数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

. Z8 U. D: a' e  N5 vd [(a+bx) ] /dx = -kc +s
# D: s/ @! u) I2 X& J+ Lwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
* O2 X4 ^& {  }% Q
[ 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:
请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
- z- ]8 }( |- ]; v8 L7 ^6 Z* Ywhere: only x and c are unknown, others are all known,  requiire c = ?
2 z( a0 [+ _" L' k
& }1 |8 {2 |, w7 ?% }1 C( X. {1 J多谢了！

it is too easy:
$ R; g+ M3 g0 Z/ r& v7 h/ w. j- d
3 o# }; |/ n4 d6 g/ Y* t& \  g. h! Jd(a+bx)/dx = b
4 C4 D) B# z$ j
so

b=-kc+s
# v# O( X; k- m
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
it is too easy:
5 M  H) M. S0 X. H
d(a+bx)/dx = b

/ l' |5 l: I, b8 cso
* h/ ]  m/ X: h4 C: J' @
b=-kc+s
/ N( Z2 [, x9 Z: |+ N0 A! s% u
finally:  c= (s-b)/k

2 d' x  _/ k6 Y0 {9 j( `the 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:
3 e: ]! A* f, D( K0 r5 ?请教大家一道微分题，多谢！
: f: U9 s: _8 j' `- c8 |  U6 ]; _2 ?
) e% m- W4 A; q7 dd [(a+bx) ] /dx = -kc +s
4 }* v5 Y& p1 [. U% z8 |4 Qwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

6 {. ]" c# T  P- }多谢了！

[ Last edited by 醉酒 ...

: X; U# E- M0 ]2 b0 S
! w/ j9 ]) V; N1 }6 ksorry, did you post another question? your statement is still not clear.
& J4 E- G& b  @; K: Q/ k
a, b, s, k are  constants? c means c(x) ?
# g3 P& d0 n5 O: w3 y4 s
  u2 `4 v6 `/ ^0 D& Iplease 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) ?
# q: H2 g  o" `! ]3 \# r
please tell me.

5 j' `$ e3 N: r" c# q3 a7 c对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

" B8 s& R3 w4 F3 E多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
* C2 \6 l5 E$ N3 t1 z' @对不起，写错了，忘了变量c,应该是
  l4 a1 v* s' L9 x' f. }* pd [(a+bx) c] /dx = -kc +s

多谢

* r: h5 Z# V: D9 U0 s
do you mean it is:
3 n% ^/ U9 D. N% C. T
d { (a+bx) C(x)}/dx = -k C(x) + s

- \) l& G1 B, t3 o, M' @, K% r" Vwhere 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 _2 n. z; i% `$ r- w; rdo you mean it is:
' J0 a4 q& J- K: h/ P
  N1 l2 v$ ?( l, F2 j1 Nd { (a+bx) C(x)}/dx = -k C(x) + s
: X7 Q5 Z2 I. t# f3 Q& o% u
  U) ?4 M5 z) Y: K5 T3 \where a, b, s are constants, you want to know what is C(x) ?

0 k# t8 O7 T9 @% Q: o! Gif so, this needs to solve the differential equation. please comfirm it fi ...

* A5 u, D! S9 _  c& u5 O7 [6 D
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:

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

十年移民路_

this answer is the good one.
/ B# n+ I* C1 [: W  A
/ ?7 s9 _8 ~6 K  A. _
+ W$ V; a' G$ [3 y% ?$ Jprocedure:
3 i/ v2 c! H9 A0 v+ E& I
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
9 G* b* j% N# l7 u6 Iso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
2 e3 x2 g$ n0 G
2 u6 \6 _: R7 P: F1 d(a+bx) dC(x)/dx  = -(k+b)C(x) +s
; w0 Z5 a  W1 |" l3 Y% y
; h& a% V7 f2 F0 J, q
$ y# A; K* y, ]# Z. Sintroduce 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- M. ~1 G1 q; W- H
7 i& g, o4 c( H3 c% u1 i- q$ O{(a+bx)/K} dY(x)/dx=Y(x)
, k! j' G: [) ], }) a7 Q% X
, t- `  o8 ?  gfrom 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)
$ r. A; ~! I+ f# ?' ^- [2 O; ]4 H
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
% u+ |* G7 C% a7 Z/ x- H2 I
- _! @* o9 t$ j$ a6 c" Q-(k+b)C(x)+s = (a+bx)^(k/b+1)
0 O% D% C8 v7 W, S, K8 r- U
finally:

