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:
9 D/ z, s1 y* j3 Y. P; i5 W3 m请教大家一道微分题，多谢！
: n$ |3 Z! W, a& ?$ H: y0 H4 N
0 c4 r) x/ J+ S, yd [(a+bx) ] /dx = -kc +s
! w& k4 p: E4 Uwhere: only x and c are unknown, others are all known, requiire c = ?- E* s! z7 l' n1 N! e, ?
& [4 K7 w6 [' ^3 {1 E7 V2 |/ 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:
$ N  e! a% }! Hit is too easy:+ F" w- k4 I* `: |/ U; K. ?, t
5 I" j' b" P5 Q+ v5 c  C
d(a+bx)/dx = b
! n9 S4 f9 u& E& c5 R7 B% I1 i2 H. s, O7 ?1 i' k4 O) k* G
so# m* ?. t  [* ^. _$ [$ j
+ c+ p# v- _% U; C
b=-kc+s
- _5 G* @- a# h& x; u) i. x
4 ^8 v$ E) T: j& m- |- Sfinally:  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:
& X# c; g4 p! u2 b+ j$ n请教大家一道微分题，多谢！
+ R- H; a4 D* Q$ P* X! i
. l4 Y  w- A4 |  Gd [(a+bx) ] /dx = -kc +s
/ k' G7 R+ ^/ B, \: g0 f- Qwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
" p& @/ q2 n; U  }% E( `2 ]+ [1 U' @  U% a
多谢了！2 b  n' s7 ?  {8 U4 e) T6 o0 g# I

2 D/ }  e$ n9 y1 X[ 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:
/ @3 O9 g! d5 W6 Rsorry, did you post another question? your statement is still not clear.
r6 x" J( x$ n+ I) w8 n( G+ v( j, C- ]# W9 @8 h3 J
a, b, s, k are constants? c means c(x) ?1 r! n$ M" K0 ^5 `" \3 N4 f# ?
! T3 y3 }- \( n% s5 o
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:+ Y: L! d! l/ o( K7 z) d7 T
对不起，写错了，忘了变量c,应该是+ z' h! c! z5 t* e2 {
d [(a+bx) c] /dx = -kc +s
- A/ } N2 ]6 o4 k! k( K5 d3 n& M( a% S# L7 W9 X1 R% S
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:
/ @1 t4 z2 ?2 L* C1 p5 }0 m# P5 Z! Odo you mean it is:7 F1 @) ^( I; C3 n
7 U, |5 U% l) ^5 I2 e0 R
d { (a+bx) C(x)}/dx = -k C(x) + s L7 a5 |1 p* R9 A* R3 d
! @+ ?- Y) ]: b3 x7 Z% Q' B
where a, b, s are constants, you want to know what is C(x) ?
4 ~5 `2 f7 D- d, L
8 i$ k6 u; o) L% E5 R, V2 y; x" p) Hif 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 14:33:9 {- h1 b1 @/ t" u) c
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 I% y# x: d, }9楼的答案有点问题吧。9 X6 t$ F. V3 z7 M; @4 g
您说：
# i  w% W& m+ B8 U, X, ^+ j. s/ M7 n6 ~5 O8 n7 t( Y, ]% u
d{(a+bx)*C(x)}/dx =-k C(x) + s
+ B: ]0 m; ]  _) F2 Z# i/ D; T( Dso:
$ S- T0 A3 y! DbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 k8 d) U6 [1 Q/ G( o4 X7 r& v3 b1 q, E3 }
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx: w9 _2 l8 Y2 c( j2 y4 [
但这是不正确的& J- d3 Q  m. ?/ O6 O2 |, }. ?, `' a
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5 s4 K2 q6 J$ B+ h: {! t
....
9 R/ k7 o9 N* ]8 Z* r, jd{(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:
; Z0 d% h; e* H5 W4 K/ C! a这样算混淆了微分和导数的概念。
% U+ O' C, Q3 j1 F0 p. T b$ x8 J& q: c2 M1 w' F- Y
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! G( j3 h/ H; Q+ ]7 |% M" s" _

z- Z6 S- c$ b" X! TIf 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:
' v" x9 o7 m; x3 r7 p+ F4 @! ?这样算混淆了微分和导数的概念。# }+ G* C) v$ c( w
* k& L8 ^3 V& u/ a6 ~9 `9 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
# @+ ?9 J. J; _& f
) i" T. e! T' B/ k$ j' U' eAh, you made a mistake: it should be:
; D9 _) X a. y/ N4 I/ e# p0 h& x7 Q h: u
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 20:51:2 }# R% P7 c1 b/ K9 P; v' P
Yes, there should not be " ' " after "f" and "g", it is a typo. :D8 A6 S4 a0 a0 y' K& l; H9 Q2 }
$ e7 d: ^, _+ u- Q- X; v
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:9 j" R7 h8 G# Y3 i' c0 `

' V" d; \- Q6 X: Z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。
1 o' L& f5 c1 R2 x/ U& J V! D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:
7 p" I; R: S4 v* g& R( q3 ]; M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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