数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
+ x/ u3 l9 J9 V: F
d [(a+bx) ] /dx = -kc +s
- g( Q. H1 o4 D8 D- O0 t4 gwhere: 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:
请教大家一道微分题，多谢！
4 m" Z6 j" \& R) J
d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

it is too easy:
1 V( }3 x! g8 J7 m8 k0 k
d(a+bx)/dx = b

so

b=-kc+s

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:

! ~' H6 g# m" l- C! |" p& l1 od(a+bx)/dx = b
$ @7 m! M" O- }( p: D0 h
" w# v' y4 o) U) U) ]; G# Tso
, p# k7 a2 R0 }$ z! c
1 f9 I4 i# J0 b9 t) C( sb=-kc+s

finally:  c= (s-b)/k

" z# s# i8 D/ @9 _. V& d" P) M* ithe 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:
请教大家一道微分题，多谢！
- u! i) f: C9 ?
: K1 s) _1 v1 I9 V2 Z/ R) ]# hd [(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?
- G3 z5 W3 ~9 l6 m
多谢了！
) {7 p* e9 x) z/ T" [
, Q2 |( ?) o" i1 z' u& N, B[ Last edited by 醉酒 ...

% h3 N/ ?# S3 Y" d+ B0 P2 V# k$ y1 Nsorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
# {% a" Q4 R# p7 A( T: o3 F  K
  K8 d) f; C2 x6 G  u! L% jplease 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.
2 I" ^/ y1 }1 H/ O  y% F
" e, i7 P3 ~" w9 s, Ba, b, s, k are  constants? c means c(x) ?
9 X4 D# P0 x. T1 T
8 q3 H4 w6 w3 `0 Mplease tell me.

5 b  j  r' L. V6 V* d+ K7 x对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
6 g  z6 \3 w, C) G
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
d [(a+bx) c] /dx = -kc +s

多谢

" e" |' K% p) t9 {2 D5 i$ ldo you mean it is:

  I) M4 c# P/ g& k3 s9 X0 Z. ~/ V" vd { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?

  {2 o: |/ S6 {4 h' ]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:
$ d& s; e" I$ Mdo you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
  Q: S/ u! N& F/ t* R' N
" H3 t9 W4 T( [where a, b, s are constants, you want to know what is C(x) ?
0 z6 p7 S2 h6 j/ O* X" P
! @$ S  {* L; n. f3 k- {if so, this needs to solve the differential equation. please comfirm it fi ...

! h& D  n$ D. j2 k$ ]2 H
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:
. s7 Y! e: Y" Q; \+ Z
" B5 L8 s7 ]0 h- K8 k' tC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
4 M2 s6 m2 Z. c0 P! r
, B5 c8 o4 B1 W. f# Z. G% n1 E
: r) \7 A0 f- o0 x: L# pprocedure:

% p: {( j0 E. v1 a; n/ H/ X- \From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

; x3 D# |& h( Y1 mbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
4 ]% J+ Z$ j% r5 b# Y1 W. n2 A" g9 T
: t8 I% V! s/ W. B' G: Y; ?(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 u8 `( K, d- ?) a6 \' g& u1 k

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
$ X0 a+ T; l; itherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
% C/ K2 c  x" O# K7 P# p/ y& r
$ F  J  M1 b1 F+ N2 D: yfrom here, we can get:

; X/ [4 Y2 u5 x$ ydY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

8 ~: P* C( B' A+ r% @so that:   ln Y(x) =( K/b) ln(a+bx)

8 {' M/ {  v/ F& kthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
, @& K: i  H" L, n! f# ~
0 K; D; o: z1 e. [; M: r0 `-(k+b)C(x)+s = (a+bx)^(k/b+1)
2 i+ }9 i# G' M: W, k2 R- r
$ ^# l7 `3 E% g1 V/ Bfinally:
& p: O/ U/ a2 a  _- s* f9 c8 q2 X
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:
+ u* S) F  l% o6 O% x# x+ c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！

) T5 U1 h% e& @# S5 Q6 H
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：

) g2 \9 @$ E0 ~1 f5 Q/ L: m6 D8 ed{(a+bx)*C(x)}/dx =-k C(x) + s
- w( A  e9 n2 nso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
) X: P# Q- w6 V) e. G
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
7 j8 B* x% {3 \2 Xd{(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。
4 M" e* z- ^0 Z; H4 D) Z比如说   dx/dy = (x'y-xy')/y2
' S: T$ a. u% r# nx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
! P( n! a* M% W( A5 Y因此：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
( A  m: H$ l' ?- ^最右边是x的平方。
5 s6 C& I+ N0 U5 H$ b' L+ x3 U
% g7 b: O4 n/ g- I7 `所以您解的有问题。
- }$ J4 o% `9 l4 ]% e& W: A
[ 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:
' w  T/ |, ~! S- B: Q9 K9楼的答案有点问题吧。
7 I; [3 w% m$ D$ ~3 f$ Y您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
7 j+ b4 @, y# ?3 b7 W+ q. [* ]6 N
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
: {" a" K0 V+ y3 od{(a+bx)* ...

Please note here:
0 U6 J! [3 K! k* S% ^1 V
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
* z% a9 c" C& sd{(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
+ V- D* ^% |; b) ^) g* Y2 b....
7 `& H, A1 Q5 }) L- i6 x% g; bd{(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。
" n. `, U! ~. r/ G! {, A* Z, A  L
$ f) B' C+ g  e. 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
4 s; P& p* m. M# q* }: P" b, R! F
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)

9 X; ^4 l' G# A' q+ [* Z0 \[ 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:
$ J% Z( \; b- w0 `; e6 \1 t9 F1 m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
, e( t4 k) L1 Z3 Q
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

+ `, ?6 S9 @6 u0 _1 v2 e
6 H7 Q: _8 `4 _& b8 j/ B. I& N这样算混淆了微分和导数的概念。
8 q# o+ `4 a% k3 o
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
/ O6 r4 }+ f& x
Ah, you made a mistake: it should be:

4 I* j1 _/ k) y1 f3 rd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

% ]* h1 g! `: n5 D/ p
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)
9 q0 ?' h" b4 |$ c0 M4 e" z
  O6 z1 t' Y6 M5 kfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

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。
1 m! H+ _1 f5 L$ F& v. A  I3 y
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
( A4 a# _& p! K0 ^! s$ o
Ah, you made a mistake: it should be:

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

Yes, there should not be " ' " after "f" and "g", it is a typo. :D
% L+ H/ u1 Y8 q0 @( t7 B: p
# A9 l! R/ K' H7 s  hBut 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.

. k5 e5 q$ u* V& ~, a3 y# A# h5 I
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 ...

9 w0 b+ X1 n$ C+ v8 i, U- j
: R" ]( [1 A5 k$ n4 U# U2 `( F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:
- F5 w7 D5 F4 U4 E( B5 [. q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。