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
6 N* [: [( i, S/ j/ X. Q; ]& `1 v6 Lwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
, `* {* [$ }: W7 P
多谢了！

! k& R1 i5 ~& G3 x/ e- Q[ 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:
* P- [9 R$ F6 }- J6 t/ g请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
9 q9 a7 k: q4 s+ t( |+ ywhere: only x and c are unknown, others are all known,  requiire c = ?
3 O# V+ }: ~' O; i8 A2 w
多谢了！

it is too easy:

! W2 a9 G+ i- m6 E' rd(a+bx)/dx = b

so

0 F1 b0 G3 K' ^+ t7 a, ~! lb=-kc+s

7 n: m" D, R; H3 R' hfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
: x# x+ \" S3 C1 Oit is too easy:
2 J( F, _: B: @
& S7 R! M! o9 \0 p3 E. t5 nd(a+bx)/dx = b
' ?! Z3 w" U7 X# D( A1 a2 N- y, |% C
% y- ^3 C; t) l& @5 R7 ?so

b=-kc+s

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:
) A# G% i8 C3 c# f3 d% E请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
- b4 L$ `8 z. Pwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
6 T9 R. z! i  z* v4 ^' x  V/ T+ Y2 v) L4 w
多谢了！
+ }/ X3 U4 z# u) u! z
[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
, X. s& `$ i4 Y  f9 W6 k( @1 A
( [% z2 z  ?& P$ 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.
, ~( k# e& r. _: x
a, b, s, k are  constants? c means c(x) ?

please tell me.

对不起，写错了，忘了变量c,应该是
2 s/ n% F/ s, B0 U1 \, n, Sd [(a+bx) c] /dx = -kc +s

( E- k1 [: c2 R5 x, H, M& f4 K多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
7 x6 O. J5 _! _, e# F( b; u对不起，写错了，忘了变量c,应该是
; t0 J2 k- J* V& B% _- Qd [(a+bx) c] /dx = -kc +s
4 f# N( c( H- P4 n
多谢

; W! D+ \  y$ T& w- j" Z
do you mean it is:

2 V1 m+ j! m9 {d { (a+bx) C(x)}/dx = -k C(x) + s
* m5 f! f, z. k& S9 g
where 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 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:
3 ~5 z- p" V% Y, H" k
d { (a+bx) C(x)}/dx = -k C(x) + s
/ M# a  w0 x7 r2 u/ s$ K" {+ N
where a, b, s are constants, you want to know what is C(x) ?
  t' S% H6 X1 o: d" [
( {6 A6 R- E) h/ w1 qif so, this needs to solve the differential equation. please comfirm it fi ...

- d0 I/ @2 h: a! }
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:
5 c( i3 T3 z/ a
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
, u! o8 {/ M* T! b8 o

procedure:
8 P$ ^$ S3 [& V, A
, h1 M8 S2 {0 e4 @From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

$ o% n: Y- [6 GbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
9 c. W; F6 J. v$ Ji.e.

& A9 Q! u$ N7 _7 ^1 k(a+bx) dC(x)/dx  = -(k+b)C(x) +s

" v- T+ L7 J& a+ V# o+ x
/ h, l  Q9 l7 ~% o' ?* G( lintroduce 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
therefore:
, u- h; d( r  f& p& G/ O$ [
* h7 C% x1 G. i/ W0 |  i* F{(a+bx)/K} dY(x)/dx=Y(x)
. w$ h4 W9 \! G, ~4 G+ d1 D
from here, we can get:
. S; ^" s; l4 P" d/ a( e4 d/ Z3 ~" q
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)
/ [6 |" W/ P& m6 J0 Z0 @3 ~
' `) X! c8 H" }& B0 jthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

/ x, H2 N2 J& A1 K- C-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
4 M# |$ B2 E% \8 Z* c, `& d
5 U' k. V6 ?( W* NC(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！

- m0 L# [7 O+ K! n2 G6 w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。
  j) H2 w% A! B; f- a/ u2 d% e! [$ i6 G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。
% ]% ~6 |/ `# x您说：
6 I4 L) J0 g2 H) ^; g
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
( F8 z) Y- X. m' m* KbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
9 W7 h+ l& n  v/ f; V7 ?0 d
3 r& d' |! t* f% w% l4 O2 f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
. o3 S! K- J0 u+ E; @$ ?d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
! t, W4 N" ~; S- j, |6 S( O+ q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
+ a) x' l! V9 ~. M/ b5 }! Lx' 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
最右边是x的平方。

所以您解的有问题。
6 L% A: d$ H! D6 J
* O- f9 w* j" 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。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
1 g; E6 Y( u$ C) cso:
5 Z* A1 n9 c5 K3 ~$ mbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
' ^/ ^4 j( x- N. Q. c
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)* ...

5 |4 ]+ i. w, s. s! m
8 l4 D% i+ T0 W" m' z( SPlease note here:
0 }" R  W, c/ W6 o- {
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
& ~# D7 y% J, a" ?# r( ?it is a multiple, not division. Right ?

十年移民路_

但这是不正确的
  e1 {4 C) r3 j: c) M& N: V* Bd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
1 p- n9 V6 ~/ v$ ?7 I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
7 [2 m5 X; x, q! y....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

7 A9 {- u/ F8 H/ w; B2 A( W这样算混淆了微分和导数的概念。

$ a/ }* q7 {3 Ud(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

8 V7 V6 |% F! t- {; ?6 f  vIf 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) X% Q9 r9 g; w( ?$ ^+ I! c1 |[ 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。
8 X6 E; ^6 x% {0 n
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 ...

' L+ J& q1 `8 @, j6 t& [, q
* l, n1 o0 M( k这样算混淆了微分和导数的概念。
8 Z6 Z1 J  |  d. D# i, w
! d; Q$ e, v+ Y9 P4 M0 }+ Md(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

+ e2 |  ]$ G4 ?  ^: h0 `Ah, you made a mistake: it should be:

- _- T# K0 {7 N5 L* v' N8 \3 c. Td(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
0 j$ O$ ~9 r; W( g  B3 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)
4 R2 W" C; W/ C. B
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
$ U3 `6 ?+ _7 [: x5 z9 `* H
* ~+ P+ ?" v3 ]- V0 |h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
7 P$ A9 @! l% W0 E' 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

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
* o8 K" ^# [; X
8 r% c6 M% q( bBut 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

& T+ K7 A0 ]" {But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

1 P! {+ D; U0 E; l9 g! z# ~
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:
) `' H/ U) I8 ^7 H5 z
' R6 ?5 Z; B+ P; Z( M7 T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。
1 A/ `, |* B5 D; E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 ...

) t% R& u6 u" m& `) v: `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:
* r0 a' `! m: F! b1 j" U. I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？

6 G, m& S4 D2 Q6 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。