数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
" r' G% w: _- [6 _8 r! s- Y' T# g
* d" N# _% j2 ^d [(a+bx) ] /dx = -kc +s
3 x9 t8 j0 X- t- T  T7 vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
8 }: S  Z5 m- _; e; c3 D
: B6 U( J) n4 `( N! E- ~多谢了！
4 B1 u* b7 K( {% ?0 D
[ 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
& w! \; `0 a& P3 ]7 n  w% rwhere: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

it is too easy:

d(a+bx)/dx = b
" w- R, r$ O$ _+ g
so

+ J0 h' K" ?" h3 e  N% t4 u; jb=-kc+s
  x* x/ k( B5 R, O, C
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
1 g# P, _3 l3 W' ?0 s
  l% f$ E0 i1 ~6 l: j: i# E. eso
4 ?  U/ T8 B, N' r/ q
/ k! s; d0 _" n( `' Nb=-kc+s
" b' Q& o1 I3 d7 |5 s# e+ h0 C: @
finally:  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:
5 K, k- ^; }) d8 [$ L6 P请教大家一道微分题，多谢！

& Y  k3 Z4 F/ x7 D3 cd [(a+bx) ] /dx = -kc +s
7 i+ A5 p3 z! }  Vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
$ w6 T) s% i9 E( q
! s: m) b3 C6 t多谢了！
# U6 i1 a! F+ x; H6 A/ C
1 X" g2 y# ?- h[ Last edited by 醉酒 ...

* {8 H+ G/ f/ W& y0 usorry, did you post another question? your statement is still not clear.
) L3 b3 ?5 B" Y0 s! p
a, b, s, k are  constants? c means c(x) ?
0 j" ]! d7 d8 F4 m* |$ s. K
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.

- L$ X7 j( L  wa, b, s, k are  constants? c means c(x) ?
5 d& b+ K4 q# o; U# F
please tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
- w, x* @; P' O% \2 H7 }4 R- G
; K1 C5 M  Z6 h4 u多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
# V, T8 R6 b" y5 g4 z8 y对不起，写错了，忘了变量c,应该是
5 \% I0 E4 l& j+ Bd [(a+bx) c] /dx = -kc +s

/ z% I" D- Q- d" E多谢

  N3 q' ^0 z  q! X8 j
do you mean it is:

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

' s' v* N1 V9 ^+ \3 b2 a- U) g0 i1 awhere a, b, s are constants, you want to know what is C(x) ?

3 B  M: l; L# y' v  cif 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:
: O5 u# s9 [1 i8 Y# ]% R, g+ cdo you mean it is:
/ x7 y9 P& N+ q* Y0 m' z
d { (a+bx) C(x)}/dx = -k C(x) + s

" |, H1 }) I1 p/ z+ 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 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:

/ @) `" U) c9 O# H8 UC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

3 G" Y# p" F. g( _
! ?4 I1 p0 g8 t- \procedure:
) R# z2 |  e( q# ]# u3 N' k
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

* w; a2 G! y9 ^, B$ E3 r" F# MbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
6 j) L6 \9 Y* `2 g0 j
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

0 ^7 Q+ ]/ w: T1 Y' o3 J# o
& g. U1 Z, U% D$ Q2 dintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
# r7 e- C/ H4 A. H1 M; Xwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
0 B3 h3 |# }) E$ d
/ I8 j: e# j. p{(a+bx)/K} dY(x)/dx=Y(x)
' }# k8 y  J" X
8 o( j! }9 F# I- Ffrom 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)
& |( p( T" ?) Y5 d0 G7 v; h) i! z6 e
8 r: z3 k: ?0 o. D; Xso that:   ln Y(x) =( K/b) ln(a+bx)
! ^0 X9 s8 x5 t8 @6 p
this means:  Y(x) = (a+bx)^(K/b)
, l; X9 ^5 S5 m6 V3 ^5 k- f' P' F$ Aby using early transform, we can have:
8 S' L. w) ?9 N& g& R* k
2 m! H6 i; @: r% d-(k+b)C(x)+s = (a+bx)^(k/b+1)

