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:/ W. I; V0 I& J5 L
请教大家一道微分题，多谢！8 ?3 P; L/ M6 ]7 r1 l: i

, c8 ~, R0 F" d2 G$ f5 ~8 ~d [(a+bx) ] /dx = -kc +s 8 J t5 I- Z! z) X8 n& M
where: only x and c are unknown, others are all known, requiire c = ?, K! }8 Y/ J" B) K& j6 D

1 f! U9 b6 ?3 }, c6 r) g/ Q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:
$ |, c# k* p  X7 v) e9 Qit is too easy:
* h( k* T3 Z  I0 g; ]8 Q
5 T4 `2 B% p' O/ t1 n  g) U+ Bd(a+bx)/dx = b% `# G1 r, Z# I, `3 `
- l7 s* j# D' Q! Z4 x
so
% g! o, p' W: O" b5 U% l2 |4 k8 C0 `
b=-kc+s
: f7 T% K% k! q+ x. O3 M3 `
: Y2 C! ]% P- ~$ vfinally:  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 20:41:1 L" P# h/ d! G3 L
请教大家一道微分题，多谢！
( v1 t3 d5 P/ e! l. A1 K* h/ N* F, y0 v" x% q
d [(a+bx) ] /dx = -kc +s ; o) l! z% u# ^
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?+ s" k+ p; C) A* k5 g2 H

& {8 ~ n5 M' U5 F! m多谢了！, x) }( ?7 d$ ~1 S

; i2 }) T. z& h2 M2 }/ A( z[ 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+ D1 `0 J2 z& W
sorry, did you post another question? your statement is still not clear.
2 l' \) k7 Y7 T- H8 Q$ y# p
* y9 o* n5 L5 |/ L$ p/ A( o6 Qa, b, s, k are constants? c means c(x) ?
( T1 J* @: G* C; {
' a& C$ ` O( f4 H! g3 ^4 E1 uplease 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:$ H2 Y9 d: R. `, i/ I O1 G7 k3 U5 m
对不起，写错了，忘了变量c,应该是
2 o1 V/ ^, J/ [+ Xd [(a+bx) c] /dx = -kc +s ' C8 n# _ i V) D1 w7 ] X- a# A7 ]

" [* h+ L. m! I; O% i. L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! A3 W+ Q; D% ]7 D
do you mean it is:
' r0 P* c: @9 P  N4 I+ l* F
5 g5 C( `- P7 R* Z. M' U% bd { (a+bx) C(x)}/dx = -k C(x) + s
; K* V$ y+ X; D4 }9 i4 ]! q4 }; k# |7 m, J, o( P
where a, b, s are constants, you want to know what is C(x) ?7 m$ O% z+ V  E& n; y( n

: y1 W" _' @0 U  vif 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:
, c0 [) m/ _' Y! O: D" i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 t) R$ x6 p) r
9楼的答案有点问题吧。- V" X+ g$ G. ^0 E- J5 L
您说：
' I; L; n* n, A J5 {8 ?8 w- H' l' C a
d{(a+bx)*C(x)}/dx =-k C(x) + s
1 T, d/ q6 t+ p; A& Qso:
3 @3 [$ N: V) F) W) xbC(x) + (a+bx) dC(x)/dx = -kC(x) +s! u9 R6 }) j! f& t
3 O" W6 J+ x+ a9 s( I7 @2 F
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx- _# _# A8 ^5 O! F3 z) k6 t
但这是不正确的
/ p9 }" P. M7 s- G" J' N4 m3 md{(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 _% A5 A4 ^! Y* o; K( a
....2 P4 y' K$ q7 z4 ?; H
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:+ A* c, S7 {1 _
这样算混淆了微分和导数的概念。4 i( D* d3 h+ V5 P) W4 ?( d% y/ u

6 l M. w+ m- m* Ud(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- q" e! _" M; K

. o" v& L M6 {: ^3 lIf 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 {" v3 ^: ?7 L' f9 D这样算混淆了微分和导数的概念。' Q) {& t! b5 \, K: s4 L
+ X: R6 @' D( D3 l( a) {
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& S v% E' r- F. T, H b

! [- M& I+ k9 g& EAh, you made a mistake: it should be:5 ~$ v6 i. {- P# f' \4 i2 k* H

f6 [2 L- y6 ~* q4 id(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:
& B) J9 Q S- W, M& [Yes, there should not be " ' " after "f" and "g", it is a typo. :D+ J6 ^3 _. W b; P- ?5 m; Z% V/ D

: y* {5 j7 w) M- DBut 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:- |* X4 N" H+ W* z
" g. n( @( h$ d! y0 ^
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。
* r6 h% K; V. Q4 ~8 p6 C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:/ Y5 C) `+ f5 B" Z- H; ~
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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