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 O! j# ~3 r9 y3 O  R2 c
请教大家一道微分题，多谢！
+ z; {. S  m/ Q$ c. q5 ]0 G
! f; @' k7 a; k* R% f, Cd [(a+bx) ] /dx = -kc +s
- E- S( t+ Z$ swhere: only x and c are unknown, others are all known,  requiire c = ?
; q7 P8 K& Y! ^. P/ x( K; h. H+ G6 {4 |
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:5 w( s, @3 d. H
it is too easy:4 d$ N5 }' g5 a( |) x$ W

! P( r; j' ]0 z3 `% Ad(a+bx)/dx = b" D+ m9 ?: ?' t9 t

% p( x# {$ V' G" n6 n( n3 a' @4 Lso: ^1 o: U: n8 r

# L3 r y6 U! ]' o/ Tb=-kc+s! g; }& m" w% Y) N" y

/ e7 r3 ^9 F: n# \ Wfinally: 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:
% R3 u* l# L2 L7 n请教大家一道微分题，多谢！
6 |/ i# e! ^8 G
* J% @9 F, x+ C% ]7 E4 y1 U( {7 ld [(a+bx) ] /dx = -kc +s + K1 z2 k! a( r" d/ ~ u
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
, J; l$ ?% y* Y8 q( W8 F
" X& a5 R% ~- Q' E2 [; G5 |多谢了！5 D( `. X7 c5 j% t

0 L1 \- U$ J$ t' d7 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:4 \4 z4 R3 m" ?* s' W! V6 {
sorry, did you post another question? your statement is still not clear.
+ |0 t" t- b5 c& r- Y' |
& n% W) _; j/ q4 I. d; f: k4 [a, b, s, k are constants? c means c(x) ?
, |* Z0 _" T( ^* z& p. ]1 L L! s$ f& y- ^; c m4 [
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:
& D) l3 c7 [: ]' }: e2 y, v对不起，写错了，忘了变量c,应该是
6 d3 F2 c3 F2 S% ~d [(a+bx) c] /dx = -kc +s
8 l6 b G8 q) [5 v1 M- x
: C) Z5 T& {/ i( G$ R# Q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:/ k J& s5 `- E; B
do you mean it is:
/ I" v" j8 W5 F0 I1 L' a3 o
* [6 z1 G5 S/ W* @1 P* e0 ^d { (a+bx) C(x)}/dx = -k C(x) + s5 v! a+ d7 ~! u) U
: N% Z" Z7 F$ w' p
where a, b, s are constants, you want to know what is C(x) ?
3 j) O! v' h* ^& S4 B2 \8 @3 r1 w+ s5 T
5 p! ~4 f% z& ? Y" \0 ~/ T* aif 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:
. Q0 p' G [: w* k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:. W" b% v/ _' K
9楼的答案有点问题吧。
* q7 C* H9 A9 N' P0 D$ _您说：
9 E7 Z. \& _- g- U
d{(a+bx)*C(x)}/dx =-k C(x) + s8 d8 |2 W; ?/ |8 j( G) R: {5 q
so:" H* W7 p4 G. c" i& f1 x; d( l
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s& r- K' k [ f7 d

- P+ g- r5 @9 F6 A# c$ C- i也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx; ?3 ~" x6 W( l: y& @, L: f5 [, w/ i+ |
但这是不正确的! e! f- R* m3 Q/ U1 {
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" W* V8 P2 X- r* I, ]
....
3 n. A" {* {$ fd{(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:
5 Z {0 L$ z0 V/ B9 O) B这样算混淆了微分和导数的概念。$ H# `1 S& @2 ?, m& b/ E
, G# I) {4 ]+ w+ Y3 a8 t7 f
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
: X, T t% |5 X/ m$ p7 N
- C7 _/ N& ?) }6 a0 h+ }( w# C& zIf 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:8 ^0 t# J7 Z, o: ~3 b$ `! e
这样算混淆了微分和导数的概念。
: K' y, P3 h7 m S/ p3 n* H* k+ z" u, A5 r
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# ~) X' J' e( \& l
) r% C) ]9 |7 ~5 @& P
Ah, you made a mistake: it should be:
* _" |3 ]5 l3 ~, j& `0 R
* J. @6 y% G8 pd(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:& F6 c) F9 \1 a: @& `
Yes, there should not be " ' " after "f" and "g", it is a typo. :D6 f7 u- J$ R6 e: Z- _% g& G
2 A5 H, c2 w, z7 ^! H# 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:- H) B- v& T" e- F( x+ P/ g
+ y! r! d! y8 L# 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 q: [/ k& \- S! f& l% e
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 23:24:4 ~5 V& X& }; J. Q+ [- u
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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