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:( h5 ]! x5 q2 p: a  u
请教大家一道微分题，多谢！
+ i) C; y% o# t) Q
, R0 Y- Q( O1 \( U& @& A( nd [(a+bx) ] /dx = -kc +s   X+ I9 @: m  Q+ t8 @) e' h5 n  ?
where: only x and c are unknown, others are all known,  requiire c = ?% e$ n5 Q; V: D( e: U1 M% @
/ b! }# W/ u0 k3 r, y/ T* ]& l
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:
# \! Z" W, l" Eit is too easy:
6 }4 N: c3 z+ o4 |) y# J, H N! J+ M$ }- p3 G9 v
d(a+bx)/dx = b6 n+ }0 v5 I1 U# S* E( x9 y0 z

- W+ d4 O: a5 V8 Aso
, R+ z, a& Q! l. N7 A
8 U5 n3 D3 s5 l9 @9 U; ?) Rb=-kc+s6 N! `, L$ m8 Z/ v* P7 G

& y. [: {9 k3 j& G9 q7 Q0 v- E) Lfinally: 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:
, ^9 G  D' [7 m/ n4 E请教大家一道微分题，多谢！
# s. _, B6 D: Z. c7 c
( K: Z) V0 U2 J9 M5 P% L% \$ Kd [(a+bx) ] /dx = -kc +s , `+ x' u5 o; Z6 B) K- l
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?6 ?, g8 J# A- H

" a3 b3 X% [7 E  S! q% J, T多谢了！
. B, c( ]- k, e, X8 j" ]4 I& M, P2 u" O" ~% ]
[ 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:
, W- {" T9 z+ Jsorry, did you post another question? your statement is still not clear.1 T/ ]6 \% G7 b3 s# q
3 j$ w2 s2 v7 p5 J7 r8 M; e
a, b, s, k are constants? c means c(x) ?
N% S: Z& Y" F( L) @& {
+ ]% _- _; g! Zplease 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:
! Q* G. |5 l8 T' F$ w. b! k对不起，写错了，忘了变量c,应该是2 K7 n0 s. p3 Z) ?9 f
d [(a+bx) c] /dx = -kc +s
0 E2 ~$ Y# F9 p9 @( h( u( ^8 l$ p
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:
, u& u% j# i% L d' @! g# W' k2 h! vdo you mean it is:$ }; \: r M" W% q
5 [( k) c/ b/ [- W7 R0 i* o
d { (a+bx) C(x)}/dx = -k C(x) + s3 ~' ~* [* D/ m( Y( Z& C

$ g/ F2 Q2 W0 j( _5 nwhere a, b, s are constants, you want to know what is C(x) ?
& } B1 `6 s) j, g/ f
+ h+ _. ~( Y1 p. Y. F2 dif 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:+ x1 o; `- ^: P) c4 N: J0 G& 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:) m* g0 {! z3 G' N! v& u9 l
9楼的答案有点问题吧。  d6 }! f  p  g3 J, U4 v
您说：
. X, s) ?! F8 g3 ^; X7 B5 T8 l! b9 s& A$ U9 @. S( B% q- I* o0 F
d{(a+bx)*C(x)}/dx =-k C(x) + s
, F3 X; s2 s6 U+ R8 Q5 n: c- Oso:3 {3 B0 q! L/ f+ }
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s7 V0 o; [5 ^3 W% \

1 J9 `* G$ G( H( L, x" p5 U也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx1 r0 e4 R" e' D% `  y
但这是不正确的! T0 {0 q; o/ _1 t  Q% q6 U
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
, S5 \: k x# W& ~$ P4 A....
% l$ h! B6 e: n9 }7 ?9 @% \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:
2 j9 d+ v) g4 r% w这样算混淆了微分和导数的概念。% S( ~! K* |) n6 ~1 o% S& |; A
1 e5 F0 G! J# S9 B1 l4 e0 e
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
% p8 M9 h+ `% c! t. W
# i' s, o/ Q; Z- P. A% Y [; [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:
1 T* D3 }! S& J8 I3 k6 p. o这样算混淆了微分和导数的概念。/ i$ P" f, Z+ Q m

. [* t8 D/ o* O. Fd(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 j' V, `9 {6 @# G/ V3 w Z8 L
8 F2 P+ N, B( x1 P7 A6 w: fAh, you made a mistake: it should be:1 L3 U& k" v. d' o

! U! m. ^" B6 u+ t- e- `7 wd(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:( g& u7 h' G2 b( b6 n- D
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
$ q& ]% Y+ i9 ~: p N' _7 u/ p3 t2 c4 j: \) {
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 · 发表于 2005-4-6 22:24

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:+ w+ L: m& T4 k- L$ I

5 y0 S; S S* C6 Q- K' I8 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。1 J4 q$ h5 ~$ i9 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:
2 @, }: A( t9 R; i1 j1 E9 s. l1 h. N; L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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