数学高手请进，多谢！

醉酒当歌

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?

- `7 \0 A5 g, T& H- Y1 {) G; q$ u8 A3 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:
% m1 g) I( G$ _9 z: _请教大家一道微分题，多谢！
7 T5 i. ?5 x% ~8 C
d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
; h( R9 i* @3 k
多谢了！

7 i6 u8 z- j2 t) U$ _- K# F$ ]% Q
  _7 b, h1 n) q. A$ P( t# r$ \  ?it is too easy:

' s, d: V1 Q+ c, J5 s! dd(a+bx)/dx = b
/ a* \) d' j5 j* ~& x
so
8 j, ~3 r6 ^. }5 W# A, n% @% |
b=-kc+s
) x4 z+ q: Q( _* o. w
5 h" L; o$ p; t2 ~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:

d(a+bx)/dx = b
4 W. S, N5 D, {9 l* J" }
% S. ]( E) M2 Kso
% N- R, l% _3 e; N& T6 k% C
b=-kc+s
2 D! Q# V: R2 L; i% D
finally:  c= (s-b)/k

2 e- U  f  ^+ V7 @+ S) {& \4 D
! ?+ p/ S+ F# O) e" _' hthe 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:
' c/ d1 Q# ?+ h7 M& G  S' {请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
% K- A( U7 \" t9 |5 uwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
. r/ h' s/ O' b. }) Y% N1 l. w9 x
多谢了！

[ Last edited by 醉酒 ...

; B* {; ]& U/ v4 m1 N  p
sorry, did you post another question? your statement is still not clear.

- p, A+ D7 A( d# d$ E8 Da, b, s, k are  constants? c means c(x) ?

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.

$ x$ w+ l, E* W' \% Aa, b, s, k are  constants? c means c(x) ?
' O2 ~1 g: j% ^" `/ x8 g
please tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

1 q* _2 [- J' E- o" e8 Y" D$ p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

4 ]8 s' H# }- x9 S多谢

do you mean it is:

" y% y; f9 H! Nd { (a+bx) C(x)}/dx = -k C(x) + s
9 y7 z! z* `# E* ~! Q$ }
where a, b, s are constants, you want to know what is C(x) ?
* b9 T$ c" i* \9 p
5 ~0 \* d: u0 Mif 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:
" s$ v: `( T' k0 l% ?
d { (a+bx) C(x)}/dx = -k C(x) + s

, v; b1 q: r' z  `7 ^+ _7 u- r6 iwhere 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 fi ...

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.


* J5 C* y3 u2 L" _5 Nprocedure:
$ @- A4 L5 j% j& {* Z9 c0 U
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
8 L$ b/ Y' Q: h- V9 h8 Lso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
4 |; F0 @' R/ r, ei.e.
2 t  P% W5 Y3 ~2 l
5 H) N) {7 E' Y1 ?; l. K  O(a+bx) dC(x)/dx  = -(k+b)C(x) +s
8 f2 }; T1 `9 _# \

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
" v/ g; w: c% iwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
( p1 S" K( P. M! U" c; w% ztherefore:
3 V8 {- m6 s3 B  ^- A# r) Y3 {
2 v, T  ]  |3 P: q( ^0 i{(a+bx)/K} dY(x)/dx=Y(x)

; T3 q) W  k: A* o( Ffrom here, we can get:
- [! }: S6 x2 h
1 F& j# x8 ~/ n/ I6 x2 }+ J' YdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

2 e" V% [8 U) y+ }so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

4 O/ m1 t- \* C% T6 g' v( i- ]: y-(k+b)C(x)+s = (a+bx)^(k/b+1)

: ]  s& c% t+ l- l7 ifinally:

. h  b* T, m& f7 C0 vC(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！

; g2 j) z& S% ~4 K
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。
: {/ l5 f/ ^# n+ 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）。这种题真的不是难题。

龙舞九天

9楼的答案有点问题吧。
; I* w1 t1 B1 V6 B" \6 h" D您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
& J" ~$ U+ Z9 zso:
- z0 M2 ^  N- R0 ?. k2 mbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
' V. Y+ c; Y- M
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
d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
& M5 q4 V! F7 \. L$ `# `; _+ X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
4 N7 k6 s# F* v2 n5 Q4 @5 qx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
" Q8 X$ j  X7 C+ b
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
/ B& ~  Q2 E1 j+ a  o" o5 _最右边是x的平方。
- Z9 x( E- t8 ]3 x; ]) L0 S, V& R
所以您解的有问题。
5 \  z# ]! s& _  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:
" l; \: v- `) j: S3 K9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
$ A, `# z0 n' A; b% M* Kso:
1 Y) w& x) I; F( \" z/ ~4 ybC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
- e2 t  `1 S0 D
# @7 i( W) Z- o6 g: q, b$ B. 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
d{(a+bx)* ...

! u! S/ |* K/ C
, }- L) b0 ^( v/ OPlease note here:

$ a& K' z3 D+ I+ cd{(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
1 h( G7 w% q5 h0 r; ?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。

8 W6 i1 e5 n. f+ \+ n8 Yno "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
3 m( L& Z& d2 W  A) M( _....
1 n+ ^0 }8 p2 c! b( h; pd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

- t9 J7 y9 r! ~! F
这样算混淆了微分和导数的概念。
( ~. K" h0 C" {' v/ I( p
- r- d7 A! 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

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)
* ^3 }% n8 v% v  h% p) a# b, @
[ 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:
& m: K2 P! Y4 b# `9 V( r  f' }3 \9 H这样算混淆了微分和导数的概念。
, ~/ C* I) Z5 C# Q
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

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

5 [1 D! y" r6 \9 @, E  @
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
! g. `  Q/ N. F
Ah, you made a mistake: it should be:

' S% I$ }" V' w+ q8 R2 A( Ud(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

6 C) M! }0 t; s" ~" C
2 Q6 V8 q5 L' M  TIf 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)

for 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。

5 n: }' [- |" D/ Cd(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:
, n+ ?) @4 s/ I7 `' j4 o
d(f(x)g(x)) ...

1 C1 ]: W& f% `/ @# p+ J  r4 m
& n7 i. W# a8 h" ]* kYes, there should not be " ' " after "f" and "g", it is a typo. :D
) X- @" F) |. [2 M1 r& m
% J: b; b4 B/ t; O0 I' |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:
9 h! {. M4 ]3 f( _, u, ^Yes, there should not be " ' " after "f" and "g", it is a typo. :D
: {: q  J% K4 ^6 c
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

! q4 m. Q. r& a8 A% ~# }/ g* p6 vThanks, 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:
  |" W3 F$ ]( H, F) g; a+ L
" o5 J1 h; u) G' M/ e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 ...

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？

& f) v6 r) X8 k; h
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。