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
/ o; p8 ?9 z6 P  s3 [where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

) f. `+ d" ^0 g& [0 U3 s' L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:
* ]: f+ `$ {4 K) Q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 = ?
. |. {: D& e1 ?  ], o* d
多谢了！

+ c. {  N8 _  C% t( o' G
it is too easy:
( H  P0 p1 X! V+ g
- `1 ^5 ^8 i. e/ Zd(a+bx)/dx = b
, L$ }6 p. v7 d! f: ^; {& t
2 y% w$ l; }" M4 Lso
! `- X8 {7 {* n; Y% w0 W2 i+ {
b=-kc+s

, g3 m) e  Q9 D8 X+ F. b. Kfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
8 b1 I$ @4 G% T5 `4 _! ^it is too easy:
; m9 _( R# H1 G, u
6 [' {# u0 ~/ y$ z) Z  W* K& P; ~0 Rd(a+bx)/dx = b

so
  U/ q! l7 z: X4 I% N4 x+ V; l2 W
- l( N6 ]2 W1 x; [, ~) G7 ^" r$ }8 Jb=-kc+s
, I% K" D* H# U* r2 ]
finally:  c= (s-b)/k

, F$ h+ I) C1 E+ _6 o8 \) g. R: d
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:
5 g2 d9 F" s& r/ J" V$ V0 V请教大家一道微分题，多谢！
+ }" J1 V, W# L# o
  Z3 E3 R  L! Ud [(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！

$ @7 M3 }6 m! u6 P! ]  X[ Last edited by 醉酒 ...

9 j1 r! O# c$ B6 X2 Q
. k& _5 s9 O  u& @0 W* _sorry, did you post another question? your statement is still not clear.

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

  f; e- Z/ `: H) N4 G( |please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
% x: o4 S5 i0 b& Q2 e% T  Esorry, did you post another question? your statement is still not clear.
4 O' `8 N2 \, R, ?5 ^
a, b, s, k are  constants? c means c(x) ?

0 |4 i4 m6 ~. I' nplease tell me.

4 w; w) q$ _! E4 Z; I0 h4 a$ O9 h对不起，写错了，忘了变量c,应该是
/ n" W7 q+ u: m* U* K9 y7 yd [(a+bx) c] /dx = -kc +s
5 Y. `' n: e6 O# Q3 @/ D  R4 D$ K/ I
) k0 m! K2 r$ r$ Z! Z1 C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
1 z; p/ {' _- ~  H+ ad [(a+bx) c] /dx = -kc +s
) L, O4 d. I* j/ ~, c3 J- j; K, M
! U, ^8 G. n# j8 {/ d0 X3 a多谢

, X6 k# y; I" I
do you mean it is:
' _$ Y7 h  a% i, o( N3 ?
$ }( d7 M4 q, i; `' d+ O3 Xd { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?
" b3 ~6 m4 F# J  C5 Z3 M- {' r
8 k4 p: }, w. w2 p$ {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:
! o& Z1 E2 E6 y, O/ D1 r
5 a% `1 S$ H, N# J. [d { (a+bx) C(x)}/dx = -k C(x) + s

, F8 v1 ?4 M# U+ }where a, b, s are constants, you want to know what is C(x) ?
: Q+ M7 d1 E2 i8 U/ b
if so, this needs to solve the differential equation. please comfirm it fi ...

6 Z" J) k! A8 O1 p& c2 O. l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.
, P  i" }: e  d2 J/ y& W: ^

4 I. F( O4 r- Oprocedure:
  i6 s- l- Z- U3 X) d& ]! @
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
' o! D* ]' o) E% ?so:

0 Z* @7 B! Z2 N6 z+ ~+ c, L8 d: U! zbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
. a/ `  I/ {4 ?, w1 Q' _& b" {i.e.
# {' |# F3 e' V. Z/ {
6 ^, Q* x7 i3 s& a; D; o(a+bx) dC(x)/dx  = -(k+b)C(x) +s
# A8 a1 v( h- O0 l$ x; S# C) x! |
/ f# O- }: r  s/ w
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
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
5 X5 A$ k) [( `$ U
# u$ s1 S) S- A$ O4 u) zfrom here, we can get:
9 z: A4 |" i' p2 u& ~) ?: P1 N( \
5 R$ D: P  _, n8 y( T( g: R: ^dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
8 r2 B0 T2 T" L4 X9 C
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
/ c  V0 \+ |7 A7 a6 c5 ]by using early transform, we can have:
* \- ]$ O, s8 X7 M! B9 y& g
-(k+b)C(x)+s = (a+bx)^(k/b+1)

' A! X( J! p( H2 I4 D5 e) A/ y1 bfinally:
: k' y' D' \' i, K4 X
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:
' e. c, x8 s4 A1 K( j5 y+ r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！

. O, i  h- [% i7 O9 P5 |4 s# p# ^
, a* q* J6 l, @& l# M; w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。
# k$ R5 z) J4 B; x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：
5 A/ R6 b+ p7 l! b8 u( w/ R
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

  H1 P9 M7 N' v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)*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
1 t8 q- I! F$ K% |1 |" Ix' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
) \3 k2 D# ~4 u- q) B2 W
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
' i) K% q5 {5 C7 S1 L7 I0 k最右边是x的平方。

- Q4 f% h1 e1 I5 j, E' _所以您解的有问题。
7 x; B  M" C' C7 a9 g
' H% e5 [0 Q; w[ 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 |: E' j" u$ K9楼的答案有点问题吧。
您说：
: |3 v8 Z' J% }8 G4 V. s# z
, }  E, h# @) j1 W% J; q/ jd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
7 `/ m) r0 z1 T$ UbC(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)* ...

8 S' h; w* {3 cPlease note here:

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
8 g# c# i' S2 ^. \% Xd{(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。
* X- ~4 @3 C6 M: y  ~
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
! T" F) ~. }- [....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

+ R4 X" Q7 l( W. O3 i
这样算混淆了微分和导数的概念。
+ \4 v. b& }) j7 R: 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
% Q& X! H( ?9 j8 l
; A' ~$ K  I. w4 g7 nIf 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)
+ @% a3 l  C# z8 m/ f, q" Y
, @: F0 A" f; U8 R[ 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
. I, Z# N& p# [8 C; T$ e& V/ j
* E/ Y# K- _$ E+ DIf 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。
& m* o- J3 x, e) ^& @
5 `0 q  u" X) L1 a+ v% K5 Ld(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
( a# a# {0 f0 t, K
& ]: g2 w9 m* m; \4 o4 y5 @. L7 EAh, you made a mistake: it should be:

d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
* s/ {0 w0 Z+ C% p9 i) J* s
  c8 V, e3 {+ s0 F
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 j5 I$ C/ h8 u$ r8 C$ K8 {8 V
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
, o& E6 q4 |& Q7 o- |7 N
" n" x) T" [8 }- c7 h1 w- A! S1 Ah(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。
! B! Q& R# \' C+ 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
+ t1 [" Q2 n0 F3 R; n
Ah, you made a mistake: it should be:
7 Q& O6 M; [, Z
d(f(x)g(x)) ...

" Y) G- S* J& E' l- g& |) E4 k4 U2 h5 FYes, there should not be " ' " after "f" and "g", it is a typo. :D
( l2 ^1 q/ r3 U+ F: h
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:
7 p" ?- ?  N) s% U0 }4 cYes, 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.

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:
$ r/ K& \' n- R
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。
5 L9 _, a  L  S. M  E8 _) `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:
& m& s5 H6 X# m- B4 ?/ J) F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。