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
4 Q3 Z+ ]  e7 F" @where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
1 N$ L8 J, \% t0 W* @, u9 E
[ 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！
8 {) T  h1 E, O% f7 p1 I
d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
' v" O+ ]8 Z/ m2 C5 _  h% Q
多谢了！

/ i; I1 G2 P( j
! W  R# q7 i. c& w* @6 E) hit is too easy:
& n! ]' q% }* g' ~
% L5 T: c2 E% d5 yd(a+bx)/dx = b
" @( y4 F5 u; @0 A) ^
so
* C7 T' x. t: I! N( X  n
6 p+ f; G, E5 x, G( [# z2 eb=-kc+s
! M8 j9 I/ T8 S( r! @
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
0 G7 }- y; a& s4 k* rit is too easy:

d(a+bx)/dx = b
! b3 H1 r* ]  `3 h& _$ i
7 M* M, s5 n9 h$ @$ v& Jso
) V9 [: x3 Z0 s4 b
7 c7 ]$ u' I, F9 }  c3 n0 Db=-kc+s

- o  p+ c4 Y; e5 E6 ufinally:  c= (s-b)/k

+ k0 M1 @: c! f* L
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！
- `4 r% q. q; t! Z/ l3 m3 ~" R1 t3 B
d [(a+bx) ] /dx = -kc +s
- k* [9 i1 i' r8 t9 I/ ]0 y. j) C6 owhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 K8 d" j5 h% f0 A
多谢了！

1 L5 t8 W8 g! D& p& N% o" I6 `[ Last edited by 醉酒 ...

( r$ P6 Z, d7 M9 B4 O1 z& X0 x7 tsorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?

" p! `' C' @& Q) a( oplease 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) ?

6 c* C# P2 f, J/ A0 c6 @" ^please tell me.

0 V8 k8 O- z) q$ ]* S1 ?/ }对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
  _& g( y7 w' V3 M; }. ], J
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
6 ~, g+ I; ?5 A1 f+ P5 |# k3 f! Y
多谢

! o$ o# ]% T3 c* k5 I# B9 @8 R. \
" o1 E& L; v" U$ {/ K' gdo you mean it is:

% B% e# [, a+ Ed { (a+bx) C(x)}/dx = -k C(x) + s
+ `" H' e: U; j$ {: w0 k3 u2 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:
! a& {; m  Q+ |) }, g
1 Z7 b# V0 i+ t. y! z# ~, Y! yd { (a+bx) C(x)}/dx = -k C(x) + s
% @4 {& U" {9 q4 m. x
1 I: u- i" @% p6 d' i- Jwhere 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 ...

4 s7 V7 q" C# j7 t: o) M8 _9 N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.


" \$ }! S. @5 E; eprocedure:
: ^  x! M, }' K" U
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
: i6 Y" C9 J( _9 |
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
6 L4 s2 k( U/ A8 ^1 R8 m
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

; ]- j/ N* Q! B: |
9 d% m: X& e5 S9 H5 t5 D  ?# a# r) W* ~introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
# h  V( Z: E* O$ s! ]6 v7 z: Pwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
: r" h( Q+ L+ ^/ ]therefore:
* r3 ?9 b( ~% N5 ~1 x5 c' P7 Q3 }
9 C$ a* v$ Z2 s( R* ~& z{(a+bx)/K} dY(x)/dx=Y(x)
0 q/ g6 d- W, ]( ?2 }
$ v. u9 M8 P2 {1 Zfrom here, we can get:
- {4 W! D; `$ F3 N8 C: ?1 `- Y/ b. S
8 E% [5 f! u( k* H- c1 NdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

' Y; n' Z6 T6 P- x* cso that:   ln Y(x) =( K/b) ln(a+bx)
0 e; c2 v" v5 ]; [  m# V
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

' o6 V6 o4 N- ?-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
( C( t6 l; b# ^/ r
+ F: ]5 H" {* B& a* R+ z  yC(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:
5 y: _! O1 Q+ M2 k( U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！

, |6 B$ h; q( r% D3 {0 p) N
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。
( Y8 n& q7 Z  m. m您说：
' `8 h% J: K% Y6 m# r2 c( R( R! [
# o7 i6 n- A, D' |: g3 }+ Id{(a+bx)*C(x)}/dx =-k C(x) + s
so:
1 z0 ~' X' o0 h" Z- ?5 wbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
; `1 K5 J1 N5 G; i% R$ y" M
" N5 k/ c# U) F* ?4 H也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
0 l% y# \8 ^# f7 Q$ e但这是不正确的
/ G6 I6 R6 }  u; e' n3 c" j8 K3 R( ld{(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
* S2 X- f" m: A! j1 X- i; mx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
$ J, z4 g( A0 t" ^, h因此：

7 x4 |9 G  O& A+ G) e- j3 t8 ]: gd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
! z+ r; [4 u- @: N) j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。
) ]$ t4 I5 R. O8 P) Q您说：

5 P9 f0 c) \5 e, c9 \+ z# Cd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 m" d3 F! Y8 w& E
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. O( L5 F6 ~( J0 F3 v" e0 B但这是不正确的
d{(a+bx)* ...

Please note here:

" g6 O& q. b7 b7 Fd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
' u0 m7 Y3 |& K! a8 [3 d  @2 sit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
" j4 J9 V$ l& d' r) G" Ed{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
1 I2 i4 I& w/ j& O' \) T: F并不是   (a+bx)*C(x)   的导数除以   x   的导数。

0 U  G4 D: i* M6 {  ]5 g) ~7 J( Yno "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。
! u% Y; h# R3 C$ y0 U- b8 J/ W
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

/ M0 }# x$ @# |! z0 `0 lIf 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:
8 _6 K" T, S1 X& [* `  O这样算混淆了微分和导数的概念。
6 S$ W5 z/ n% W' ]0 L( u
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 ...

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
- }8 i1 k0 ?* e, Y; E
' a2 f5 J2 c) ]; r. LAh, you made a mistake: it should be:
9 X) k( S# t7 {
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

% I% [, H+ k# u; W# u
. u# m: a8 f9 ^! y" OIf 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)
' K( e8 Y  G( o( {: r: t
! U) [3 p0 L7 j* Y# K+ Pfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

, K% H! g8 d7 `: u" I8 j8 C5 ?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。
6 m: l" U, W( |# T3 ?1 Z4 K6 D9 p/ M* `
( [* F9 ?+ Q* Q) }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 W; H. w! F) G+ g  {7 c6 y
Ah, you made a mistake: it should be:

' F9 ~% H4 }6 dd(f(x)g(x)) ...

' ~( ?. J, D" X/ wYes, there should not be " ' " after "f" and "g", it is a typo. :D

* r1 W( s& P  a# M3 i$ e8 \" iBut 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.

% \6 R7 A3 V/ f! Y8 dThanks, 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 q1 G3 ~6 z$ }
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 ...

: X1 ~, x" g8 y" _+ E; ^
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:
7 X1 ]6 ^6 Z# 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？

& y: {. S. t" \" R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。