数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
- }3 N5 x  A2 o! J# W
d [(a+bx) ] /dx = -kc +s
% N9 T  W+ l$ d; z* uwhere: 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_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
9 _; H7 G- G5 G, b3 m+ E2 h1 A8 }请教大家一道微分题，多谢！
% o+ i4 w3 g! i) E
: j# l3 K2 `3 O  w' L$ {d [(a+bx) ] /dx = -kc +s
" }9 H& \. y7 a9 a* y4 \% X4 N' Hwhere: only x and c are unknown, others are all known,  requiire c = ?
2 r* c! J* V! o* e
6 U. c4 ~; Z; b多谢了！

1 |1 I. q4 m3 {8 G9 bit is too easy:
5 t1 p" F+ q( ~
d(a+bx)/dx = b
' P+ l  X4 R: P1 M( E
& l1 G! D* C0 U' w& V, Wso

b=-kc+s

: |1 {1 k6 s# s) q! [" ufinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
$ K0 M# u) k6 }it is too easy:
) l; o$ K+ a* v  Z% c9 D
d(a+bx)/dx = b

  |% F4 @' ?  Yso
# A7 g+ z- m1 R) F8 \; G
b=-kc+s

4 e* u( H9 A, R% t& V1 g9 F1 [finally:  c= (s-b)/k

' L4 D# e1 {2 `8 L* v- Q$ cthe right side cannnot contain c, but b contains c, so it is still not solved.

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
* ~. k) M! X9 R" f, w请教大家一道微分题，多谢！
9 p4 Q, f" A$ R0 P3 J
2 i8 U% i- w. {! Yd [(a+bx) ] /dx = -kc +s
( v" O. k9 L9 @) Iwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

! l" q  a0 [! \8 f( k, K  C: }+ v多谢了！
" t- N2 T, B9 }# e  F
( i8 _5 y) c  u/ z& i" m[ Last edited by 醉酒 ...

0 |, ~2 K7 ^' J5 Z, P& W3 \+ q4 ?: G
sorry, did you post another question? your statement is still not clear.
1 ^8 I) k! W% w
, `$ I0 T, a/ @a, b, s, k are  constants? c means c(x) ?
; {1 I$ S, F; o2 _
please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
sorry, did you post another question? your statement is still not clear.

% T2 z% A8 c& R7 g( W/ l' e5 j4 Ia, b, s, k are  constants? c means c(x) ?

please tell me.

  w& p- x$ A5 G: |  C* |对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

) o; C/ E* {0 x, z: ^多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
对不起，写错了，忘了变量c,应该是
6 X! t' B5 }# G: Z4 z" ?d [(a+bx) c] /dx = -kc +s

多谢

do you mean it is:
  Q& H3 G& Z! N; x
* Y+ n1 ^$ Z( td { (a+bx) C(x)}/dx = -k C(x) + s

* }5 `# ~* I" |; A  ]0 U! F3 Lwhere 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

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:
8 S& R1 I. C, E1 h3 ~do you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
2 I7 [, h, T- Y* y* W- i
, v# I4 l. E; g% `where a, b, s are constants, you want to know what is C(x) ?
. ?( T. _, P8 o3 G  I3 y
if so, this needs to solve the differential equation. please comfirm it fi ...

$ o) U1 y- q$ m" K: Q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_

answer:
. f5 i9 ^. F, u- E/ g1 p
& w% F* ^4 {7 K& t; D* rC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
* L; A9 J8 x, {3 p- L* F; k

; L2 t7 Y& i: {# |- s4 mprocedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 P. |- o7 j1 _  y% E. W/ P5 lso:

5 r! J2 c5 W) f. c  Z# y7 ]/ G4 FbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
: r& h  A7 s) y- m% P7 Y9 J; Bi.e.
% P# b' ^! ^0 s- f
. ~' \( ]) G0 a2 I7 |+ q) a" \(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
9 R& I8 x* _. @3 ctherefore:

