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
7 A1 t: T' r7 _where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

. Z" S& v! K4 o5 o" y- u- p多谢了！
. ?5 f% @" q5 W% M* \2 P
[ 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:
0 V  J5 w% f1 S7 m+ R! X6 _, [- ~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 = ?
4 N  x! [/ R0 ^$ K: o
/ f) r& q' d- x$ `' f5 g# {多谢了！

! t7 S( d& H8 t' {it is too easy:
3 o! _& \! ?% ^5 E
. B0 M8 d1 ~4 N& ^* Cd(a+bx)/dx = b
9 j1 B2 A/ p5 _2 {2 {
so

0 n  }! G: X3 P3 n& Z5 s' vb=-kc+s
5 x  z6 P, N3 E0 ^
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
( r1 u. l- @- V( Vit is too easy:
% n7 ^" N" ^" E* f% c; _
d(a+bx)/dx = b

so

b=-kc+s
8 }6 o% S. Q" ]  m
! t0 _' v* k5 I  Tfinally:  c= (s-b)/k

/ w) G% U( w% l% m' Y3 z, y8 v6 Pthe 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:
2 g# i! J% `8 q" G3 P( g+ g/ A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 = function of x, what is this function?

多谢了！

[ Last edited by 醉酒 ...

4 r  ~/ @  K& Z* U  _* S0 Vsorry, did you post another question? your statement is still not clear.

' Q: G- k4 @1 Ga, b, s, k are  constants? c means c(x) ?

( W$ ~/ @' Q: m" g, G4 X" `4 f# bplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
8 ]7 P& \: V5 x5 o" Z& X- y+ [  Zsorry, did you post another question? your statement is still not clear.
. K$ B% T. J! a7 J3 t1 h
1 k+ X8 \$ G* R3 u7 Da, b, s, k are  constants? c means c(x) ?

8 P: \, ~! Y2 O; bplease tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
& G8 q% E! @9 \1 L
( b, o. e. k, s; d% j4 M+ m. i' Z- `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

多谢

% H' E/ Y3 ]: F1 U1 V6 v5 ~do 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) ?

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:
6 X0 |- ]7 B+ m" ~  g9 c
5 J9 L. g) m* }& ^1 D$ wd { (a+bx) C(x)}/dx = -k C(x) + s

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:
. _' Q$ |( V( s1 Q# b
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
: y4 M# d! k+ p4 a

- G- T2 h; X7 Y2 v4 T" H# rprocedure:

0 }0 v7 A8 Z8 YFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
+ D2 L" U0 Q+ H" v1 y$ @% Gso:
- D# E' |) g; U1 K
# p% N6 G. ^  Z6 G& a7 FbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
" v# o+ {& J7 c2 Q# z6 C4 Di.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

2 ?$ [  x% W* M# I3 ^. V/ d' a
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
! S2 |! Y6 v0 Q& C1 W# [which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
7 H- a9 O% [" F! s! ]1 N& i
. K. V8 R$ }- E' ~{(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)

so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
/ Z  @+ _; D# d" fby using early transform, we can have:
; }$ }; E2 x1 H' p, z# H
: r+ P* `2 c0 k1 I- I9 P1 y-(k+b)C(x)+s = (a+bx)^(k/b+1)

* H; B( U6 c1 ^% N4 dfinally:

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:
( \1 u/ z* s& z. l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。
您说：

! L, G6 r1 z. R) Sd{(a+bx)*C(x)}/dx =-k C(x) + s
  J) y6 C* }( R+ }4 v; Jso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
( X5 w* ?9 |6 @
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
, {8 b3 F3 ^9 z) @, i& m9 vd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
$ \# o# }* n" T$ x- Y+ _7 l并不是   (a+bx)*C(x)   的导数除以   x   的导数。
& d! v  e, G/ t( q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
  I* A# ?, c- U8 q& S. J最右边是x的平方。
2 W( B5 o6 H7 x3 H/ x" S8 A# S
2 \. e! |1 \! z5 U# A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:
6 s1 E5 h+ y7 P6 T% C9楼的答案有点问题吧。
您说：
0 d8 H" |% U# P+ c
; w8 y: F' N0 g" _% s' nd{(a+bx)*C(x)}/dx =-k C(x) + s
: ]6 }; [: y& \2 O% N# n; |+ bso:
3 a4 C$ U* z1 C- x6 UbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
$ d- d& I) `9 j* s0 G* S% X) Y1 @7 N. h
7 w# N8 Z8 r* h  I% b% Q( D+ X  B+ 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)* ...

8 S* B/ ~& ?8 u# d* |8 G( C: y
, s1 L% s+ g- g( s% NPlease note here:
# \2 _# r% x3 C& G/ w
* j# J: B* D0 b# V3 Y# w7 Vd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
+ S9 l# a0 s4 oit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
; j* F( h: r' K) c; K+ qd{(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。
7 G, Z& `0 f8 r! K
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
! H" q- @" B6 j* S....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

! K. J' r& j3 J" z6 ?* C, J) L这样算混淆了微分和导数的概念。
9 N/ c8 b. y; R8 |: w
, F& a* N# b+ j. \9 [, J' Yd(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'(x) = [xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2 (应该对吧， 我也没验算)
, ]3 _. u" h- P# l% @
[ 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。
% i6 w0 {" h* x7 q
. F5 N2 j; n$ S# S$ a5 ]8 N4 jd(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
) e" [) |! C( A4 G
% f+ o5 [5 D  B: x  A% kIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

& q4 f7 [. o  N3 U3 K' Q
# O# C' U" H2 P* X这样算混淆了微分和导数的概念。
6 b0 ]! T. Q( `+ R
% t3 ]* `8 F) q& B9 V+ J9 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
6 r7 l3 s: V" `  d" v8 r
; K& Z- g1 V# L9 L/ y0 V2 CAh, you made a mistake: it should be:
" l: ~& T6 e% ^7 y4 |4 @
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
4 N+ a9 f9 Z  l9 f

* s# B. v# I; u8 zIf 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:
; A4 |5 ~# q) L# z7 H* V' m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:

d(f(x)g(x)) ...

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:
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.

; x$ s, d( P( s; y7 VThanks, 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 ...

# a( g# n5 i2 L; o* F/ x* B3 P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。