数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
" t( y- h( @, _$ d! o3 t# u+ M! G
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?

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:
9 `* Q$ H% {- c+ x3 p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 = ?
% N( _+ L- ]1 R2 V7 F% A
多谢了！

2 d' q8 d' I' Lit is too easy:
& K4 n: k: z  L( e6 V% O" Z. c' Q% |
6 S  l$ J9 v( v8 m% Od(a+bx)/dx = b

: d( f/ d3 E- F9 dso
+ ?" I% K0 e# R* _/ b
b=-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:
it is too easy:

0 B) W( v8 @; q4 i- A5 qd(a+bx)/dx = b

* ~/ F  P1 o( n& L% j$ gso
6 {9 y$ B- ^3 G1 @. R* N1 G& Y' r
b=-kc+s
% o' |8 `5 h/ m3 a
finally:  c= (s-b)/k

. P  S6 T: f7 d3 _6 |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！

+ X9 M; S+ \& H* xd [(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?

多谢了！
) a  X+ j: C2 X2 M
[ Last edited by 醉酒 ...

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

) h! F; K" M1 Da, b, s, k are  constants? c means c(x) ?
0 O: V1 }: z8 {# f7 F/ @" J/ Y
please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
) W) G4 n  r6 K5 G3 Lsorry, did you post another question? your statement is still not clear.
; O& B0 s7 q! n4 M3 n$ L/ {
a, b, s, k are  constants? c means c(x) ?
0 `! G- b1 S+ i/ k
% n5 V1 Z. L" Z2 I) Z, ~, z( b- ^9 aplease tell me.

对不起，写错了，忘了变量c,应该是
6 ?: {: B) m, F" g; N5 ?& C+ x$ sd [(a+bx) c] /dx = -kc +s

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
" Y/ [* Q1 L! g, vd [(a+bx) c] /dx = -kc +s
- C) X) E# [% v- U2 \
! e2 b/ C" I: {多谢

) l  e: J  J* {  f. {( n5 T: W' R
do you mean it is:

9 q1 H3 U" a* S+ \% Qd { (a+bx) C(x)}/dx = -k C(x) + s
" u& c. q3 u$ `. h% t
4 D( {% D! n7 J1 ~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:

d { (a+bx) C(x)}/dx = -k C(x) + s
# |' F, g, `9 ?, C$ s5 p$ I/ d
where a, b, s are constants, you want to know what is C(x) ?
' m# [8 e* {* v' o4 k
if so, this needs to solve the differential equation. please comfirm it fi ...

% ~" X( L/ `( G% T" a% l/ R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:
! m6 a, z/ v: ~9 [# {  B
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

/ M- C# z* u1 D: ~+ L* L4 {8 \: B
5 ?0 r/ Q+ N, h1 \# I3 k" fprocedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
) o! O0 V( G- R$ t8 p5 s
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
/ V  Z: R' A  l5 J' R% e
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

% m5 d& \/ B% w& c2 x- d
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
  L' v$ h, w# i; n. ewhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
4 X: G; S; V4 {) H) qtherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

; D) b5 ~( Y4 {& v2 pdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

0 \. k9 D/ _4 M& b# Kso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
8 T% `! {& k1 N7 cby using early transform, we can have:
! I+ `  j. H& z0 X) A+ D
-(k+b)C(x)+s = (a+bx)^(k/b+1)

$ l! q) K* e# m" I! p* @( S2 U. M8 \finally:
' W7 m% `0 x; h' [9 m
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！

3 J2 c, P7 b- C4 ^( d3 Q2 C$ x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。
; s4 n7 y. Z0 A. t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：
% {8 g. k* ?7 y9 O( r' ^. @
d{(a+bx)*C(x)}/dx =-k C(x) + s
# R5 E3 y/ z  d7 B9 N9 i' iso:
% D/ w+ M) f3 |, ?* I8 J) a; nbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
. B) ~7 u' M0 r# n' o8 m0 b1 a
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
并不是   (a+bx)*C(x)   的导数除以   x   的导数。
4 `% D* @& q/ }! p" Y$ T2 f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。
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。
) N- N$ V# V5 `& R4 q
3 u/ x+ l# M9 `2 F9 i所以您解的有问题。

1 c% q1 B8 {% @+ C8 s[ 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：
* g0 E; A# ?8 P0 {
$ @+ U, x3 i: K2 t8 P0 Y: Vd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
( A; E- J  k% k" {; q  H; R  b5 \9 hbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

8 w! T& E: j% s& O# u也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
+ B7 m4 }5 l' }但这是不正确的
# n$ a$ _, t' N$ E5 ^7 F' \3 J) m8 _d{(a+bx)* ...

: a8 a, y, ]6 |5 N  C4 V
, B! X7 t1 Z  |- I) P" Y3 k/ HPlease note here:
) N/ F& p1 `0 T7 w
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
7 {6 y/ \& l# sit 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
* W% t6 d5 A/ Z1 [并不是   (a+bx)*C(x)   的导数除以   x   的导数。

  ?: T2 }/ l, h! t" j; |) D$ Lno "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
# C5 A+ E; }0 h, p, @* M2 }) d4 T....
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。

" j9 B- K) A& qd(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
' U$ i8 c$ w- _2 L
, L, q9 l" b" Q( H( eIf 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)

' ^6 v4 s/ }/ Z. E) R4 c[ 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。
3 I& W( I# l# ~( J/ J; 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

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

! M7 v/ k% {8 ~+ G
# r) E# _, F7 y: |这样算混淆了微分和导数的概念。

8 H% c9 g5 l  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

Ah, you made a mistake: it should be:
5 u) X3 ^( _; Q0 ]: |- P" ?
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
7 `. D, Z* }+ J  m+ z6 c  \% H; w  |
+ ?$ o) \% |6 r5 p" V
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 `4 [4 Q! B$ {! X* R) E0 A
) o& q) {4 L7 Z" K8 Pfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
4 D( }; s1 Z2 s
& i0 g) @, g. Y; kh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
6 n+ q! }. s4 L$ k这样算混淆了微分和导数的概念。

) ^# P8 H; N8 E( ^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
8 }6 w' S6 B0 |, ~  {* F9 J9 Y
8 W1 L5 M6 ]* o) f6 N$ u. JAh, you made a mistake: it should be:
# J8 t3 j% z+ }/ z( h9 m
+ O9 T. A: Z4 r5 l4 t8 md(f(x)g(x)) ...

- V  ^' X2 }& }7 {+ r  q
Yes, there should not be " ' " after "f" and "g", it is a typo. :D

, g1 f! e; `! M3 n& L+ q* \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

But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

8 `+ Y0 Q4 r$ l- _
) n/ a2 x' d3 j9 N9 I! m6 H- MThanks, 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 J) w: w3 M$ r8 l' T8 ]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 ...

, `* s# o# P0 D: n% A& m
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？

3 i) B8 [! [% J% T. ?6 ]; R( |
+ a$ `7 Y  Q! d+ m) ]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。