* v) J3 i; r9 C1 ]finally:
, _" }9 }$ ]! J+ W' ~4 _
5 a9 A. S' P. ^9 o$ \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:
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！

+ J8 i6 E3 m. x+ D: @& X. z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 a1 Q% o' x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。
8 i; X6 K% w7 |4 ^# {您说：

( B8 h  x' E  K  {8 r  bd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
# `; C) L: \( b% s# cbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
% q3 R( G0 z' A# D! S9 L' ]0 B( ~" W
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
, \1 V' Q8 Y. \" V# I但这是不正确的
9 e$ _; |/ G% F" F2 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。
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。
3 C- e+ a6 S0 |) p因此：

d{(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。
- H& V% ?/ X4 L# Y- F
所以您解的有问题。
- x4 b, [- ~* k0 r4 u' W
0 {0 R- Y$ N4 C6 p[ 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。
1 o" X3 u% h' @3 i. r您说：
) X1 M* G& r3 J. ~1 C
" ]4 U2 E6 `1 y+ w6 R& j1 O7 `d{(a+bx)*C(x)}/dx =-k C(x) + s
$ P; C) Y& S# L/ ^/ x8 F; M$ i% J: Jso:
1 D' D2 }' O" O: f7 j; obC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
. h+ D$ k9 g) o7 S$ h
3 A! \3 I3 h5 k& ]5 R也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
* ]: X. k: J# \, u/ y( D但这是不正确的
* z8 v) o/ l: b4 }! Sd{(a+bx)* ...

- l+ _! E) e1 ^- {4 p8 Q) R  s
Please note here:

d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
. D2 U4 d, j$ U& V4 M( S) {9 Kit 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
- }: u1 u# R; q; D- S. u/ _并不是   (a+bx)*C(x)   的导数除以   x   的导数。
; g4 X2 ]$ [" H' s
1 F, O, P& f# v1 l% Eno "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
( r3 P8 C% ^- o. {
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)

) t! U+ f/ R  B# _( f- @[ 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:
$ |1 Z% }, u: Z) C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
; p' C! e, S0 ~- p
* e( m, _/ `, j+ |0 JIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

+ t5 I' P) ~3 G$ K; I! ]& U这样算混淆了微分和导数的概念。
  c/ p& M, l+ X( {" L* A
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
6 i$ |2 H) }2 \+ ~/ s. Y
8 R- a1 v/ g- M2 Z( |6 N0 W7 \2 k, mAh, you made a mistake: it should be:
) o4 l! X0 |$ a. F) [/ B5 w
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

. E  W- Z* @. o6 W* u
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)
. |' K: ~0 {3 k$ O' v
) G2 t  h8 }& G% a' ufor 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:
8 |: K7 c) q* f这样算混淆了微分和导数的概念。
% N! x; V2 i' I2 F5 r! b
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
  Q/ T9 M6 a1 L3 O& F1 n- L
. d! j, H# K5 N5 E( @4 p9 J( tAh, you made a mistake: it should be:
5 u0 y2 v. M, i0 e: D% j
' q$ H: T% ]: sd(f(x)g(x)) ...

Yes, there should not be " ' " after "f" and "g", it is a typo. :D

0 z4 q8 M& L8 W( e0 I6 X  W( JBut 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:
7 A9 N, O- T4 _Yes, there should not be " ' " after "f" and "g", it is a typo. :D
4 R  D0 J9 u4 V
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

8 h9 X1 r; V( E, r* L
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:
* {4 f+ U5 S, ?- G
* i& N; C4 r9 F% x% z( {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。
9 a( C( Z) C0 C. P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 ...

. w0 R& i# w% @" q+ ]* ~: I, 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:
0 Y  W: z$ M' g* Q7 E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。