数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
- N4 {4 \# z' r- T, J/ d
' t# _& x& {; ^8 A. md [(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?
# l( B; E  d+ D0 a- h
' y3 o- f& h8 s0 T: q1 _7 ]# n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:
$ O  {( k1 b* Z, p请教大家一道微分题，多谢！
0 Z+ u% e" w% Z/ p% _. Y! Q$ N
; _2 H4 b4 U! F# v$ s4 qd [(a+bx) ] /dx = -kc +s
5 H! q2 U, _1 q; b: s3 ?where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

it is too easy:
' N& D/ \6 h' q4 M) M6 `
+ _- p% R( ]5 Kd(a+bx)/dx = b

8 G5 k# P2 K8 U# Iso

b=-kc+s
+ s- y$ j( h, G
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
, L! Z4 ]* B. t6 eit is too easy:

d(a+bx)/dx = b

so

8 d! x3 R" m; p9 _: \$ }5 gb=-kc+s

finally:  c= (s-b)/k

# t- o/ s9 `# y
: U7 o( d. t# U* ]1 B( u; S  _; F* Dthe 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:
( J9 F* {& |' A' ~# G3 H请教大家一道微分题，多谢！
7 T# t1 G  Z6 |( ?( Q
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?
# I% ~1 W9 x; ?- l: [7 x* ?
4 n0 ]( |+ o! }3 D3 N多谢了！

5 R1 M# }9 z" U% I( `+ X$ ^[ Last edited by 醉酒 ...

6 R# h. p5 T  C5 m- zsorry, did you post another question? your statement is still not clear.
6 a# K3 h7 Q& g0 V/ w
7 b' x/ C  h) Y- `; N+ M; ]a, b, s, k are  constants? c means c(x) ?
8 K* U+ v4 e4 {0 L/ T5 ]
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.

a, b, s, k are  constants? c means c(x) ?

6 m6 y' E0 u' d6 e) u5 Dplease tell me.

对不起，写错了，忘了变量c,应该是
0 p7 o- N2 ?" @) @6 L1 y; xd [(a+bx) c] /dx = -kc +s

+ C* f9 F8 s; F; N7 E  Z4 x) c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
. v; h# C, z5 y; v( Q4 pd [(a+bx) c] /dx = -kc +s

7 p+ b# t. j/ P) u3 \* E$ e多谢

3 |* q3 e  ?, _do you mean it is:
$ K' c6 ^( y  M8 ^0 c
d { (a+bx) C(x)}/dx = -k C(x) + s
+ }- w' y+ M, p% w. x/ p
where a, b, s are constants, you want to know what is C(x) ?
" g1 e7 `' C+ C: X' O/ K
6 d7 j# r3 s% {) K3 @0 Eif 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 `- E0 l# ?0 D4 \- K+ V" v+ G3 ado you mean it is:
9 E  B5 k' S) R3 c3 q* t1 B
; L! Q+ x4 z% m$ R0 b$ ~5 C- }" id { (a+bx) C(x)}/dx = -k C(x) + s

* d* E4 e: u5 O2 fwhere a, b, s are constants, you want to know what is C(x) ?

0 I3 Q- P$ ^# Z) Pif so, this needs to solve the differential equation. please comfirm it fi ...

% @2 r; y. \6 o3 K3 ~
4 B. O$ s- A. v1 t4 c. U( I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:

" ^) I7 C& B2 x) qC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


  d1 t5 d  Y; C/ U. [0 E; Zprocedure:
' y% C7 G& u# Q- K! B
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
. L3 `. \3 Q* }8 A3 S# dso:
$ Q  s$ B3 `( G* t$ p
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

6 J& {. }9 B6 C7 ]; R# ~(a+bx) dC(x)/dx  = -(k+b)C(x) +s
6 T% O' G* d& F6 G
$ i; `5 e: Z1 B) b
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
# _8 o1 |9 l9 a" f) o( C% Ktherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

; z: s" x) q, G% \* L! pfrom 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)

) V# m' Z2 Z3 z! Q6 w# C3 L6 k3 fthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

# K7 e3 W: v0 p3 ^9 Bfinally:

2 R1 R  W; E  _9 L; dC(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:
2 f$ }8 f2 V! B" e# D0 R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。
7 X5 t3 N- ^7 B2 g8 f; g2 I- Y9 f2 a9 w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：
/ s. Z" z( i$ f- a, M) x# c
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

7 r$ j! h+ ?! }# Q也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
$ v9 }3 y4 K' O9 m9 c9 C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。
; g" [, b/ z; ^) {7 o0 i因此：
$ }" M5 B; u# Q$ T4 x
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
" |3 K; K! c1 W; Z3 S! x; O- U最右边是x的平方。
4 V2 Q0 k0 m7 b- D1 r) [6 t* ^8 ?
! L4 q  F% b; N5 g所以您解的有问题。
7 t7 g8 N3 N! w( p# Z2 w( y- \
% G2 c  h& l$ Y$ \/ U[ 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:
  z( l( T. `3 Y( e  }  r- p' E9楼的答案有点问题吧。
, \+ O) e* Q) }您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
5 ?' p" d5 t6 z9 ^5 h1 p3 Mso:
' u2 d$ N! g: JbC(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
9 Y5 N8 e0 f1 R6 a$ p* E4 P9 |8 {& F* E但这是不正确的
d{(a+bx)* ...

& z$ P% n  W: u4 Y8 B9 s
Please note here:
/ n; i! ^, Q; P# r
9 c4 [8 N$ @% P$ c+ T% ad{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
; W6 A! @( |" p/ P, l  w1 m  Tit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
3 g) V& j0 _! P; r$ E* }5 \$ Y/ Vd{(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。
, p+ X* t* C; |
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
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

2 k: R/ W' t, d4 O
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
/ _* n; v% O  f! 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)

" L- V# [7 T0 j1 P[ 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。

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
: ^& R, ]2 |0 g/ }  R& S/ N
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

" v# }! u+ T7 _! }  Z
' r6 A' t/ _4 h' ^这样算混淆了微分和导数的概念。

+ ^: I, y1 ^5 M) y/ G' v2 _+ Hd(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
" ?% R) j) m0 @; W9 `# c" G, ^4 Q' h# 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)

- e! z6 h' y1 J, T/ j8 Gfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
; x+ X* b) j* D. l
h(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。

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
$ t7 q' `6 r% I; U
Ah, you made a mistake: it should be:
3 D/ E6 m$ y* z: n& ^0 ?$ i; w
d(f(x)g(x)) ...

% l% x# V6 D7 j1 z3 m1 g8 w# |
/ M  n2 Z$ f7 w; y5 E( B6 [) ~. ~/ NYes, there should not be " ' " after "f" and "g", it is a typo. :D
) g6 d: P- A$ M% `3 X" m
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:
! x1 O9 q! l: U. k+ G. ^1 qYes, 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.

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:
) O/ v; t& ?+ f, X9 A' f' ~5 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。
$ P7 f% a# j, l( X$ S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 ...

4 i/ E* P0 K; w) i
5 s- G( l  o0 S8 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:
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？

. U9 d" i3 }# D' z# F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。