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:( B$ F& U, a1 B: B8 @: w7 r
请教大家一道微分题，多谢！2 n3 P9 B( B% b0 h

9 s6 j4 c, k, l! hd [(a+bx) ] /dx = -kc +s
) d( B( |, ?$ g3 iwhere: only x and c are unknown, others are all known, requiire c = ?
- K0 A, [! K- E m8 z$ m l# O, z2 o) i- Q
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:
+ T# Z/ U6 t1 l- X# |it is too easy: j; C, K- W$ @6 a3 B' }$ p! W

0 w/ O( s# P, Qd(a+bx)/dx = b
) `9 U. M5 ]4 x0 Y7 e: K5 ?& j1 s1 a
so# s2 }+ p+ r3 C" z& U

0 D/ }+ j% J0 @; B: b( nb=-kc+s
% Z( H: n3 B& q, ~8 x( Q
7 t% S& \0 G! wfinally: 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:/ \  u$ P; |* Q3 ^
请教大家一道微分题，多谢！
- Y- c1 ?* ]- }
- H7 B  D/ B2 V& H1 j! Nd [(a+bx) ] /dx = -kc +s , X6 b7 ?' o: ]
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?+ J  |* X& M) D6 t

1 g" V- l4 Y$ W1 _* X7 E多谢了！
4 v0 l  b9 ]) M7 N' g/ I
  X' F) ]- r3 Z5 @[ 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:
2 M+ p7 V& c5 \& L" I& G8 O1 R7 tsorry, did you post another question? your statement is still not clear.3 Q, E) Y% V. i l6 }, r
/ \5 ~9 J' V; I
a, b, s, k are constants? c means c(x) ?6 |" f: _, y; C& ]! g# c2 h
0 i# `: w* y4 f. ~( w
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:# r; k3 o4 ^+ @& M
对不起，写错了，忘了变量c,应该是( y: n5 `0 J( }; ^
d [(a+bx) c] /dx = -kc +s # ^: G5 Q! O0 R6 s
9 ^+ o0 ^2 b5 t p% M- j, i* U g
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:
% G0 ]( x1 S: p; ]# ], G7 {do you mean it is:5 d% H& [$ `1 B* W
0 w& p- P9 x5 y3 ~9 T9 O: ^1 D
d { (a+bx) C(x)}/dx = -k C(x) + s
2 e! a" c; B/ N2 ?9 o
5 \1 }. A, N4 a, e4 X/ nwhere a, b, s are constants, you want to know what is C(x) ?
/ j- N$ j ?* ?8 L, ?$ c- h
! S; w9 O; K7 a! f( ^0 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:$ ~( |) Z {# y4 l; i
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:
, ^5 j/ \& z% }6 p, S- d J9楼的答案有点问题吧。
* c- u) f5 x1 v; q1 U# Q% G" p; g您说：
- g1 B% p0 G: Q* H
8 b, y/ {5 ?- U5 |* I& fd{(a+bx)*C(x)}/dx =-k C(x) + s8 y6 }+ Q3 E4 P+ M6 N
so:
! Q$ ^' y8 K+ d' u- HbC(x) + (a+bx) dC(x)/dx = -kC(x) +s7 L8 \6 ~" m7 G1 H) C' ^
7 [& a& [# b: E9 Z0 j. o% C
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx% w( N& v. y: p4 l
但这是不正确的$ d9 [2 l. v( I
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. N; v0 ~% E; Y4 R
....
0 @' q% k7 T! O2 J$ v* T: jd{(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:
* s" R: ?* G8 |( B G. r+ \+ s这样算混淆了微分和导数的概念。
- j( P" ~9 {' g% `" `8 a1 `4 Z* A9 Q) w) A5 i( l' q
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
. ]1 h- Y% x+ v$ M: H3 D5 O) A6 ]/ ^4 f* {1 S
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:( z% A0 r9 A* d( V- y! ]& ^
这样算混淆了微分和导数的概念。, }: J6 W- k9 |. `: O2 q
k+ ^8 g+ Z' ^
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
1 Y2 _4 i2 x" p' [2 u5 P! {- s) E4 v- k# X% Q4 @
Ah, you made a mistake: it should be:
, X% V* Z1 f$ b- X
. d, N4 w$ f) V. j Z( fd(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:. p" D, G; V* j! ]' X
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
3 }) o6 Y& ~/ w, a+ {& x% m" t6 Z0 _
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:
+ B6 E ?+ B; @* c4 C: O6 K! q; w: z& _7 K' F, J8 W& v8 H( ^
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。5 {6 h; V: A* T5 B3 u. [! \
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:/ w, _) G6 p9 ^1 [! }% q) M
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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