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
3 \! a5 C4 g# z# Twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
: u; q$ r; k5 @! t: 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！
6 k* {* o% Y$ l5 V2 s  g' L4 b: G
d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

: q  ^. ]- h" Bit is too easy:

4 L; \& R$ Z* @' y; h& a. U" R! f/ Vd(a+bx)/dx = b

- @2 |1 ^( t7 Jso
0 E6 A7 y1 h# V0 B
b=-kc+s
' }" s* g" u6 ?/ t9 p3 n
1 I0 I) O5 H) `; x2 qfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
8 w7 v0 T# r- F  o; W  }% uit is too easy:
( [0 q* Q. n. j  `6 K
" h/ k% U4 {; _5 ]* }2 \d(a+bx)/dx = b

so

b=-kc+s
) M0 `3 z1 j: t9 x$ Y/ E. Y2 I
finally:  c= (s-b)/k

. b* t. G: j0 k) b. d4 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！

" s- S7 D) L& b, ud [(a+bx) ] /dx = -kc +s
2 x! A$ w, [  m3 Y( h! Swhere: 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 ...

+ a: K) k! b# i
sorry, did you post another question? your statement is still not clear.
1 X1 {6 @. M. e% E
a, b, s, k are  constants? c means c(x) ?
( s6 I, S/ a" }9 Z- ~
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.

a, b, s, k are  constants? c means c(x) ?
8 {  {6 Q0 L0 n5 e* X/ m# T
please tell me.

. n  `8 e6 P5 W# y  F对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
$ k* x& J( _1 k
. @! B8 {% M1 T" r多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
+ s% h6 K# w+ h! M3 `对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

8 i! H1 s) F3 a0 R8 @5 X6 \* G; _多谢

/ y9 _* J2 t! \3 I
do you mean it is:

$ y* o& o1 g5 l3 d3 md { (a+bx) C(x)}/dx = -k C(x) + s

3 [4 J: y8 H8 l; N  m' K. Q- lwhere a, b, s are constants, you want to know what is C(x) ?
, |/ ?  o+ M- h7 B
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:
# c. S. M& R5 N" N3 vdo you mean it is:

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

where a, b, s are constants, you want to know what is C(x) ?
! w; B6 ?3 W2 J1 U
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:

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

十年移民路_

this answer is the good one.


procedure:

, e5 Z3 \; C& I2 U+ J# w2 _, B" zFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
# [+ Q- B; u5 ]* x2 ^so:
% s: i+ [. K) B& l) b0 q; I5 M# F
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

0 m  Y+ O4 i, M8 G% @/ U& z
introduce 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
3 H' U5 J( I% e. s7 X) t9 ?therefore:

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

3 p+ v- I4 `! h& x2 g; y% E- R" c' ?from here, we can get:
0 L% w* l5 M! L4 T  N2 r
% m# {$ H7 G: m( o: v8 adY(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)
% f$ B9 V" F, l+ ~: G
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

7 m  N: U6 ~( o; e9 K* J-(k+b)C(x)+s = (a+bx)^(k/b+1)
2 `7 C% n6 d1 X* [4 O
finally:
% G. n+ T8 l8 |5 |
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:
7 J& N+ q1 @6 {, C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！

: E- z, \( F- T, y0 O
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。
& Z+ L0 P9 C+ y0 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）。这种题真的不是难题。

龙舞九天

9楼的答案有点问题吧。
' _; E8 }- e; {1 M: p  ^5 ^( ?您说：
1 @& s) Y+ E( Z- B5 d: x! g6 V. L
d{(a+bx)*C(x)}/dx =-k C(x) + s
; Y2 }9 L2 p0 v5 f+ I* R4 W& D  g5 W- Z% uso:
; q2 U% m3 [: a4 o: B7 VbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

$ l# M; [/ S! }5 e* w也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
; C; _$ r6 k8 V" \. W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。
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：

  f; `* o, j0 Yd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
$ q7 y) B7 K8 ]0 k/ ?* \+ a6 ~最右边是x的平方。
) _% |* ~5 E+ }+ x
" V2 R2 u6 b7 g+ W% }- l所以您解的有问题。
9 P4 [- P( j( a* A
0 W! ~2 X. w& `[ 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:
% t& C- `- ^# B9 v7 C$ F- c; H' N9楼的答案有点问题吧。
4 m% Y  m' ~! k& [7 ]0 w您说：

, z8 h! L- V; R; L3 t- F% qd{(a+bx)*C(x)}/dx =-k C(x) + s
# q3 B8 K4 i# [  S$ |, Qso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

5 ]8 {( Z: g/ ?5 Z, g# J! x) _/ U' P也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
* ]" j& I2 S  m* h; m+ x, @& u但这是不正确的
8 r1 M* `+ \" k& O9 y' e  fd{(a+bx)* ...

7 |/ c5 l" `4 j" M/ o# U
Please note here:
! `  G6 w' G: ^: V0 O
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
0 y3 `. g# p9 C  I, ~9 S# 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
并不是   (a+bx)*C(x)   的导数除以   x   的导数。

0 w6 S% j% r5 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

2 B' Y) Q" m! o& P+ g
8 c* U, \, r  d" `2 X7 m" \这样算混淆了微分和导数的概念。
8 U* z, c5 s7 ?! `' ?
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
* w: c; w. r- _
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)

[ 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:
  B' m) {2 Q3 ~2 k: B: A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

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

0 J% S$ @8 c, i! D) D* L
这样算混淆了微分和导数的概念。
) L4 B1 s# k* |) Q( b3 k1 K, 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
+ v* n% a, _5 z' f* T' I4 u' q
Ah, you made a mistake: it should be:

+ B0 F8 i. K7 p2 J0 c7 Y; rd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

# g0 S4 p  d5 e- X# t
1 N. `; |6 N+ D$ x- I, N& h# 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)
) m0 t+ p+ `8 Z4 r5 k0 y
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
$ G+ v  _+ S( C& F9 R6 A5 B
$ O5 f. @1 \7 A/ n* b& z$ W( I5 Wh(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。

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
5 G. B0 R5 o7 F% V0 u/ a  l
Ah, you made a mistake: it should be:
  t8 w0 O6 f+ A5 K- b. b1 F. x
  n$ F+ Z6 n$ p1 Y* W1 p$ xd(f(x)g(x)) ...

$ U, Q* P) G7 b" ]" k3 U$ t  j
+ G$ g" X* }: @$ X' i/ nYes, there should not be " ' " after "f" and "g", it is a typo. :D

1 N& V  p' B7 V/ ?0 X: T" c" pBut 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
( `! b, ]9 O  R& W9 _$ c9 {9 I/ ]
7 V9 d1 e" S' \1 r- |* R! GBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

+ u2 L3 E  O9 h8 U) L+ T# z* m
! e! {" _/ z' |+ XThanks, 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:
8 Z: ^1 V4 |- u& j3 `$ }
& \6 w: s' z5 l9 H% X% ~  r: K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 ...

7 C& v8 G: ~3 _$ k) g
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？

% [) q9 ]) \( f; @  r# x2 c: W
$ O9 x7 {# C) Y: @$ d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。