数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
0 i- S/ `- a$ g" p/ f" B! }
! r* D; d6 O3 u3 ?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?

5 b" ^) ]1 v7 o6 |* M4 R! s多谢了！

: M- i! V. E# X  ~! P9 B. c  ][ 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:
5 \) d, D1 N% I5 i, E$ i% l7 p请教大家一道微分题，多谢！
- W7 |$ _' t/ T$ Z
2 w* b7 ^8 F/ F; @6 k% b$ B2 Wd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

' o$ j2 x9 G1 [% O4 D9 [3 V
it is too easy:

d(a+bx)/dx = b
* n% C; j( l$ j) Z
so
. }& P# q& S  U9 i4 R
, L, C! i3 `3 Gb=-kc+s
  q5 T0 K4 ^* |' q
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
  s: o& F3 t# s" @" Hit is too easy:
7 M4 V# b3 c* X% o7 e/ t5 w
+ w- W* t1 O- ]" zd(a+bx)/dx = b

; u- c1 _; ?! m+ z+ Z+ h* s2 _so

* r' p9 z, }8 x2 hb=-kc+s

9 {5 @# S7 D5 m* M6 ?finally:  c= (s-b)/k

/ q( g! J+ k& ]7 D; h( X
: S' L- r* k0 f0 t* 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:
2 W5 C( |1 L% s4 y2 [- D请教大家一道微分题，多谢！
2 f* z! |6 |" h  K. U
1 u9 o2 o6 R' Jd [(a+bx) ] /dx = -kc +s
' D& M; \4 ?% Z6 C: x/ u3 X- Vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
- _4 U% l" T* `' T! H% O
多谢了！
6 C# p' m, d, D7 L
[ Last edited by 醉酒 ...

* R8 D  i3 _! G2 \% w/ L
sorry, did you post another question? your statement is still not clear.
( Z1 Y% t3 C2 a
* x. }5 n2 C. s% w4 Ea, b, s, k are  constants? c means c(x) ?
* I3 i) m0 ?, J, Y" \
9 ?7 x! }, ~4 \. ^" `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.

2 g" d1 x! F" i! ?  o9 _. W& Ha, b, s, k are  constants? c means c(x) ?
( T9 u) f4 z( Z& `2 c
please tell me.

; G( x, T$ h. W' d对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
* w. \* q$ }0 l, |+ T* z4 R" ^
) |7 G! E0 B5 {5 ?. i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
& e- l5 Q7 }/ X) t: [
- |9 @6 n& A* [2 N; U; ^0 ]9 w多谢

$ e+ U* c; J" O+ ^3 A- Pdo you mean it is:

7 n1 o) w/ p+ [+ ~. L+ U! F2 j5 i3 t0 Od { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?
3 b  a2 [( [& b6 I' c3 y$ ^3 u
, e; B0 v- G3 Q% p( O  {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:
( b1 o' O1 Y! A) b8 g( Ndo you mean it is:

$ f% k5 J! ]# e' f1 ?  ]& bd { (a+bx) C(x)}/dx = -k C(x) + s
4 R) t) G( r! h' ^7 q6 ^" _5 ?
where a, b, s are constants, you want to know what is C(x) ?

# s% U9 a, s7 E! p, \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:
, R) S4 q2 n8 |  j. j6 g
- I, T) v8 \! Q% X4 C' }C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
, N1 Z* H+ N! b9 j4 d7 A* |( x
0 n2 V( M+ F  `* v1 R- F
procedure:

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

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


0 t8 y: h! a) G- D) W. A7 Yintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
5 M/ C, H9 ?4 N: `$ ]9 ?which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
3 t6 r( k4 [9 b9 F! B
{(a+bx)/K} dY(x)/dx=Y(x)

9 W( r6 {  y/ H* O% `3 i! Vfrom here, we can get:
5 ^, W% I# L9 n1 e( _# F/ Z
$ A9 e  y3 ]6 ~1 qdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

