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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
2 E$ V* c6 j5 f
多谢了！
2 K1 F4 f% k8 E% A2 s
# Z% ?8 L: B" e* Z3 x[ 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！
# Y4 ?5 ^. Z; O8 b& C% x
. W. {( o% u+ k/ ?1 dd [(a+bx) ] /dx = -kc +s
) I/ b+ r1 m, E+ jwhere: only x and c are unknown, others are all known,  requiire c = ?
4 O$ j* H: E  i! E. @1 X  R5 }
0 {. a1 C  p4 d% F- R多谢了！

; C3 F! S: L8 h# Wit is too easy:

! |, T, I7 C2 ~" [5 r/ a7 f8 nd(a+bx)/dx = b
8 E# f% s. h8 c9 p
so
& {0 }1 N5 q; ]- e/ F1 g
b=-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:
  W8 \3 s  Q) g( u) fit is too easy:
  Z* S* V- i) i6 {- J
d(a+bx)/dx = b
, _+ r, j' z$ a9 v8 \
0 ]: G% W" P, G2 U# L" sso
( s% T! M) M0 H( H: v3 }3 M3 w7 e) \
0 V- S/ `' c2 F. }$ H3 p' Sb=-kc+s

finally:  c= (s-b)/k

# m, W2 G5 l& M3 v- ?5 f
) T( u1 I! M" 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:
3 b9 k. @7 F- c: s9 U' x) ]请教大家一道微分题，多谢！

5 Q' _* T- H8 }; d% ]d [(a+bx) ] /dx = -kc +s
: E3 M( B5 K6 t; c, f+ G' m1 i) gwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
1 Y. m5 o4 W9 k2 e2 p; ]. z  R
1 H4 Q4 ?. G6 }9 E( y多谢了！

- `. S* Z5 U6 V6 J[ Last edited by 醉酒 ...

5 z% Z5 E9 o$ o8 i5 x/ ysorry, did you post another question? your statement is still not clear.
! h" ~, L7 d0 c8 B2 G
a, b, s, k are  constants? c means c(x) ?
7 f* t+ Y! M2 L+ ^8 M
please 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) ?

please tell me.

- Y6 U" M8 U, b  a0 W对不起，写错了，忘了变量c,应该是
d [(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:
对不起，写错了，忘了变量c,应该是
% [- c9 ?: R5 L4 i& @9 pd [(a+bx) c] /dx = -kc +s
9 l2 W  \1 G* {  m7 l4 E* H2 Y
多谢

. p& l) L9 P6 P/ T4 f, f0 fdo you mean it is:

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

+ j( E; h: K  ~, y2 P: ]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:
4 \' v4 V& p& T4 b1 o/ c# Pdo you mean it is:

$ g$ b, @: m( w, P$ Md { (a+bx) C(x)}/dx = -k C(x) + s

' D$ t4 p8 r, B2 J: _where a, b, s are constants, you want to know what is C(x) ?
5 D% j6 X, r5 [7 |7 e# l# ~
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:

' h8 D4 ]0 F4 r( W0 r4 c) m- XC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
; \. G( F9 W) g/ y

procedure:
. p: H; X, Z$ ~% a7 R; O/ `" V( e/ C
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 S0 x0 ^6 I) w! c, `) Y" Vso:
% {7 k# _# V! r$ A
0 j6 }  _, G5 W1 Y3 q3 J" ubC(x) + (a+bx) dC(x)/dx = -kC(x) +s
" X3 L8 M, R; b1 P  {  X( gi.e.
; r  y. B3 `% o4 F& G6 X6 w0 Q
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

' C: o5 C% W% s* p
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:
1 Z  W: b! B- ?- Y
{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

( M& d( k  `/ w/ t5 z1 u" O0 udY(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)
2 h6 o  A! Q% a+ a' ]% z
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
7 A$ q' G* R; m# y( G% R
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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:
3 e2 T, `8 Y8 G! C& {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！

. r" x9 }  H, }! P, P
2 w" e- G6 u- 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。
您说：
" q0 [: F3 A0 \; r6 g. p2 r
d{(a+bx)*C(x)}/dx =-k C(x) + s
/ ^8 S, Q  s- F* c5 Eso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
# X  h8 h3 a1 Z; r- G! c6 _. K9 n0 G- M
) j1 D+ I0 M, A5 d也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. E0 t4 m8 L2 Q, H但这是不正确的
1 H3 u/ u! {5 u* \8 ]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。
5 s* f8 C% Z. n& C; W比如说   dx/dy = (x'y-xy')/y2
: H6 m& q, Q  L0 U5 i) d, R& v7 j, Mx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
& l, d! z" d6 c" `* A1 I. o2 [" n% T
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
. G/ x8 O* X1 f7 O3 {- a) [最右边是x的平方。
& b1 a- p5 W; A0 }1 d8 L6 g" ~
8 ]2 ^1 [, m% A) n7 c  }' [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。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
0 \' m: K' E' V9 P' I$ m
8 _! s: e9 H$ _1 I4 y8 i+ E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
0 V- \/ U7 s: }# r. Nd{(a+bx)* ...

Please note here:
) R9 C/ ~  t/ }: C: w# c
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
5 ~# q. C+ y1 J' g( U并不是   (a+bx)*C(x)   的导数除以   x   的导数。
& s3 U6 W1 N' P: Z8 g0 o4 Q% J
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
1 J  u, a0 a  z+ T....
" @4 t7 v+ M3 l9 Md{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

9 u, h( d7 J' m& D9 n4 `; e
* y2 n' Y* }( u9 J( O9 O5 ^2 n7 z* b2 d: a这样算混淆了微分和导数的概念。

  M; i' d) E9 x: H3 Bd(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

* S0 r; y0 I- c8 BIf 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:
, \: }( b4 H, S2 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

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

0 d- e# G2 A& p1 `4 t& p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
0 Z2 O) g  h0 o( i2 o3 m/ ?" W
3 G( \! }* `9 N' v( O4 tAh, you made a mistake: it should be:
( ?" A8 Z' T& s( F% Z! d2 x
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
6 |; \" @4 }1 N* a8 [: m
, G! E' ?- c1 p1 x: Z* N8 Y
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)
6 A) s1 @+ W. e* C8 [% A5 f( O
. f8 y- a4 B; ifor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

' m7 i& w% q% l. Z0 m3 l* d) Z  ch(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 j: P5 B8 u5 `  G! `$ j5 c* dd(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)) ...

# J: o. N  h  Y) g0 |/ H0 y# cYes, there should not be " ' " after "f" and "g", it is a typo. :D

: z9 x. O* ~  P* Y& w2 G9 U( |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

# O1 j" g2 O( M8 oBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

. K: y$ p% g% J+ A/ {$ S$ j
2 T+ C, z- j) v. t- M* PThanks, 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:

9 s% k2 ^4 l/ {+ [5 M+ 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。
+ A" b3 g: y9 {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 ...

, K  m$ l  J3 e4 `
1 A/ N; h# J  h; w0 g  y* W! v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？

  D! r, K+ a4 W9 s' c- U+ k2 }: N3 o
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。