s
& _/ p) ^' M, x$ `0 ~- ]( ~) l! h( f" g
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:
6 I( j) y/ O6 V1 Q. @) }9 _* d, Edo you mean it is:
% _4 ]3 g6 F4 \0 Z% f
) h9 G% n/ |0 w" `+ n6 _d { (a+bx) C(x)}/dx = -k C(x) + s+ b k" `; Z( l- C h$ x1 I8 d% v
# Q, b, W6 H B7 a' R( s! K
where a, b, s are constants, you want to know what is C(x) ?5 t$ ]% {4 M- b2 T

: ]6 g- v& Q# 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:
$ y, t" l, h: q7 [8 z w- _- |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 ^2 h5 K: [3 j; @: L2 n2 M' d' b
9楼的答案有点问题吧。
, Z" C2 Z* d7 s# Y- L7 w您说：, E" p7 g4 U' K/ }* `

6 ^) l; }; T, U& R# W3 vd{(a+bx)*C(x)}/dx =-k C(x) + s" K7 s+ o1 x, v- T+ o2 t3 w8 j
so:
/ ?1 o( b3 ]( N2 \7 _. z+ \4 abC(x) + (a+bx) dC(x)/dx = -kC(x) +s
4 A5 v# u3 }7 B8 q! w3 V# G. r3 [. E% G% H5 a* _! M
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx- R+ j7 Z: e8 [0 E
但这是不正确的- Q' e) \# m6 x' G8 J" r" ]
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
! y; i& P3 d8 l% Z- d& d....5 b" @5 G! M. b8 {
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:
# S! F( D- I( W/ s( [4 O这样算混淆了微分和导数的概念。
: I7 C# C) M: R5 ? v# H
9 N7 L! k* E. ?; a# E# k0 sd(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: n z( w9 K6 j8 k. F! T/ \

5 R" ?7 p) E3 T7 w+ 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:
) W, o1 M2 P4 T; X这样算混淆了微分和导数的概念。  U: ]& `5 d' z4 F% t, G4 ]

3 ~8 l; t; w8 G/ I( jd(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
# x3 q* z& T" N6 o
% O5 i  w) C* J7 g; u1 E1 N' ~Ah, you made a mistake: it should be:
* h: j- y" j( b# P( K& i7 `& j8 ]( |
6 {  ]; s3 ^$ F$ y& R: S4 Vd(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:
" X7 K0 r3 T _, T% {0 v# U9 vYes, there should not be " ' " after "f" and "g", it is a typo. :D
, h9 d6 t8 k& v7 P3 F0 A7 A0 z. K: e9 Z% L% U
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:# y& X5 Q' c* K) a/ @' R% W2 L! Z
" X( F6 s! x( x7 O, [4 o3 C
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。2 k1 V4 t' V# w1 i% s1 A6 d- m) W( 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:
/ R6 Y* H# G, W2 X$ K. S' B9 i0 q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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