' d  k5 Z5 y7 E2 F( Yso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
3 ^5 S; V* O, l% b9 j( `! U3 Oby using early transform, we can have:

) w7 I* W5 S2 \  ?+ }, L-(k+b)C(x)+s = (a+bx)^(k/b+1)
1 S3 ]$ j& a' O7 g0 D- ]
' J9 R% r- g' d7 |( w' B9 r* gfinally:
* Y5 ^4 z% v+ m0 Q- G2 }
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:
5 @5 S6 a" Q0 e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！

" w8 o% s7 L0 @0 e0 v  p8 \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。
/ A* h: O% t+ _& W7 m$ k# I您说：
! N# t( y4 \  j) _
d{(a+bx)*C(x)}/dx =-k C(x) + s
0 j( D6 P0 ~! uso:
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
/ j; N+ t0 s  p' A4 g但这是不正确的
% q9 y0 R4 H( c* L* Ad{(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。
* {) o2 I4 C0 D比如说   dx/dy = (x'y-xy')/y2
8 Y& O( U3 W1 d& T! V# \4 _% _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：

2 m. U' h% m- B" K' A" E5 nd{(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。
, {2 ^- x6 f& B/ g! R$ w
所以您解的有问题。

[ 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:
8 g6 F4 N$ o# N" z; e5 `* d+ t9楼的答案有点问题吧。
9 o& p# ]: ?0 A. O1 E) f您说：
7 l( E: e( m( |6 g6 ]" N8 E! c7 C
d{(a+bx)*C(x)}/dx =-k C(x) + s
# i8 Y* P# h' m$ {% Mso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! o, l' L8 p  U. v
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)* ...

! y6 o- _9 D- A  k& u7 p# O; MPlease note here:

" x. ~  _* v/ N# d; Y! J/ y( C9 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
" H: H5 W, ?& J, m& |6 Nd{(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。

: k. @  P2 d3 v. 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
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

" e9 s5 M# q6 @5 n) w/ S/ R
- y/ a  m: S( v+ M& _这样算混淆了微分和导数的概念。

. K) p5 \: |% h3 m  |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
$ C# Y, H8 L! U+ ^7 R$ `# t0 I  P, ^
& n0 k% j/ Q- x( p# v) MIf 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)

* c& f3 r4 H: D0 M[ 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。
% q& L& B( x& j
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 l8 {: v$ u8 Z* N% ?
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

/ {7 F! p; E$ S3 `  v这样算混淆了微分和导数的概念。

9 }9 Z7 ?1 w2 i/ e# }" w. s+ 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

& A0 V9 \. a: s7 J+ l) A1 kAh, you made a mistake: it should be:
9 h5 T1 h, X! M$ j. P( {4 Z5 X
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


  l5 P( K1 O/ A7 d. }  d  q5 nIf 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! ?( A8 w% rfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

, |7 S( J; z+ x3 n/ k4 C$ S' Dh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
4 Y% h, y! X. w+ k$ n% @% c/ ^这样算混淆了微分和导数的概念。
1 [5 _( k; L/ I6 k4 y: o  D
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

Ah, you made a mistake: it should be:

! V( V' J4 ~/ y) [d(f(x)g(x)) ...

- B/ [3 x& i5 b, ?. p0 o
/ O% _' H6 p6 n* S1 j) y; [1 QYes, there should not be " ' " after "f" and "g", it is a typo. :D

9 I  F8 V' i2 ?8 z6 ?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:
, ^; e/ H4 ~4 d0 g* B( d' V# wYes, there should not be " ' " after "f" and "g", it is a typo. :D
' z0 E0 Q1 @! S) y: N
2 i/ b# J9 A" ?! lBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

5 V$ e% U# F9 K' D* j' d' o
: H% o/ \4 l" AThanks, 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:
% `- {$ V7 u. ^" w4 _1 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 ...

, V7 V2 N8 F$ F* V% ~" S+ d& c
' R! _. f5 |% {( n' 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？

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
3 u& W7 ~1 q/ d/ K) Q0 D. k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。