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:, I6 E3 ^2 {0 }5 Q* }: v
请教大家一道微分题，多谢！
6 r! u. U2 Q$ ]% a$ J* z2 |- V. k
d [(a+bx) ] /dx = -kc +s
4 W8 d0 H% I/ S1 R% I/ l. `+ E% swhere: only x and c are unknown, others are all known, requiire c = ?, r B$ F7 P6 L( N1 c0 B
/ L7 E F! _ ^, ?
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:" o# x' E5 [& P
it is too easy:
5 O& b4 I0 e7 p- G/ ^* F8 X
- J8 e+ V5 ?( `3 ~3 @  ?; ~d(a+bx)/dx = b& @' ^; v$ Y/ S0 Z

) S; x% O) N- G- r/ y% B) hso4 q% S( j! [  l8 |8 \6 }

7 }# ^  j' |  _* U8 v+ I7 Nb=-kc+s  m( b7 H# v8 l$ d* ?+ m$ e$ J
# f7 K. P7 q5 ]0 J$ f" c
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:
' k5 V  w3 H/ [; C请教大家一道微分题，多谢！
0 U% Y0 W5 Y8 ~/ E% W% o8 Q2 s% Y3 o8 L8 r; |7 M% Q" v. Z# d. a1 m
d [(a+bx) ] /dx = -kc +s
" m0 N9 ]6 U( B2 [" fwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
2 O7 E1 P/ `) u( c! k
  Y  h2 l9 V  [9 L多谢了！7 f. l6 o2 e/ R4 J" [6 ]
. z8 v9 h( ~, l7 ]* |# n
[ 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:7 \9 B) g, Q. t
sorry, did you post another question? your statement is still not clear.# Q# {8 @4 U# ~% s& Z

& Q' r/ i- t3 b% o: D/ \a, b, s, k are constants? c means c(x) ?9 e* E. z* t! R8 c5 {% v" t$ X' K3 n

7 u2 U: H& x1 D& z; {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: Q, m8 g4 n6 I' w& H
对不起，写错了，忘了变量c,应该是
) @) B$ x" q2 |" _ kd [(a+bx) c] /dx = -kc +s 8 w8 n+ G) R9 ~- U% x

& q. N; e0 U0 T1 D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:3 \1 p8 y9 v, @: I0 K7 e
do you mean it is:
# O7 N* F# M# {3 g+ Z4 {
5 B3 _& P* V! y; X/ Td { (a+bx) C(x)}/dx = -k C(x) + s
6 Q2 i% [  N5 N6 p5 K5 z
8 }" m& F  @" ~  X4 Nwhere a, b, s are constants, you want to know what is C(x) ?" n* W# j: r' J

2 s1 p1 c* \9 j- a1 \  |& gif 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:
+ g+ e3 `% C. R: F" l9 Q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:, p3 K, \; R* d1 C& U$ V4 e
9楼的答案有点问题吧。4 {! ?- r( E# y$ V
您说：$ Q/ T2 W5 o* U2 @/ P, t( g
8 m9 G! S9 B% }% l4 x) O
d{(a+bx)*C(x)}/dx =-k C(x) + s
# h! a3 `" {% i/ `" Z2 X( Q( ]/ bso:
7 b7 J- R2 _2 m! a/ C, z( t; d( g& A' KbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
. L/ y. ^( Y6 t7 U( k7 r9 d- i5 I. c, _
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
6 X$ M4 r: Y1 }4 J$ g. ^但这是不正确的1 Z" }" d7 U7 V' U- d& Q5 R. e
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3 X/ ?5 m+ J+ T( t) Y h- j
....
9 @; w' Q8 P$ F8 J) T/ A( R9 Qd{(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:! L+ C" n Y. k
这样算混淆了微分和导数的概念。3 ~& J+ g6 h$ [% i+ M

) w; S; \3 x( n" g1 N7 f8 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
) k" M4 Y O1 C9 G7 ~+ x" }4 o4 s: x4 U9 d% O" C1 e) j
If 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:
' l) \' z" }* {- g  o0 C这样算混淆了微分和导数的概念。
! a/ ?; B( O' L) O
- Q: e  q6 P" C6 x, u& i) ]$ i# 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
$ ~4 L  U8 {7 }- y8 r+ a, k
3 k# @. i7 Q! _; M) p( tAh, you made a mistake: it should be:1 `$ A$ S7 l4 A$ y
/ M  l6 B7 i8 Q% g
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:' h( [4 z0 ^* X# y. u
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
; ~ h' B- r, D) W9 s! r
8 u6 |$ z" S* D' c6 V- L# H9 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:4 ?0 v; Y: W4 f) P3 g
1 K' [9 _" a, ~$ Q4 J
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。: A( k! c2 a# o! K2 `7 q1 c5 H
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:: O* F! o# g$ J0 d: n# S. W
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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