数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

  t: G- w& {5 D9 V) w/ Ad [(a+bx) ] /dx = -kc +s
- e& k! U$ u* _9 o& n7 iwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

; Q: S8 ^& ]1 k[ 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:
4 a# @; A. S4 T+ F/ M9 C* a; i请教大家一道微分题，多谢！
( C' F$ z9 `2 k: S0 C: P
3 c) e. Y" s) f. I" Z* zd [(a+bx) ] /dx = -kc +s
4 q  @4 |; f5 p- t$ J( cwhere: only x and c are unknown, others are all known,  requiire c = ?
, m9 {( n8 R' o
多谢了！

it is too easy:

' g7 \8 I/ o6 A* O! Ad(a+bx)/dx = b

2 @0 |; u8 ]1 C6 N. V0 Pso

# x+ `% F. ~* ?, s1 d1 y  ub=-kc+s
6 B/ l1 m" L$ B
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
3 @; u0 g/ R2 ]it is too easy:

d(a+bx)/dx = b

# ?/ s) @& c7 N$ L/ h) S6 Oso
) i0 x  x) T5 _/ K8 K
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:
请教大家一道微分题，多谢！
% K( o3 E; [) Y. J8 I
6 O8 V. I% C9 a7 D7 _( s, D% vd [(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?

4 Y+ I% w* L" o3 H2 O0 F8 E多谢了！
& }# E, o$ B& q5 X+ u
" `0 E# ~# O7 c5 g[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.
$ O! Y8 n, I# N4 Z6 M
a, b, s, k are  constants? c means c(x) ?

$ w7 ]& g: t$ k7 Cplease 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) ?
! t# ^! U8 G' Q) U& l# }0 l
: |/ _1 }6 f' H) Aplease tell me.

/ i! F5 A" m+ Z对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
% `% F7 E5 @' h/ s
" C' X/ Y* }; w! U# V# D# l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
9 v' u; _( X" G4 H6 Vd [(a+bx) c] /dx = -kc +s
% ~" V3 i+ o4 |- O- ?
: @# U5 x6 {! ?多谢

do you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
( v' S" Y$ L. X8 Z0 J9 u
where a, b, s are constants, you want to know what is C(x) ?
5 e9 c- N9 B- b, j- `
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:
- b) i# O2 b( H) w( j8 K1 Zdo you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
; k5 v  _' j* T8 \5 ~7 @" k
where a, b, s are constants, you want to know what is C(x) ?
- `6 T; T  w  e+ J2 Q
if so, this needs to solve the differential equation. please comfirm it fi ...

  a. g3 B- w  S! z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:

; r7 d: A9 t) \# y5 l! vC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
- k8 ?8 G; x2 \8 `; H
# v+ w5 P" s1 T0 f
procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
% w" h; F" u, N# t9 `
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

4 n0 K% c  z# m) g2 E4 a
8 l' e! m- u% F$ kintroduce 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
$ A# f1 h. [' Q2 ^/ t- Otherefore:
. ~8 A8 p0 D6 C0 s0 ~0 w
{(a+bx)/K} dY(x)/dx=Y(x)

7 O' G8 `& u7 t1 Xfrom 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)

1 ^) U7 n$ F8 ~6 Y0 x- k& D4 J4 C, vso that:   ln Y(x) =( K/b) ln(a+bx)
( b$ H+ b" @& \% C$ q
this means:  Y(x) = (a+bx)^(K/b)
# O& h# q  k2 D2 C4 m& b0 \by using early transform, we can have:
) C  y) R( _; h
-(k+b)C(x)+s = (a+bx)^(k/b+1)

  ^( E. A  c1 O2 i" }0 e, w: Dfinally:

1 F' W& u/ F- V$ T, xC(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:
) X7 d- M1 o6 o$ `5 F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！

& m; e2 V  U( ?" V6 q- G# z; S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：

d{(a+bx)*C(x)}/dx =-k C(x) + s
1 t8 S9 k' ?7 }7 kso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
$ o) ]7 b# f# R+ Z  X: l% Y6 j
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
) B5 G4 e6 ~, {* f  L8 I7 Y3 t1 o* H+ Z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
' I; |6 G0 B7 N, c, k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：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
* c/ ]/ n7 C1 k4 A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:
+ j5 L+ B$ x) W/ k+ w) s9楼的答案有点问题吧。
您说：
6 s1 a8 J' @- k  r3 H) T1 f' s$ T8 n
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
' ~& y. o/ H6 ]! f, H# HbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
' q& V  i) }6 z; o; {9 |5 k但这是不正确的
d{(a+bx)* ...

Please note here:
, M8 @. ?! e" ~. l( O9 O/ k( A: I2 B
d{(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
d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
" n% }  p4 I9 Z* y9 ~' \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{(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。

6 |! h$ @2 U6 U6 td(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

1 [. w2 z8 b  m* F' `1 RIf 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)

0 X# o/ n4 y! X' a[ 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

; q# X; U: h4 y6 yIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

1 u, r0 ~7 s3 y+ g( g6 b: u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
9 C- J$ @# L( p- R3 u& J. J
Ah, you made a mistake: it should be:
1 K7 V4 k6 h9 d
7 {1 ^% p7 `& O" u8 s  Jd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


. h% {# y9 {7 b% B: }3 iIf 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:

/ T0 G& q8 n& {- f( Y! Z6 g0 `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。
" }& f3 N$ ~% F# i, Z+ W6 k
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
. x; X( R2 P" N4 ]- N8 L! ?
Ah, you made a mistake: it should be:
0 l+ K) ?8 {2 I6 s3 i! \* m1 r
d(f(x)g(x)) ...

& }2 L; l- ]  L# P, \* g. Z
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.

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:51:
* W) t. i2 u- w" HYes, there should not be " ' " after "f" and "g", it is a typo. :D
8 P$ \9 ]  m! V
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

+ E$ G2 v2 J5 z8 {: J/ w# }1 J) \
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:
& F0 V* f+ f7 d! j; x* X
9 }1 l( e  Q2 n2 i+ z$ q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。
- Y5 R+ e& G/ [8 U7 }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 ...

! f# f% d2 ?/ D2 }3 l9 o) q+ T
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？

2 H3 S/ j  D3 C
" `& x) t2 X  A2 J0 e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。