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 _! [6 r: x9 I1 p请教大家一道微分题，多谢！
* d, b) T  j2 ], k$ k- [* L1 _, A" z" B3 r8 j
d [(a+bx) ] /dx = -kc +s ( F, Z5 z! [8 C0 e6 t! A7 h
where: only x and c are unknown, others are all known,  requiire c = ?
9 `0 T) |8 i+ [" O" d- x6 ~! P3 G" u/ ^7 Z  I
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:
, S3 R" \/ {% g# _" I' git is too easy:
/ m( D, W8 s4 C! y  p
6 E4 r+ B, J" a' `" |4 v* Y8 sd(a+bx)/dx = b
, o- f: G) D5 A6 X% ^. u
' u  f# Y% l7 s5 k: k9 eso  S3 w% s: V* J. a: i5 `& D4 B

: h; F6 e- T8 F7 {% H% Ib=-kc+s/ A8 I: g: N+ r! a8 e5 {
) k7 H) n; _% t% 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:
# M* p8 E9 A$ j" G7 H请教大家一道微分题，多谢！
1 ?; q) V+ t# t3 t
& N9 q U, W. Zd [(a+bx) ] /dx = -kc +s " _% s* c0 o8 ~( H& f+ T9 w
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?* y8 C# @- J: Y
9 t( I, M& G3 t' H3 X
多谢了！
5 f1 q3 v3 G8 _& Y) Y2 U% Z7 ~2 _! l7 m9 C/ O# m+ |" O+ ^5 N
[ 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:
; V. u. H+ @+ o: o% [- Y% msorry, did you post another question? your statement is still not clear.
& V+ ?+ g# `" X
) ]* Y- ]. h0 ua, b, s, k are constants? c means c(x) ?& F) G4 X# V7 I: l

/ J0 a3 T: k% g2 wplease 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 12:30:4 f; M/ [& K, Q
对不起，写错了，忘了变量c,应该是+ W7 a: r6 l! e8 X
d [(a+bx) c] /dx = -kc +s : _2 H3 C6 G" a# P) U2 [" a1 i

: x% g7 Q# y/ ?# W9 c4 B9 a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:$ e2 i$ Y6 \; V+ Y+ L1 ^3 X1 \2 s
do you mean it is:
) v# @) Y1 q* a4 J6 v1 M$ T" f! ]- {' l) X: l+ g
d { (a+bx) C(x)}/dx = -k C(x) + s
$ G) I9 x5 S% W2 A6 P0 N/ ?$ |5 G4 {3 H) V% U) H) w& l3 N
where a, b, s are constants, you want to know what is C(x) ?3 M" H# e0 ?. P5 S

0 G5 J7 H5 }9 u$ N; V( h0 u Hif 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:
. ~; q/ A" w. F2 b0 j8 W0 E/ 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！

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 19:34:2 j6 }& N& K, D4 ^
9楼的答案有点问题吧。& Y: l, ]0 J7 A3 G
您说：
' W0 J: ]6 z1 s3 \. b9 c
% c2 y5 n2 d) gd{(a+bx)*C(x)}/dx =-k C(x) + s7 Z" z0 F" K; Y8 w8 v! N$ j
so:
\+ R1 O$ D( Q) v6 T# y5 s, GbC(x) + (a+bx) dC(x)/dx = -kC(x) +s/ K5 Z: I% e" |
; F: b; C/ v9 _( y4 Z2 c4 K
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
, t2 V2 n2 u4 D* j+ ]) E P但这是不正确的
- Q7 p. s6 J* X+ xd{(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
" k9 N) a G# n5 C....3 a. F q0 i; _9 w, `9 B# |
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:  E* s: R$ Q+ H
这样算混淆了微分和导数的概念。( f; z5 @- h  I

/ Q' P2 R' e+ c9 f' |: {# l. I6 Vd(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! z  [& i, S; f% d% N3 S

  A9 r3 V: {9 ^: q' W4 n; JIf 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:
' i7 I1 w6 F) H这样算混淆了微分和导数的概念。
7 @ M, z+ m5 X3 X8 @. ?5 _8 ~1 I7 q0 ~( ?5 N
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
8 n4 E8 `5 d: W" d+ a( s- {4 {& c+ k* `! N, T: ~
Ah, you made a mistake: it should be:
- p; f3 ?2 ?2 I+ ^# D
) v% T4 h( q# f. Kd(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:
" K9 y, ~: B8 y4 zYes, there should not be " ' " after "f" and "g", it is a typo. :D
- F5 V' a e! l- W& J5 ^4 O1 `; s
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:
2 h7 }4 \4 n- F* W! w% U N
" l! O9 t' `* ?" h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。# \; E7 j7 [6 H' B0 v
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: j+ F9 x) c. I6 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？

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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