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:
7 t* a3 ]$ A+ I请教大家一道微分题，多谢！
% B  `7 j* g6 ^- r: U# a* _1 h) w% ~9 i0 O  y( c
d [(a+bx) ] /dx = -kc +s 1 v! `7 |: y# W5 A& Q( r
where: only x and c are unknown, others are all known,  requiire c = ?
. v' ^" W, U* D! |# y; T) ^7 K" K2 Y  ?" @" ~
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:2 {% n2 p3 z; i
it is too easy:
& U0 e  s$ e5 W* \5 ^5 Y& o  ]! A
" D0 p) e8 e$ U6 X& s% id(a+bx)/dx = b
; ?7 |4 [/ _" }# l+ p" m
4 {# ]( q* H. M. f! [9 t6 sso& f1 G6 X9 R" u1 R% T

4 J  F  n/ j, T, k9 D' Cb=-kc+s
+ U# Q- W+ V* d. U, S/ `0 k& S: f4 r$ m5 V0 I
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 20:41:4 M" e$ P$ u3 p# M: _
请教大家一道微分题，多谢！* g: f S U5 `- e+ j
5 I9 a5 n. G: e6 x0 \0 S# J2 Q
d [(a+bx) ] /dx = -kc +s
+ d( F0 H/ b6 [5 O+ wwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?" {' u$ F" }- f7 Z0 w5 U

9 J- \. |9 z' X! X" I多谢了！6 D4 ?* p$ d# t9 y+ b7 S, p& n" Z/ c

( K( V5 H& O, T& R& d3 T% S[ 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:; o* \0 Y, N- U: M
sorry, did you post another question? your statement is still not clear.
* F' J! y' r/ t* n2 W" ^/ I0 x; d+ F* r
a, b, s, k are constants? c means c(x) ?$ x+ S/ [. ^! I3 r

& V! ^8 p) I8 f$ aplease 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:
$ d: V$ G$ W) N- Q2 M+ [对不起，写错了，忘了变量c,应该是
7 V' e* ~) y4 \! Td [(a+bx) c] /dx = -kc +s
' {! F, L9 M7 y" I, }, {% x/ G9 a! ~2 ?9 m* ^% R4 y
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:: x( J8 w+ D3 k1 [
do you mean it is:$ ^% R6 R ^, J& G1 p- f2 L
, O; o% P6 ], @8 ~# m
d { (a+bx) C(x)}/dx = -k C(x) + s
0 w5 m2 O a% Q: {% k. V; c2 Y# |0 y6 U
where a, b, s are constants, you want to know what is C(x) ?( h0 x) {) [: v1 L% T5 i3 j
( o, _# o& L( i4 q8 y& S: L- ?
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: o' K3 W* s7 U( X. t3 x1 U# 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:
1 n4 ~6 E. i" c! C; j1 A9楼的答案有点问题吧。
4 u1 G8 ?/ M( I2 E8 F8 U2 B* u/ `) G/ g您说：+ S8 X  Z! p9 p3 |

. ~: I4 s" w8 d5 u1 C$ f5 j( gd{(a+bx)*C(x)}/dx =-k C(x) + s
+ e& y0 `% x' i1 X& y% Q' sso:/ O1 P4 g% V# N2 z2 w
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s7 ~1 c) Z# [$ t3 K, B+ m

& a/ a$ p$ f$ b5 ?/ _也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. A6 N5 m2 t0 S% F, u) y5 t3 v7 f但这是不正确的
: [& C  A3 R" q# Zd{(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1 `7 |" K9 O% a
....
$ ~$ G- z) t9 _, nd{(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:
$ O4 n! ~- ?5 y6 Z% j8 Q: F这样算混淆了微分和导数的概念。' x0 v- z2 B- W3 F5 N8 m/ a

2 D1 l, o- G% F9 O! h0 q3 v& y3 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& b% N. f7 X$ z4 M

8 F9 z4 |4 a7 A2 v+ u1 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:4 A7 p/ [0 f) J3 l. i- G7 [
这样算混淆了微分和导数的概念。- Z) w$ E6 A/ ?2 q+ R

3 V ]& \ P. ^0 Y& Fd(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0 M0 Z9 \3 D0 w7 L" L
3 K j! ?* T+ j! e: [7 s
Ah, you made a mistake: it should be:& [0 i9 V) y" g; a: F
' I- m; q' M8 K
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 20:51:6 {1 [9 r9 n2 R* m
Yes, there should not be " ' " after "f" and "g", it is a typo. :D/ C0 e: H6 x& \+ Q
, ^ c H) @; n! F
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:& n* X5 M3 E/ j* h9 V

5 w- y Q* E& y+ X$ p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。
7 C; C4 E; S( k9 F" j: w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:
6 L$ P( ^) _3 h7 c/ a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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