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 c+ A7 ~- x5 t+ }) a& f请教大家一道微分题，多谢！% O. k/ R; i3 x, Y
' L1 E: J% e9 a9 F9 \/ W' O8 I& @
d [(a+bx) ] /dx = -kc +s
5 ^2 e/ M" y* f9 @where: only x and c are unknown, others are all known, requiire c = ?; W# g4 B. E3 @! Y" m v1 Z
( s. _/ |* D+ _* Y A
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:' v% B. S, \" _4 A9 |* }
it is too easy:8 n( D" w5 V# ]9 T6 w& ^ m! ?
- j, ?; t- T/ Q- ^" h
d(a+bx)/dx = b
. |' ?8 B; v; f' V
, e2 }& Y7 i) G# s, Kso
$ M! K2 u7 @! F1 v4 A* F& z% F3 ~3 y+ ^* l( ]2 ~
b=-kc+s
3 o6 i8 H! T7 s+ K% X: y# e6 s
, L, y7 T& N: G X3 ~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:
/ Y" G, S' k( u6 ~请教大家一道微分题，多谢！
: j3 Y6 U- s+ N0 R7 Y, e/ w; Q! ^1 ]& r0 M+ d
d [(a+bx) ] /dx = -kc +s
4 Y' y1 K- v8 d3 A. h7 w' pwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?. z/ c7 |+ k6 j

, l3 e- M; \! Q5 g; H/ l4 d多谢了！
# x8 }. v2 @! H2 }+ J- T: ?
9 y8 W3 r; I. L0 t B) w[ 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. h) S# N3 l5 L( O
sorry, did you post another question? your statement is still not clear.
: Y9 J S9 j/ Q: |" A; Y0 P! x* k0 F$ G; [- S. t9 L( g. v
a, b, s, k are constants? c means c(x) ?
8 G5 c/ N( q9 ]2 M+ c# `9 H' q; x* d4 i. E. N# v
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:
. m- b$ ?! b- N0 k: |# }/ ]3 C对不起，写错了，忘了变量c,应该是8 L! b9 M' Z P5 k( v$ p7 D
d [(a+bx) c] /dx = -kc +s
* z5 {% X. r# z" v9 H! R
+ J5 v4 J5 y' f, t" l+ N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:
! k; \7 J& ^, D2 s9 f0 p1 L* Fdo you mean it is:
. b, q# N1 \; A) _/ {& v& u( C( V- ~" t: J
d { (a+bx) C(x)}/dx = -k C(x) + s# l, }6 q1 _. p! o# z
7 C$ f6 r) t% g4 `( E
where a, b, s are constants, you want to know what is C(x) ?
v A1 |7 k8 T( [ t# j# t4 U
1 _6 L* w* N* xif 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 14:33:6 d( h# a6 q1 o* G1 L* f T" x3 M( G, ~
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:
, ]! c' a: x. e! W9楼的答案有点问题吧。
2 O9 J, w: u4 I1 [您说：
/ \- u* P9 X; U- v6 h' ~: t
1 r: ~/ _% C, u2 H3 f4 ?! ed{(a+bx)*C(x)}/dx =-k C(x) + s' I# m  x" A% q9 |: u
so:
* |/ H$ ~$ P. ]+ V) K3 AbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
" ?0 l$ X7 d0 U
  L! x" C- H, K4 |$ f1 Q2 ~也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx2 r& H/ w6 T0 a+ |) z+ w
但这是不正确的
  D, F9 r0 s# D+ _9 j- `' g; J' h- Hd{(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5 ^7 j0 P! a1 {
....+ x1 A+ Z$ p; G+ h5 l. y5 g
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:
$ U# V: H# a- O4 n+ S4 {; R. X这样算混淆了微分和导数的概念。0 H; [6 Z' j* A1 I3 ]+ f) S6 ?

# S( H9 L. R5 O# }4 b. f1 Td(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4 i h1 U4 o) f2 \2 l/ j. M4 m
9 a: L( f. j" ?' r* Q5 M
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:& w# h6 ~- `8 }. Y
这样算混淆了微分和导数的概念。
2 [0 M& z3 Z, i- p/ |) F5 K6 A! i4 R0 i7 e8 m) c; Y$ 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0 E1 U6 G! U) m0 B& M
( r. v: W% |' S, {1 R$ N8 N* s
Ah, you made a mistake: it should be:5 t/ h' O4 D( z
6 S. ] P. l0 V. N$ F' w) Q
d(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:; _% X) \9 o$ W/ q) ^% T
Yes, there should not be " ' " after "f" and "g", it is a typo. :D# P1 s E8 t) Z/ P% w, w9 @8 `9 L q) f

3 x: Z5 A; D; OBut 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:( o+ I, A" p- E; \- t

! d3 V$ p" _: [: p* z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。$ F! X5 s* h/ _4 I/ u- E
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:. j8 [' K) o! T
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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