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
" W, C, K- Y! C9 f5 T; zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
- M* ]) Z2 W  s5 Y. Y' J( D6 q9 r& E$ Q
多谢了！
  L8 |# y1 Q/ Z: 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:
请教大家一道微分题，多谢！

/ o& l1 y0 n6 }  m: |d [(a+bx) ] /dx = -kc +s
7 P( ^9 R( s% }/ v, n9 z3 `where: only x and c are unknown, others are all known,  requiire c = ?
, \: M" ^$ M* m1 E: m) M
5 Q" N" {$ i4 {) u* [0 b多谢了！

) `) w# b/ l* T$ V; {- ~# U7 Jit is too easy:

d(a+bx)/dx = b
+ d) K4 ]$ O. y2 r# i0 `
/ n  z9 T) K2 d- {! R! q3 Kso
% j( b# C$ y3 n$ C. Z; T
; h& f, z. y; s1 c+ h8 Y3 Kb=-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:
; E9 W$ N& F( \& hit is too easy:

d(a+bx)/dx = b

+ J: ~6 c2 _% B1 R% E" K" E8 C* P7 d" wso

b=-kc+s
: p% ~- |" _, M8 B$ k8 K& J
. _+ E0 ]3 |! X' u9 ]# ]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:
请教大家一道微分题，多谢！
+ h! R$ r; h4 A, @+ u
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?

8 p* R" n- ^! L& u多谢了！
7 o: A6 L$ S0 s# |7 b
4 v' t; D* J9 ?/ _- d% H! z' o7 G[ Last edited by 醉酒 ...

3 p% H8 k& }2 i+ R' Gsorry, did you post another question? your statement is still not clear.
. g1 f8 v; Q- m/ _4 \
a, b, s, k are  constants? c means c(x) ?
- Y8 l* W1 W$ M* `$ }
please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
! h0 b* T, S  gsorry, did you post another question? your statement is still not clear.
- V. h3 c% v+ F' W% y1 o
8 v/ l+ I* @2 i7 P& h+ G. ?9 c$ Da, 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
% G# P, ^# f& i: u% vd [(a+bx) c] /dx = -kc +s

2 H* [5 n/ f" T* H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
3 D$ g9 l( a5 s% W" B9 _& e. ~: z
$ N1 c6 v, s/ S9 C/ ]多谢

do you mean it is:

  g  g" U* B. }8 `- |- f: e; dd { (a+bx) C(x)}/dx = -k C(x) + s

+ [# B& \& q. N, Q+ [+ Zwhere 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:
; N: K9 R/ T2 `" a/ }3 zdo you mean it is:
2 M2 v  c' |: L6 F
d { (a+bx) C(x)}/dx = -k C(x) + s

; u) y9 _2 d2 H* L9 a6 T" }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 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:
# ~4 I7 f7 q! h, X6 u
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
0 R2 T9 j6 h4 D* P- m5 `

: ~! N% j4 g/ k9 Fprocedure:
: A: j6 {3 z5 x/ |
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
+ F, T# D( k# y1 B
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

  u/ j. h) X5 v6 T. O( C(a+bx) dC(x)/dx  = -(k+b)C(x) +s
6 m6 ]) e4 j5 w" J
- h# R& J4 \% n5 \! ?2 J# `
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
, e& w' k% K& Fwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
% d1 G/ ^, Z6 Z5 g( E8 ?6 xtherefore:
1 N/ ?; P' |/ D& R) J6 J
; [% Q9 f) _* Z/ E{(a+bx)/K} dY(x)/dx=Y(x)

% z6 ]5 d! {8 H+ `+ yfrom 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)

so that:   ln Y(x) =( K/b) ln(a+bx)

; `" W. g' D8 E: k3 h0 b5 \this means:  Y(x) = (a+bx)^(K/b)
/ {- R+ a1 N- C' E7 C( u( Aby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
' e; i& J, G5 K
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！

- O( ]9 w  O) X' c0 B
, W9 N; s' g; U. s5 B9 C" B- 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。
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：
" ^. J1 K' ]# J0 W9 h) A1 F! d
d{(a+bx)*C(x)}/dx =-k C(x) + s
7 ?6 ?! w( D$ Eso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 }9 `, `6 k; E% p3 t$ g8 v
2 w# s3 P9 o8 u7 v5 r4 i, Q9 Z也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
7 T8 w/ n3 x# b3 U" P( M1 E/ g4 `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
并不是   (a+bx)*C(x)   的导数除以   x   的导数。
( e4 K8 L" G9 t- A2 n. J' O1 L- {比如说   dx/dy = (x'y-xy')/y2
! R! i* }! L/ e6 ?' _$ ~6 Sx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
, {1 i9 y6 e+ @0 ?因此：
$ Y# \% v2 `( L& s2 o8 y1 i
9 L7 M" {5 m5 p1 S+ e  Z& t6 ud{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
3 g/ i: e5 k. _% K4 m6 j5 o最右边是x的平方。

所以您解的有问题。
* Z: K; P0 y" g' \, W; J1 J% l
[ 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：

: E% e: @4 W) Y) a% x8 zd{(a+bx)*C(x)}/dx =-k C(x) + s
  R5 U% Z0 o9 x# Q8 m3 l0 Tso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

) |* S4 {5 K" [! q( x也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
& Y. R+ A% I7 c  G- s但这是不正确的
: D& Q8 @1 m$ D( [d{(a+bx)* ...

Please note here:

3 [. K4 ]0 `8 P; F3 Od{(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
5 ?5 ]9 m# W7 A# Hd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
8 Q9 w( L" I" b并不是   (a+bx)*C(x)   的导数除以   x   的导数。
: k, G( [- g7 f8 p
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
....
/ L+ {/ Z' y" _5 {9 F* a* bd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

+ D2 h- h% l9 t
这样算混淆了微分和导数的概念。
3 M- Y8 v* p; k
" |6 T- l& q  H. [4 Z  F/ \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

) J9 x; U" T: V8 C: q# i$ ?  xIf 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 R$ W) ^: \  V& [) {/ J  [% g3 U9 O
" P2 D+ b6 n0 V' s9 M0 S# 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。
0 C# w" N. y% a: S; p' [& r
3 k3 i; d, |' 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
' Z$ I6 l7 f3 G6 b2 s/ S
3 |* m0 }0 P0 y* Q, Q5 U# }8 DIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

$ }6 T% |: C2 G" P
% z5 z- U/ z3 F6 Y2 j$ o% n, 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
" i! r3 F/ w' Q5 P
Ah, you made a mistake: it should be:
% J5 S! _4 ^. g4 t
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
0 Q2 E8 Z6 g, I, }* q3 e

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)

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。
$ x5 D" D; Q- f
4 }$ w3 n" `  Y5 ^+ h1 f1 ed(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:

( p: X" ]4 @3 R. a: W0 E' p( ]d(f(x)g(x)) ...

Yes, there should not be " ' " after "f" and "g", it is a typo. :D
" r9 M2 \9 ^2 e8 {
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:
Yes, there should not be " ' " after "f" and "g", it is a typo. :D

8 i4 x. r& s7 w8 O9 n# T1 KBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

( A- \) i& [1 X
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:

. d$ S; W$ U4 g: M4 d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？

: d4 Y! f' o% l9 c8 ]
  R) E1 u& n  V" }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。