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 20:41:0 J# Q% i& [* V7 G
请教大家一道微分题，多谢！1 c$ M Z/ |3 h( i1 P$ q
. P/ z- X* x; |, c
d [(a+bx) ] /dx = -kc +s # J0 q6 ?, W3 F$ M- `
where: only x and c are unknown, others are all known, requiire c = ?0 m- f) \' g3 j) f3 J
; i; S( B, `& j( V; y
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:
8 z7 V3 o* I8 F" W( G+ r6 t+ jit is too easy:4 c- I, x2 V* Z
; b2 F2 ]* E' c/ T' u; n
d(a+bx)/dx = b& t& B% L7 p& T  _" V0 \+ R$ o( s4 v
8 g! d- |+ D" D/ K& k. ]
so
2 q  i) l8 O3 s+ y! z' ^; g) _9 T# Z7 q) Q& c3 m
b=-kc+s
8 Y: l/ `* h* v6 {! V6 M/ \; c) ~7 |
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:
' a7 P9 _& q6 g& ]# z1 E" a请教大家一道微分题，多谢！
4 @5 a( C5 g5 f/ V6 P) _- Y* G0 U, p/ Q4 I& o
d [(a+bx) ] /dx = -kc +s
6 m6 H! E( x# s! D2 P d7 P4 cwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
( N0 [# d# ~+ V6 v; R. O2 L# c/ n
多谢了！8 x& a; s+ `! h% Q

* a3 d9 x2 i _* O0 F. i& |3 |% l[ 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:6 e0 k! ?% R/ B5 c' F
sorry, did you post another question? your statement is still not clear.& {* z5 t, B9 {; B
# m4 H2 v8 F/ {! }& a; s( l
a, b, s, k are constants? c means c(x) ?
! ]3 [8 D3 b0 ?1 T7 b- [7 t) w! p3 e6 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:" Q g! g7 \! G. x
对不起，写错了，忘了变量c,应该是( |9 ~& E) x$ [! D0 _# s" [4 B
d [(a+bx) c] /dx = -kc +s
2 \. j% I [ V2 u
) a! s. d$ S9 V4 J! w/ d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:* H1 r* v: d6 s
do you mean it is:' l, Y5 |) G: \) m4 O; G! e
2 Z3 k6 m% Y3 I4 t
d { (a+bx) C(x)}/dx = -k C(x) + s7 |9 J2 l* c( `# K9 q% Z- L
' n2 o* h$ A/ o1 J4 x5 _* y
where a, b, s are constants, you want to know what is C(x) ?9 l8 e% |2 q. \( ?, g t4 X4 l0 j0 L
3 @9 {2 g* F/ }7 m ]8 E
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:! d4 V) V8 L9 X' E" f
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:
- o1 |7 |4 n; p+ R7 `  I8 |. a: i1 z9楼的答案有点问题吧。
  J5 u& t/ x' `您说：2 V2 t2 c* y; |- [5 T
# G: S# m+ D: I% r. P2 K+ T( a- d
d{(a+bx)*C(x)}/dx =-k C(x) + s+ c1 g' V9 \' a# ~
so:5 N3 p' a  C  q& V# O9 n; z
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s6 b7 n7 D& B. o2 X' A

  D* j2 i. w8 h8 q8 m也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx  H6 L8 B  J& ~. @
但这是不正确的
) v% `' r3 Y6 o) Gd{(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 m7 ?6 }% b# h; q8 E9 ^4 a; @7 K....& H. `& p+ f* A3 B: `3 n( F0 H
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 20:30:4 W9 |# L- S; q
这样算混淆了微分和导数的概念。
7 q& X$ t, S0 M: U1 y/ A3 G2 B: m3 ^3 \% _ D
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' j2 o, j* I/ M3 o, w0 l! @! `
( S# }& a' H0 T; ]( RIf 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: Q+ ?( z* J$ J0 q* X这样算混淆了微分和导数的概念。! c9 u$ W; m( E/ w7 H# \8 T

3 R! z) m0 P& R( Q; o' h8 K1 Kd(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 c' p7 p! T0 J+ p( T3 \, Z9 K" f0 j$ J3 O, p! V
Ah, you made a mistake: it should be:
; a$ J8 }4 D& Q9 `6 {/ I7 v4 P+ I# [% B4 o, Y
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:
7 f4 R1 r% |# X! u* Z. c3 iYes, there should not be " ' " after "f" and "g", it is a typo. :D& y" I$ W- {5 u6 ^

2 ~" w& v/ ?$ X# SBut 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:# I, ^* C' g, n: D
0 A" ]; R' n$ d$ S4 |) s
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。2 G# f% S/ w" U/ v" u& a. C) }6 \; M! @( ?
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:% C9 V; R' W& V0 |- e" i( Y4 d
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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