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:& H0 `" v* I1 }) o, y
请教大家一道微分题，多谢！2 l5 B8 `, F8 z' I1 I4 _0 @- N

5 y+ {9 V" y. F3 O8 ^  sd [(a+bx) ] /dx = -kc +s " K6 l' j7 h! l* p/ f) |( m
where: only x and c are unknown, others are all known,  requiire c = ?
- |( P5 [* ^  e: `3 e( Q0 l
) Q1 h( B' Q1 _0 N. `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:  z2 A! T# N. X' P9 w/ H+ Y
it is too easy:9 D- b% m( I( N' R, m; f8 \

, L& K2 e6 n9 e( rd(a+bx)/dx = b  y5 r' V* A, z, V% m% Z& t
7 ^% y( ^, N0 m$ [1 t
so" S5 ]( q. a5 C6 a
0 v3 w" ^9 I( E2 d9 v' m. n) ?& P
b=-kc+s9 d% @# {8 \7 }' ^% @% z# e9 [
+ ^( H4 T4 |- ^' R5 v: 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 20:41:2 _: u7 ?* Z2 q* Y( p- r2 x) Q
请教大家一道微分题，多谢！0 C( W' _7 w* S) Q' L1 z. S" j

. o$ I" Z6 y6 v/ Id [(a+bx) ] /dx = -kc +s
5 y: ?' I0 S4 E" M& f7 ?: Zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?6 i+ n7 i6 Y- I; e8 i/ O

9 E. R2 s+ ?2 t1 s' l& s多谢了！
# ~$ D- R# K  ~+ [* _, m! `) _
( G5 s1 {8 v# ]  A; ~[ 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:( b3 J3 J* _% Q- o6 o9 O
sorry, did you post another question? your statement is still not clear.
* H. H% h- k' U, g6 C! F: ~9 W  _- E! h( p9 ~3 h8 u% m
a, b, s, k are  constants? c means c(x) ?
' o& _# a1 N  |2 {4 [6 T- a& j  r% a
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:
) e8 a% C. k5 Y' p9 @对不起，写错了，忘了变量c,应该是3 I. F6 B1 A `3 @: L3 c
d [(a+bx) c] /dx = -kc +s 8 D' v; ~" ~2 k5 b
1 h' e% {% S2 W+ M7 F
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:
; k5 k5 Y! k! l- P3 bdo you mean it is:
$ P: ` _6 ]; K6 y; |; P
& U: o% b, H0 O1 p& }! v- c3 id { (a+bx) C(x)}/dx = -k C(x) + s
+ g/ c4 t% m$ m% x' z% Z- Z4 `$ [3 p+ Q9 z+ S( [
where a, b, s are constants, you want to know what is C(x) ?
" {3 ] t4 `( O/ n# ?/ n4 S, V0 o2 z$ Q1 N! o8 A! \* M
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 14:33:6 b, h- e' z t) w8 e- J7 U' s
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:
8 R7 j2 R& D- l+ x9楼的答案有点问题吧。
) {, U' q+ h* ]+ _. e4 M您说：
4 K4 J8 F3 @0 s; Y% G: W* M; V* @4 T' j
d{(a+bx)*C(x)}/dx =-k C(x) + s& H( y4 w; q$ ]: }# ?: R
so:7 S1 f" }  \+ A/ y% K% z
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s- u5 m9 f4 V& I2 `2 D- k) _# g
1 C2 k, s  K2 X& b; k1 `
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx1 b, H) R/ f' X' G  u, m5 y4 g
但这是不正确的
$ p, G! ]# W: V' T+ H2 a/ qd{(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
+ H: z7 u2 O! U* m6 ~* _....
, l' e8 A( w& W+ _" [- q1 i* Pd{(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:
! `( c3 l7 T9 ^# E这样算混淆了微分和导数的概念。- M2 _* U1 E' K" H4 y, Q% x

" r3 y* e; G4 O. Jd(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3 y8 `7 e/ c8 ~0 P. G( P

1 Q0 ]9 a% c5 Y; [$ x$ h. v# ~! }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:- h7 }+ `" K( b3 X" ^
这样算混淆了微分和导数的概念。
6 ^0 y- X0 s. \  ~4 j1 Z4 u+ A
, p6 E, q! V, 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
" j5 F. {3 s( m8 @! \0 e8 S' O. F0 W0 {  q1 F7 S
Ah, you made a mistake: it should be:
4 C5 J; G( `6 {0 T
+ r8 B* ~3 Y2 c3 w- N  s" dd(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:
4 o( I4 u" Q8 i0 F' xYes, there should not be " ' " after "f" and "g", it is a typo. :D* F6 G2 _ G' p0 X
+ t6 `% I3 g! u* f
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:
l Q2 W" Q5 F) O/ L% f- s* u
3 l5 o3 U b e& w: ~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。
- N" f7 a$ G8 @* v6 }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 ]# M& M( o5 x9 j* W$ z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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