0 H. X/ h/ W* m& k' ]8 m8 U# {- X{(a+bx)/K} dY(x)/dx=Y(x)
" ?: k# N! E6 t( j6 n; f
from here, we can get:

$ F' a: R% X, V; K6 ^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)
2 o( n$ g) E/ t4 \9 U
" N* P) b6 J! H1 j3 Rthis means:  Y(x) = (a+bx)^(K/b)
+ M. m, K7 L0 Cby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
) _/ {- h; d- e, {
6 ^" D- ]6 [2 Q! D  @1 t7 r* dfinally:
. a$ t2 c3 H1 v; B1 n& K) r/ q6 C
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)

醉酒当歌

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_

Originally posted by 醉酒当歌 at 2005-4-5 14:33:
7 C( ^! c4 Z. Q. Y+ s4 s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！

; L* M* B1 J( h0 R. F# p4 z3 x1 M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

9楼的答案有点问题吧。
您说：

% y) K# T/ |; c+ T  |* c& b" T# j2 xd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! R8 e" ^. c/ Y& \
/ s  A3 J; Y$ g8 P. a) j0 ?也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
* @3 Y1 \! x1 }' g! v" }但这是不正确的
  R/ L/ J5 f7 W# Zd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
' S/ T# B5 o6 g! s; O; j* J2 g并不是   (a+bx)*C(x)   的导数除以   x   的导数。
* b$ n/ S" ?# {2 F7 l: _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。
: e3 \& C: ^3 F) Q因此：

" s+ G& @" l1 _- od{(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。

, O0 i: o$ O" x[ Last edited by 龙舞九天 on 2005-4-6 at 07:35 PM ]

十年移民路_

Originally posted by 龙舞九天 at 2005-4-6 19:34:
9楼的答案有点问题吧。
您说：
& `% {! R; F; K1 K0 Z* D) W" U, v
6 L) v8 q8 ~) L& }d{(a+bx)*C(x)}/dx =-k C(x) + s
' s" @2 C* b0 L$ sso:
8 O8 J& k  J. s- _5 _1 B7 }$ GbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

3 y4 B1 [6 Y! k6 p$ z, @也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
  z6 I  r8 u; J7 _& ]5 f但这是不正确的
- n4 V. z$ Y# y5 N; xd{(a+bx)* ...

Please note here:

% F: \6 \# G* E& b6 L% R* `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_

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
! ]4 C& G* h0 i8 K并不是   (a+bx)*C(x)   的导数除以   x   的导数。
/ q8 {; z4 {5 c$ h/ q
no "x" in that position.

gnome-kde

d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
* l2 ^6 x8 |7 W  K  F- G0 Z....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

, g) ?! k6 A4 y' V
0 g/ d- m- i! S2 Q+ q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
, Y/ s% i# V% k0 y/ x
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)
' N' |' O/ l8 ^
% B6 l( r- |# ^$ I[ Last edited by gnome-kde on 2005-4-6 at 08:31 PM ]

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:30:
这样算混淆了微分和导数的概念。
+ r, H& G  V$ s0 j
5 C9 U, f6 k  B' @" Zd(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
* X6 w+ Q0 F8 Z  a
; L7 p& O6 o3 C7 V- o+ p! o8 FIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

# e! S$ B) ~. i# U
这样算混淆了微分和导数的概念。
% q/ H. D5 k+ ~* P( m
6 s/ j  r  T1 o$ P6 Td(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

% M4 v3 o2 ?3 }# o; M, zAh, you made a mistake: it should be:
8 h- `/ t) Q% E
  y: G1 K* d- m- c" n/ J4 Ud(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
/ r2 B" {7 f! w

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)

1 F! A& G8 k1 e* X' v" W9 mfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
6 y# ]2 i! d, x
+ X0 f( W8 a5 J" K  l9 y* rh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
这样算混淆了微分和导数的概念。

/ V" e2 v# d5 Wd(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% e' f+ @( c$ S" q
/ e( p; b" d" y, fAh, you made a mistake: it should be:

d(f(x)g(x)) ...

0 x+ s5 G9 N1 u( |2 q% _
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_

Originally posted by gnome-kde at 2005-4-6 20:51:
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.

( v5 J, q. S5 W% m; f
Thanks, I am happy here to discuss math.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:

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 ...

) j. A3 j# b- y/ z, @/ Q1 h
0 y* Y! o% u# D2 ^) `- C7 G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_

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
/ |$ ]% @; m- B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？

' ?% r8 y+ D8 _8 T( o, F! @1 J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。