数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

1 H" r# c+ f' T" ]+ h  @, F/ md [(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?

! r- V6 l9 z# ?2 n& o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！
: R) |" S; d! y% ?5 U0 S; G% `. l0 K
d [(a+bx) ] /dx = -kc +s
; x5 f! s# B& t. V, H/ F5 x! j5 \where: only x and c are unknown, others are all known,  requiire c = ?
7 u$ B$ {8 R/ O) M
多谢了！

7 g9 t$ y( S8 \; d
it is too easy:
) ~+ h' l/ D! D- M
d(a+bx)/dx = b
0 s$ u4 e' h8 D8 P
so
4 w0 j" \) N- V; ~) S
+ k% `" H2 x; o: pb=-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:
, H, [4 U/ `: e* J# c: d5 A5 Z* l# Pit is too easy:

d(a+bx)/dx = b

! d1 T, `* U+ n( h* r' I# lso
8 r7 U* i2 ]6 R! F0 L  i5 X
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！
. p- `! Z5 s1 }0 M
d [(a+bx) ] /dx = -kc +s
  N# @$ v4 f: N6 K% D. ~+ y$ ]where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

& z" t/ K& L' M# v[ Last edited by 醉酒 ...

8 Q. ~: x+ L7 t6 S( Lsorry, did you post another question? your statement is still not clear.
. m" t3 T- ?0 V/ {
9 ^5 a2 u- e/ l; qa, b, s, k are  constants? c means c(x) ?

please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
; H2 j/ w  m$ d0 d2 f4 U7 Z5 gsorry, did you post another question? your statement is still not clear.
  M& J! E/ r# P
( L  S: L; |* b- U  qa, b, s, k are  constants? c means c(x) ?
5 V# L+ y& p4 w. Q7 p' Z( T
7 T# O( u- ]5 j: L9 l+ R+ q# ^please tell me.

对不起，写错了，忘了变量c,应该是
# q! p  j8 P; qd [(a+bx) c] /dx = -kc +s

多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
+ \0 \' M- G: K( v* _# M- D对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
1 t7 \8 b6 j. R
多谢

. ~- B' R5 |" J- P2 I8 X$ A
do you mean it is:
( @5 Z7 ]  P1 B: l3 y5 b
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:
) q3 F  x1 ?! K5 gdo you mean it is:
% C. e1 `! q/ D  }5 ~. Y$ n+ r" x
d { (a+bx) C(x)}/dx = -k C(x) + s
) ^2 L: M1 g% X  I
8 S1 O3 u6 u5 [2 G4 zwhere a, b, s are constants, you want to know what is C(x) ?

  g& R2 l, Z/ Hif so, this needs to solve the differential equation. please comfirm it fi ...

5 L0 B3 A+ s( x0 x
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:
% s& x" X* T  V4 \' o
' H6 l9 I$ z- G' `2 Q! Z7 n) PC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

9 i0 v! e6 l/ R" B; \& I% d  k
procedure:

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

: C% k" A  v; CbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
* n* |4 S/ {+ a0 I+ O% e; j
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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
. G2 A$ {9 X. }. T$ N( Xtherefore:
/ ~  c: M/ |7 D  g- B
{(a+bx)/K} dY(x)/dx=Y(x)
2 @2 T4 j6 T% J7 B/ U9 v
& v) d/ _% \: ffrom 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)

8 f' k( ~* J1 p* |. lso that:   ln Y(x) =( K/b) ln(a+bx)
3 b# w/ _' h" s/ Q* q9 q. u
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

! c, ?0 y8 @: _) }$ @, [0 Y- U-(k+b)C(x)+s = (a+bx)^(k/b+1)

5 d: F9 h6 L" D" [finally:
& e+ e, M, {. R; S1 n
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。
/ q4 U+ b9 C  ~* ?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。
/ ?3 e3 l2 y2 }+ l& \" Z0 }1 k您说：
+ g" ~7 ?! r& }+ @* _) I
d{(a+bx)*C(x)}/dx =-k C(x) + s
0 ^: e# Z( b: _7 S+ qso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
7 s# a; _0 J" ~" t# w5 a
+ T5 K4 x1 ?; d% P/ k+ K; q; Q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
. l% {# t3 c# w7 N& u% qd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
' _$ A. ~+ ?& L$ }( Z0 v/ U3 {并不是   (a+bx)*C(x)   的导数除以   x   的导数。
; S, U) C# D2 a( g比如说   dx/dy = (x'y-xy')/y2
; g0 T6 f& {% n+ f- v5 N$ Ax' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
% V! Z: `4 j/ l+ k3 ~! s+ Q' O$ ~
- a- \3 h% T' Y) C" O2 Y! Sd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
最右边是x的平方。

所以您解的有问题。
; `: t8 Z& J1 _: P$ ?! n  n" E
0 p; U% N7 v, r& p' h[ 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:
2 v# i, f. j0 r# ^1 b( m! ~9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(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
但这是不正确的
d{(a+bx)* ...

Please note here:
. W8 Q3 I7 M, C* M* @
- B/ _" \! ~7 d6 v( h% a# _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
# I& s/ h. i9 C! Qd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
1 b/ I" h& ?1 U& W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
7 P/ B2 j& X8 G+ |  X  q/ N....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

: J  U" W! J% w( {/ P
) q8 C  I7 S/ c1 a8 r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
; i: \2 s- k( i" q  o& C3 y2 c
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)
, Q7 |/ ?8 d7 z
[ 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
* B4 u* E& |( f3 F: s
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

7 y0 k" K* i3 E! K' d. I. a
# C# S$ W( }4 L% j1 C* C. `0 }, W/ w; R这样算混淆了微分和导数的概念。
7 P; k+ w/ n- q* o
; D' p7 y5 ?( F# @8 m% y! f* md(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
3 u; l+ i0 k( s* l; G# x# _
1 }# e& B" N8 b3 ]7 WAh, you made a mistake: it should be:
8 M3 m" E8 c0 s4 r$ }4 A
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
+ q/ O1 [! ~3 S& D  D1 Q
: S/ T! D' {  B" H+ j- e
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)
1 ?2 r. i0 @$ h: E, h2 Z
0 v" T9 u" g0 Y+ t3 \9 rfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
  H' H  ]% k4 x, q' K- Z
& v$ V, a, L4 `& c- fh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
" v% }  x% v/ U" w# ]0 L) d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
- `. x. ^1 g* R/ R" `9 t0 i2 Q
' T; K( E' n& f2 gBut 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
$ |; H8 v% m3 ]4 }
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

. n* @- ]; D, O; i; p/ u1 m
- d' [' t+ q' y. E/ |" YThanks, 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。
9 t2 P) D. |* ?, G% a3 b$ ]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 ...

2 e2 u1 s+ L; A1 v5 w" h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:
& p9 m" Q6 ?4 A( U# 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？

" z. M8 i5 @+ r$ \  X2 i+ I9 `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。