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:% Q, p) a( c& v# Y
请教大家一道微分题，多谢！; X; V$ r8 n [2 k. n1 k+ e. C
" F; C% }) B4 Z" U: L8 H
d [(a+bx) ] /dx = -kc +s
* Z1 v( ?3 k3 rwhere: only x and c are unknown, others are all known, requiire c = ?
0 e5 k) x; O' F5 O9 E+ X
( ?7 x* c4 w" t5 E, w- D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:0 o, ?6 M5 W; ?$ {8 B& S0 h
it is too easy:
  B: ]6 ^9 k! J1 S9 b& a8 X/ g) M5 y+ U
d(a+bx)/dx = b
9 f% g- Y( e, O! [& F' @
# Z  A- ?/ e# i/ R) bso/ d! T, u5 o, P/ w& c1 F& Q" z; @
& n- ?: G) j" q8 _) v+ o# E' n
b=-kc+s
/ t, p% J3 |  k# m
! m9 A$ n( @  Q0 i# Ufinally:  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:
: J3 A# X0 m: ~请教大家一道微分题，多谢！
0 f5 t2 P- T0 w3 }* \) d; ^6 ] |) y8 a7 X
d [(a+bx) ] /dx = -kc +s ! |( h2 s. T' j+ G
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
- @" g& v& C5 u- a+ U# t4 a; K
多谢了！9 A8 Y$ b' \* S% s% ^3 k( q
9 T- a- [6 b9 ~# |
[ 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:
/ n: f! e  D/ v4 Vsorry, did you post another question? your statement is still not clear.
4 s8 G& |  y5 \. e+ E0 g. e
: \2 B" R$ D- n/ c  k9 E) `0 La, b, s, k are  constants? c means c(x) ?
/ j2 ~  y& Y3 O# D/ K7 ~6 a3 j3 Y. c$ p6 c% ?# t0 u/ m1 N' j
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:
+ |6 U$ ~; r/ R9 m% ^对不起，写错了，忘了变量c,应该是
: L9 c$ z! L [4 E' M2 Ld [(a+bx) c] /dx = -kc +s 2 R2 |2 X0 @9 G) ?/ f$ A" \3 p

, S% C; v% L+ T/ P2 r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:
1 v! c" Q# A2 ?; {/ a3 kdo you mean it is:: h6 d& U* ]; k$ Z5 u6 I

& i H/ y2 m6 B& [( ud { (a+bx) C(x)}/dx = -k C(x) + s1 s2 u( |2 m' _) W4 p

0 c/ I5 w( I# k z3 P! \where a, b, s are constants, you want to know what is C(x) ?$ |+ P% v0 b3 A V& g+ |) a. n' B2 r

9 ^' c# b1 ]5 A4 u, d; Qif 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:& l3 q s0 [) ]5 z) O2 g& i9 z
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- i' e  x4 U: r4 ^- @: A
9楼的答案有点问题吧。: K/ ^; M+ a$ D+ B- I
您说：
! u! K9 `! p" i0 `/ u/ Q
+ i* D/ e& |& I+ n) I' ^d{(a+bx)*C(x)}/dx =-k C(x) + s
2 J, j+ I# k0 C, R$ y  e5 i9 K/ lso:
7 `; l, Z% L2 o5 z( EbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
9 w. y6 C  _! N
# b+ x' F9 z+ B1 w也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
& Q8 w+ @" {- K0 h) y, f0 L但这是不正确的4 [; R* b, e, h( V* _7 _
d{(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' `) z$ |/ _0 y1 {3 K
....
1 v7 Z1 o/ W' K4 N6 ~: v) b6 @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:
) V7 g4 |/ a' n% L" b% Q- K/ V这样算混淆了微分和导数的概念。2 O) E( Z$ ?+ t- E8 [: p* D
* E# w* D2 ?/ ^% p6 B x3 }3 }
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
9 C( {& k$ H' P6 l+ q. o9 w, M, b! R2 ~0 W
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! N& [4 a0 |* K这样算混淆了微分和导数的概念。
* f& B& {1 }. o( u1 g E3 x0 M
8 r' U |! a2 J- x. w( M) L8 xd(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, E: G. n4 {; N% x

5 D7 w! C& F' V- A4 G, ?8 IAh, you made a mistake: it should be:
# z; h8 t. B3 r* Y8 U9 p/ K* O9 A6 F/ D* ?* A
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 20:51:5 ~/ ~; K" q0 v0 i) p
Yes, there should not be " ' " after "f" and "g", it is a typo. :D5 N8 a4 b& r$ c' U' \" b

8 `, E0 s- {$ x. ~) bBut 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:+ X( o' s0 f y2 g) X! R( \! v
* s% j, g( [' D2 O: D6 c% J3 T
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。
$ i. T! @* P8 Q5 T% b3 a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:
; h' `- }) u2 E8 I4 b' f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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