数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
- F$ u9 }4 g) G
d [(a+bx) ] /dx = -kc +s
7 E- M5 @  L7 P# @# V9 @& fwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
3 L2 `- m- V& Y2 k4 I
9 ~7 S" `% z& H. V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:
请教大家一道微分题，多谢！

5 c* @/ X$ u+ c5 f6 k* R! t3 x) Pd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
" h* Q9 g5 f( e) L" X$ n
& ?; @+ O  v; x* j! F多谢了！

+ Z1 a- {( \+ c
it is too easy:
" U9 F5 I* s+ R) r
2 ~: p7 Y% r7 \5 y2 Pd(a+bx)/dx = b
; N) \* D6 r4 \# D
7 a  e. ~; {4 l/ b6 v5 xso

# c) ?: c: `& y1 ob=-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:
$ V3 v/ ^) _. G2 U' a
d(a+bx)/dx = b
' b. m; w" M# t7 V
; l; O& J8 }* F$ e. w5 p: Iso

8 t. o' d3 m  X) Q( j: pb=-kc+s
) ^  D' B6 ]! u5 ]$ m
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:
% ?! h$ C6 _5 k+ y4 T4 {' l请教大家一道微分题，多谢！

- J& [% ~  X& g% wd [(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?

) M% m# S2 d4 i9 S多谢了！

[ Last edited by 醉酒 ...

0 R. ~6 {/ L  k; k8 _! p, D7 j
sorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
) z8 m4 c  Z8 I, \5 X' M% V
6 I. I# P4 G0 i, R8 ?$ W# A* r+ bplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
: K' y# S7 |  x# {: r4 E! {, Nsorry, did you post another question? your statement is still not clear.

" U# C9 ]/ v1 S- U( u! k, ua, b, s, k are  constants? c means c(x) ?

$ O- j% B- C$ o; h/ v4 Xplease tell me.

% k4 M/ V* }3 O) g0 i- P' K- h$ p对不起，写错了，忘了变量c,应该是
6 I, W$ W+ k6 `& q+ Md [(a+bx) c] /dx = -kc +s
3 E0 a( o4 H8 y  M. j* h% V  D
; e& q3 z# v$ b" J5 v多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
6 @% J% Z( e# Y+ Q9 c! Q3 j对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

; k# J" Z% n6 w$ U+ s多谢

do you mean it is:

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

: u1 ?/ ?( Z( e) @where a, b, s are constants, you want to know what is C(x) ?
8 `7 U2 X( N7 l4 @
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:
' n3 j2 g. J7 L4 l( e. s
% z; u* e* D7 ld { (a+bx) C(x)}/dx = -k C(x) + s

* @7 k7 c1 Q* @$ e% ?) B8 Y" bwhere a, b, s are constants, you want to know what is C(x) ?
5 V% A. n) W+ N$ W
if so, this needs to solve the differential equation. please comfirm it fi ...

: u9 \/ k1 q# v/ U+ J( ^: {
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:

& B$ i. I: G: k% T" {- `2 {C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
4 Z& b; m; T; o; }
  p/ F/ _1 ~5 T" I9 j
procedure:
- D2 T* W1 b2 k0 g3 h0 ?
- w  R" M9 O" ]+ _9 h% L: Z' @From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 B' y0 W' k% ~$ mso:

$ Y1 x1 @, ~$ d% }bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
/ `! ^$ ~6 r- n0 d/ ^

' V* I& ~- k5 N1 M8 @: j0 o7 sintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
9 k! V  b" M4 p8 ewhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

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

from 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)

: i* ?7 Q" t! I. L5 z3 j9 r  Eso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
  O* t2 `$ \" z3 h. T0 |
7 P4 M6 E' Q) z! B. a! Efinally:

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！

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。
, Z9 a# K, W/ L* g+ V! X4 V您说：
; L; c; P# o# H3 h$ g
d{(a+bx)*C(x)}/dx =-k C(x) + s
" I; }* i1 j+ r" vso:
9 w; x0 S- H) P( @+ r& Y2 y2 F7 RbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
, k; Q' u2 ^; S, ^) K8 p
0 P2 h! n9 ?" {8 T) o也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
+ h0 @/ i% O  u  C  X1 ]1 z但这是不正确的
* e( @7 {: n  Y* D' F+ w- c8 Zd{(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。
" T0 T1 o4 C* P5 j" t9 v& h3 z比如说   dx/dy = (x'y-xy')/y2
: ]' ]9 `8 R. z* c# Px' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
, b5 ?1 @0 d9 Y) t
1 i0 e" e& d3 Y" m; md{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
4 l% I3 L, I5 Z/ P* e) l/ a' G最右边是x的平方。

所以您解的有问题。

[ 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：
& f0 j, u  b/ @$ B: g
d{(a+bx)*C(x)}/dx =-k C(x) + s
, r3 I# c3 r: Tso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! W& G& v. n6 b1 E: s
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
5 Q: t' p3 L/ V3 Y$ q! A2 \d{(a+bx)* ...

2 B: g0 q2 N7 }  V$ A' P
Please note here:
; F" S1 e( r( X( A6 B/ R
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
; @8 h2 S1 d) C" wit 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
  z, y$ u6 r- E9 `! q# z3 Q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
% ~' }% D. g) j....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

4 g2 T+ N7 N; u
% X) {: }" \3 O! u; _! n这样算混淆了微分和导数的概念。

+ t$ f) X% {/ C$ r, R2 Od(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 h3 K. y3 p9 x! j- R- HIf 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:
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 ...

1 G+ d6 Q( @* @: ]% _0 T
" p8 h9 ]% m# ?; N这样算混淆了微分和导数的概念。
! I0 @' v3 x4 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
* U. g! b  k: c) `) _- d
2 d( h$ F0 X: D0 ~" q1 lAh, you made a mistake: it should be:
8 n! I9 r. p4 l1 a+ M" \' M
/ @4 y: q# L- i: ed(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


) q, y5 n. Z, Q1 {. L7 N1 ^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)
( ]8 R9 Y3 [# X0 v( h. |( v
. g" a; G) Z% W1 r' vfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
; V2 M9 J0 o1 E( y% A! }4 M
3 l- ?0 ]; w0 m& p. K& t' E/ c. bh(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

Ah, you made a mistake: it should be:
% H* t- |* Q9 @4 O
d(f(x)g(x)) ...

4 d3 v( Z) Q9 U/ }
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
8 I3 M* x. W0 W% u
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
7 u6 m$ f+ I" A1 t5 q$ G
$ \2 l; L  Q, ^& mBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

& Z! O& j# Y9 f; z6 i! G
% p; L: V. S- [$ z6 E/ S* gThanks, 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:

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？

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。