数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

8 d  ?: R, X* p4 vd [(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！
( G* M9 ?" ~2 j+ Q
. B# L/ M9 R$ B( k5 S+ W. O1 Y3 S, O[ 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
  g+ q9 ~2 b( \% x2 z  e+ ]2 D5 _where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

- y3 @# }7 E7 H
it is too easy:
! o" [* B+ p# f, C
+ [( Z% X+ U  U4 q/ v5 b* R5 id(a+bx)/dx = b

: J" H5 X/ f, I  `, a# fso

b=-kc+s

$ `, G) I! C. c; f6 e& }finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
9 o% ?/ z) C2 |2 mit is too easy:

d(a+bx)/dx = b

$ L; J" K" c% W& W6 h/ y: y9 Tso
, l0 V7 u9 \+ P4 H
b=-kc+s

: c, ?7 K0 ?4 a9 R( S, 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_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
( n# K$ }/ A6 Y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?
: l( E) i* e% }- g3 G
多谢了！

& L6 n) P, B+ P' w0 z8 h[ Last edited by 醉酒 ...

* I5 ^1 J4 Z& ~/ v5 r7 z
sorry, did you post another question? your statement is still not clear.
8 |% y3 ?' e+ s( R
2 r" H; d2 l- P0 ~# la, b, s, k are  constants? c means c(x) ?
$ t* G# O6 \, b8 u1 n# _4 T
6 {' J7 X$ p: N& |5 Q7 i0 _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.
, U* c5 v0 `+ Y2 g% c- k6 |: k
3 n8 C- b2 U/ x$ c; c6 P2 n6 T* N  Fa, b, s, k are  constants? c means c(x) ?
; q1 i* w. R# Z( |4 O' v  I1 z
1 B0 |; e# e/ [& L! i) \please tell me.

( b$ j. ?* G( h8 g对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
- o: ]1 n( p$ q6 i  I$ T4 ~
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
8 G% h; z& v! d" B  md [(a+bx) c] /dx = -kc +s
, \4 E+ l  `3 U4 Q# |
9 R1 W1 V4 [: n, ^; q: r多谢

do you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s

4 L( Y/ z# Q7 O2 m! Y& B1 jwhere a, b, s are constants, you want to know what is C(x) ?
- U% G9 o4 Q& i
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:
do you mean it is:

. Y% v  Y8 }2 p7 s8 r% Bd { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?
& F2 f! n8 c' S& O" s
if so, this needs to solve the differential equation. please comfirm it fi ...

% D: h3 B: v4 R- y1 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:
! c7 v! b( j8 |& }* b
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
7 p' c5 \6 k, L& [2 q

procedure:
* X7 f  T. S( z- @8 z2 T5 e  N
- L  K" B( D- N$ l3 K- BFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
" m" E+ |# i3 D& b; H) C6 d
) [, ]' _8 Q4 e1 N. YbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
1 u+ h) g' t8 I6 ii.e.

1 Y: F! ]5 @; D, t3 B  u+ G* M$ B! b0 Q(a+bx) dC(x)/dx  = -(k+b)C(x) +s

: m4 K$ ]; ?5 E
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
# J9 L  }8 l; L$ a. c/ @, Qwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

7 F7 R8 N) }( T! l- u{(a+bx)/K} dY(x)/dx=Y(x)
/ P) r( T! f1 p$ b, k
from here, we can get:
2 Z- e# @9 q( c' T) ?! k
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)

$ B! s" }7 b& O( p* {this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

/ M8 G  {3 Y, c5 ?# Q-(k+b)C(x)+s = (a+bx)^(k/b+1)
! g0 p% T  X# c% K: O: [( b
finally:

2 W( P) b* a+ i# z' ?/ w: iC(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。
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：
/ B2 T0 r1 D; }5 N8 n1 s
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
* Q4 y5 x2 R' W$ q7 h
1 E* ^( ^$ K! Y: p. l也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
- ^9 \7 x0 W9 T9 A5 n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。
9 ^3 x' o% K* N: y4 }) Y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
7 e5 Z$ W/ g- A) r最右边是x的平方。
; s, |7 G/ R5 r4 {( C. r" G  p
; c/ F. _& F" ~5 @所以您解的有问题。
$ B# ~+ Q: _& B1 o5 @+ V
8 Z! H# ]9 O1 x) d8 `4 b[ 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：
( a' C* z: T8 n! g% a( \' y
, {2 ~- l  D: dd{(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
" R0 d+ C7 e; V0 V% y4 I! cd{(a+bx)* ...

8 z8 W% \% w: i
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_

但这是不正确的
6 K" g7 c" a1 k% o2 Td{(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

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

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
/ I0 ]9 w+ R- @6 @
) y! F8 a( \2 l2 c0 g# L' k  _# a) ZIf 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)
7 `$ [0 Z4 D: 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:
# u8 m: V5 j9 O# N8 i这样算混淆了微分和导数的概念。
/ B% P* g' k' L: D0 e3 t' A& S
: M; Q& C  W5 i% b$ O: bd(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 v- k7 }( a2 a0 i9 k
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

* ~; P, c3 i, o" {/ T5 |
) N# ~  j9 w/ Q3 z) _/ @; \9 P4 ^这样算混淆了微分和导数的概念。
9 i/ Q) w0 E7 j4 n! d
7 S. w( `9 `, 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

$ P2 y- I, |4 Y# ]! R9 Z1 B2 ^Ah, you made a mistake: it should be:

/ [! }3 Y- T1 S8 @d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
6 C  w& x* h+ @- J$ E. S; t5 j

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)
& ~* a2 c7 n( C7 w( `3 z
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
  P4 l) C# l1 B1 p
h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
1 Y. p# [* r5 W7 B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

4 I! ]/ f. H$ t! Q/ H1 T3 JAh, you made a mistake: it should be:
* t5 v) ^% T" D. F' ^" b8 ]
5 W1 ?- e, E( X; Rd(f(x)g(x)) ...

/ M) f9 @  J# @& k7 z
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:
0 K2 P, J' O9 n8 SYes, there should not be " ' " after "f" and "g", it is a typo. :D
3 y0 s2 G  d/ g2 l# p
: U4 t6 Q, }) p# [/ w) NBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

& t& \7 @  d' w% P
; Z% N: e2 H8 a) ~* ~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 ...

8 h: |! r8 _. s8 ^0 A5 Y
, u, g0 w+ |0 v/ 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:
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？

! V1 R( s) H! K" ?4 b4 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。