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:
" Y2 Z) v+ F8 F- z请教大家一道微分题，多谢！
3 }0 o/ A; v1 P2 O$ o  q3 y7 M* |  T) q' ~; y
d [(a+bx) ] /dx = -kc +s . m1 @) G+ f0 Y( b* R
where: only x and c are unknown, others are all known,  requiire c = ?' L* h9 j- c3 Y+ l3 _) y
8 L+ A1 U+ L$ r& ~  g0 [
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:
. N1 N* E( E2 Y$ D' s3 `it is too easy:
; i7 W2 d+ }# L( W5 p% |0 ? n( D# k# n
d(a+bx)/dx = b
5 W, M( `6 g1 B4 A: |! l
0 {5 ~4 z! j/ O3 Qso& N- i& _9 q: W

3 e+ h( A; t& M: bb=-kc+s
4 ]. `( t( A+ n1 A
' m7 |1 v6 `7 ifinally: 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:
; P- D5 p% Z, {$ l, f. C请教大家一道微分题，多谢！3 q6 h5 [' l. Y" j

' @+ l6 V6 _- k8 K( N6 _2 Qd [(a+bx) ] /dx = -kc +s
" S; b' ?7 E9 E& g" e. r; ?' w5 k4 xwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
7 j3 K& H  Q! R, I! `& i- H- K: ~$ \  c% d9 s+ M( K" @
多谢了！
, k4 o# L0 n1 ?0 T
8 |% ]  y: o1 R2 i[ 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:
" O; m* K. B: C& w' i! ^8 Osorry, did you post another question? your statement is still not clear.
8 ^8 N: Y" S# V# L8 I( a5 a! Q4 e6 N8 P: m! N5 `
a, b, s, k are constants? c means c(x) ?3 x ?1 G* Y' g( _) Q: L$ _7 n
/ `6 b& e2 F# i
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:
" b6 |0 l1 ~5 O& n3 q" J: o对不起，写错了，忘了变量c,应该是
/ T) y! `% |% s! \+ }$ `d [(a+bx) c] /dx = -kc +s ( F2 U$ F* P3 u# ~8 \4 e
8 \: S9 K6 I0 W8 E2 k+ i
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:& X; e: k8 t3 y P/ w
do you mean it is:
6 o9 t4 ~$ M' X( e& U) G3 Y: ^8 k5 e5 V
d { (a+bx) C(x)}/dx = -k C(x) + s: P# s" j4 O5 I

7 I# S( g. m2 h8 Twhere a, b, s are constants, you want to know what is C(x) ?
4 K5 a" X+ k, s& ~# X' k# g
r* l7 E0 n- x3 L$ Z% x. D/ ?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:
/ r% V6 ]; q( |% c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:
: n8 T* ?, n# \7 {/ s9楼的答案有点问题吧。
' ]2 x3 D, Y7 e$ D8 W4 p您说：
7 {. }6 I7 J) i7 e  A' |: I; V6 M- z: e
d{(a+bx)*C(x)}/dx =-k C(x) + s
' V: e2 ^6 Q- k% l* n0 i7 X4 Aso:; a3 E- `6 s. t* ?8 j5 k
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s5 ~8 ^3 [7 e; p6 N+ J3 l$ p6 X
! t  F! i/ [" m6 K, e$ v+ o! z
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx4 T+ B5 N% z% B+ \7 X5 E
但这是不正确的
: k% T- r9 `! W' O2 Sd{(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
6 S& i; e) h! q: Z4 L....
8 r' {- U) b. Ed{(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:$ p& H; |9 p: [% V" E2 l
这样算混淆了微分和导数的概念。
( m. D5 F/ W! O* [0 c. @( D6 ?' m9 A, ^% A* a2 n5 h& [$ p4 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% K* ~9 E+ X# u1 v# u; @. |
2 S' Y" c5 c5 @: q
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:
, e  i0 {+ G/ t这样算混淆了微分和导数的概念。
0 z* _$ a* V& Q( f4 Q6 ]9 \7 u" ^5 Z* 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
) m+ m5 T# H$ |( V1 s
: i+ Z0 u8 N  F2 O9 MAh, you made a mistake: it should be:
" c; N- C: o( A) N& O+ w$ ]) L6 Y$ p1 ]+ U  X5 n9 R
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 20:51:1 P- M6 {% h6 ]" ^, F
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
$ N$ l; K( D' Q4 H! P6 y
( ^0 D) G3 i4 T8 iBut 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:5 |! k% H# l4 W! _' }' K
! E% o* d5 `- l
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。
6 d9 c6 B" z2 M- b5 |$ `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:/ a0 [: ~ s% ~) t7 H
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。

账号		自动登录	找回密码
密码			注册

数学高手请进，多谢！

浏览过的版块