9 C$ s4 }$ G  T( F; w* RC(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:
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。
5 n( u. A& J4 ?5 a3 E9 j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。
0 A; r1 a7 O; ~+ `您说：

7 e4 ]' h6 V6 a/ _0 @/ Ad{(a+bx)*C(x)}/dx =-k C(x) + s
- ^( O. m$ g( Y7 S: ^* |$ Yso:
6 @6 s% e, z; ?, }  bbC(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 s( a% b# [; Q. c# l4 M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
. I7 d' V5 q6 n. A5 I5 Y! w并不是   (a+bx)*C(x)   的导数除以   x   的导数。
% S4 N( c" l; ^& l5 V4 N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。
6 V7 z2 |7 M4 m9 ^: `- z% }# V因此：

' Y4 ^# {8 w6 T$ ?: Cd{(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。

6 V7 w# e8 t, f& T; t5 j# T所以您解的有问题。
" O8 Z' t! ~& V! e. u9 e
- k$ z$ U& }; U- 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。
! J3 G- G+ g9 n3 u. n您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
8 ?4 G  [( s' B1 L0 J6 `4 fbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
& ^( n9 b* U- z3 Q4 G
4 J* Y1 ?9 O/ Q' |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
" `, b: F: ~' n& a" Zd{(a+bx)* ...

# u3 u1 I- Z* ]* i4 g
Please note here:

1 o( o' u1 b2 a* {: g: {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
并不是   (a+bx)*C(x)   的导数除以   x   的导数。

6 U0 D/ @( A6 t- ?! Jno "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
8 ^! F$ W3 e& z% p& n7 A7 q....
& R3 J8 C& t4 K" z2 f. ]" `$ s6 E' c1 od{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

. F' R* }) i( g( L这样算混淆了微分和导数的概念。
: p( A7 \0 O" P) E2 `  d/ b- h
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

% b9 Q+ @# L, N: D) SIf 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_

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。

$ h' f7 ]: _$ h6 Yd(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
3 K7 v% n$ g/ h6 N4 A0 t
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

. [4 O* A9 l0 Q. x  q5 U
9 @, _/ @1 g7 I9 }7 `3 e# N$ g7 T这样算混淆了微分和导数的概念。
, r( @( {7 x+ {. R  A7 F$ K2 p8 j6 ?8 w& P
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: i2 A; WAh, you made a mistake: it should be:
  S9 d& y4 @8 E8 G/ h" G
3 S/ `: J. q- u7 A6 s7 hd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


  @, s% L1 z' g8 sIf 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)
9 _' r& b$ [3 f$ j
" ^$ o9 ~' o. h: |+ lfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

+ o* R, s3 a; Q- m$ V8 k4 N$ E8 gh(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
3 R- e& Y# f3 ?% U0 e
Ah, you made a mistake: it should be:
' R, m4 |1 q* V6 L
d(f(x)g(x)) ...

1 J( x7 H, B8 g( c. G* c0 b
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
( I' P( j4 y. m, }- _/ [
& S1 g! Y* p$ Z2 P# x" vBut 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:
# k4 j, P% `7 E* p2 {1 O! g3 JYes, there should not be " ' " after "f" and "g", it is a typo. :D

# d' Q% `$ f0 F: c: c& g/ y. L) tBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

' q1 [: Y1 @) e3 _$ Z) x# A, z
2 R- Z0 A5 t; c: RThanks, 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。
2 r, n' G. h, n- g% K3 l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 ...

7 i! C; z% q* u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:
9 g# L% ~! ^  Z5 {2 M* V+ z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。