数学高手请进，多谢！

醉酒当歌 · 发表于 2005-4-3 19:41

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 醉酒当歌 on 2005-4-3 at 07:56 PM ]

十年移民路_ · 发表于 2005-4-3 19:52

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
" ^3 n9 h! t8 y9 G4 O请教大家一道微分题，多谢！
+ L" U& y4 P/ Z' D$ W: J) g
* b+ y( Z; V3 ]+ t& @d [(a+bx) ] /dx = -kc +s
# }8 ?9 f) u; E# y( k8 y/ a7 @where: only x and c are unknown, others are all known, requiire c = ?
6 `) V7 v; a# |2 T, C3 W7 p# l8 X7 |
多谢了！

it is too easy:

d(a+bx)/dx = b

so

b=-kc+s

finally: c= (s-b)/k

醉酒当歌 · 发表于 2005-4-3 19:57

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
* {5 c8 F9 T' Y; ait is too easy:
- m! C8 o0 k" `  B
, h& {& B3 G2 Z  F& }5 \d(a+bx)/dx = b9 C$ Q( S; N! J/ D

) w6 a& p) Y- ]! S) c' Kso
) O1 `3 X! C4 d/ k3 q& c6 q8 v/ J
; @3 p& Q; [! }0 v; o- Q. Z2 Pb=-kc+s
7 z2 }: {0 H8 P/ ?) I+ V6 \4 O3 y; i6 D
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_ · 发表于 2005-4-3 20:51

Originally posted by 醉酒当歌 at 2005-4-3 20:41:$ I& V: S( L( E, T
请教大家一道微分题，多谢！
& i' L3 R( P# N0 v6 Y" m8 i8 H* b0 z3 W( t
d [(a+bx) ] /dx = -kc +s ) r4 ^3 m1 X# |# B- y; k; I- u$ K" X
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
8 R3 Q* c% v9 `7 u' I E
) G9 X; J2 m' b- ~3 q多谢了！- O; U- E9 y9 R

! ^- ~$ Y6 j+ {# X[ Last edited by 醉酒 ...

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.

醉酒当歌 · 发表于 2005-4-4 11:30

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:1 v0 L' y( {' H; O; i( c* K
sorry, did you post another question? your statement is still not clear.( Q1 C0 X8 X- ]2 F9 ?0 `

; S9 z* e1 P* X: {& F7 R$ K1 ka, b, s, k are constants? c means c(x) ?, k, g* N: ` X& z% M e Q' s/ C9 b
% b7 {" j) O! t: P7 t- y
please tell me.

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_ · 发表于 2005-4-4 18:36

Originally posted by 醉酒当歌 at 2005-4-4 12:30:9 a. o7 k/ |+ Y. W$ `4 a: Q& ? x
对不起，写错了，忘了变量c,应该是8 |3 `( H' h, \! Y2 z/ m1 m
d [(a+bx) c] /dx = -kc +s
0 \% M, f) i, T& I3 c& L& j0 c: X. o4 G
多谢

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 · 发表于 2005-4-4 18:43

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:
% h3 d c: ]3 m4 s5 i5 Z3 M4 Pdo you mean it is:
7 B f4 u! h" C3 a4 t! {" D
! x- J0 ]4 w: D, Z4 S6 ~d { (a+bx) C(x)}/dx = -k C(x) + s
) u; }. q' Y: `" q) Y/ h4 Q9 }! O- \, N3 }; x) |
where a, b, s are constants, you want to know what is C(x) ?
; z) F) \: K; y; l/ n" h# o3 X# D) {5 j* k% R( l( R
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_ · 发表于 2005-4-4 19:17

answer:

C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_ · 发表于 2005-4-4 22:26

this answer is the good one.

procedure:

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

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(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
therefore:

{(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)
by using early transform, we can have:

-(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 · 发表于 2005-4-5 13:33

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！

十年移民路_ · 发表于 2005-4-5 20:41

Originally posted by 醉酒当歌 at 2005-4-5 14:33:
5 w4 W8 ]- P8 C( b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 · 发表于 2005-4-6 18:34

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

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
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
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_ · 发表于 2005-4-6 18:53

Originally posted by 龙舞九天 at 2005-4-6 19:34:
0 L2 Z3 y. H! |/ s9楼的答案有点问题吧。
! o0 ?7 `  J0 |3 x您说：7 n. A$ F: ~  u! q& K$ i) `

& f- [) N" Y( x+ `" Q+ rd{(a+bx)*C(x)}/dx =-k C(x) + s
7 a$ w! D. o4 N0 U" Jso:7 V" v+ U/ V  @' G# t- f1 H3 B- c. N
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s8 `9 S6 h0 U9 r5 a. t
7 e( Z* s0 s4 J4 o
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx, H8 H; o) G& {, F2 C4 S" k
但这是不正确的. N  i2 ^# h0 I& H. V& \7 G7 u4 h
d{(a+bx)* ...

Please 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_ · 发表于 2005-4-6 18:56

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。

no "x" in that position.

gnome-kde · 发表于 2005-4-6 19:30

d{(a+bx)*C(x)}/dx 意思是说对 (a+bx)*C(x)/x 整个代数式求导数
+ R, l* z5 H+ C) l....
$ J2 M w% c( _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。

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'(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_ · 发表于 2005-4-6 19:45

Originally posted by gnome-kde at 2005-4-6 20:30:
: r5 N& H# S S& x, |4 c: ^这样算混淆了微分和导数的概念。% c G5 L' D, S% S6 f( i9 o
! `0 K6 w# Q1 b* b6 _7 r* Q9 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
! \9 J) L. o X' n% [) t& R$ m" o" t9 h1 F$ M( j* P
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

Ah, 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

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)

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 · 发表于 2005-4-6 19:51

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
2 H4 F2 u f7 ~. C1 C这样算混淆了微分和导数的概念。" O: `0 q8 ^4 E: P9 d% F
5 ]' v7 n1 h" t. M- y3 R- V
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& _' T) l# E( m9 ^2 M

) Y5 S+ j7 l% j, l6 f3 RAh, you made a mistake: it should be:9 [1 [7 o, a2 G* h: \( d5 J& [% e. @' }4 H

9 a, P) D3 `1 o1 X8 md(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_ · 发表于 2005-4-6 20:05

Originally posted by gnome-kde at 2005-4-6 20:51:
' a* ]/ f v1 q; |7 Y8 [Yes, there should not be " ' " after "f" and "g", it is a typo. :D
9 N% \$ E8 B$ S, g/ Y
' K& s3 s/ \9 o7 L& NBut 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 · 发表于 2005-4-6 22:24

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:, M4 T& s& [/ N* p0 @0 d

4 o/ j9 Q! V, s* K& x$ p. c" E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。
; n6 @) [9 K( I& l' O8 Z% s' f$ i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_ · 发表于 2005-4-7 20:44

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
; G* p8 x5 T4 x* {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。

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