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 20:41:1 u. _0 n8 R/ {. k. e9 T  O
请教大家一道微分题，多谢！7 ~: e. P: [$ C. @; S% T

+ M- m' }& n4 k! A% W9 md [(a+bx) ] /dx = -kc +s
7 ^; n  K7 D. L' i0 Vwhere: only x and c are unknown, others are all known,  requiire c = ?
3 ]$ P% L0 k2 R/ k# i% R6 \) [9 E  G& v; p
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:# @: |) k1 t. J" F% m2 W$ o
it is too easy:
; T1 q( P7 _- f$ s  R! q7 x- S5 b6 }$ }
d(a+bx)/dx = b
) |0 x% k7 h. A4 Y0 g5 Z! |9 h$ W+ J, Y- K9 I: B
so
5 e! Y' q) G# P) r: s7 _' F
3 S7 ?/ J; N. N+ Eb=-kc+s
$ Y7 h$ B  A, N6 f. n& S) `$ p& T$ q$ w; A; j9 G
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 20:41:1 U' I& s m7 _" w! l- b, W
请教大家一道微分题，多谢！3 C1 K3 f" v; m7 q3 h$ B* j' N* c
1 j* w6 j5 v* M4 ]' z
d [(a+bx) ] /dx = -kc +s ! c1 d+ U+ P$ p1 j0 e* j# h
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?8 `/ s% x# s& [7 w+ r
5 L3 h, k( E' V; m5 z
多谢了！$ x/ S' h$ M' X9 u( y* E1 r: r6 Z0 a
/ p- o& h5 @$ D- u2 C5 X1 y
[ 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 [( V' ]0 ]6 j. s! @
sorry, did you post another question? your statement is still not clear.
& }( f1 K4 {. |6 W8 H1 u, y! ~* U2 _$ _( [$ X, X& F S1 O
a, b, s, k are constants? c means c(x) ?
$ y( K# N& R0 D9 [
0 I4 ]; U* \) o. t% e. Oplease 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:$ W- W5 p3 z; p* p, f
对不起，写错了，忘了变量c,应该是+ p$ N( ], h5 k- H: u
d [(a+bx) c] /dx = -kc +s
& L2 C0 j* G0 Y# A) Q k, C2 U# i, M# ^
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:& A& U$ q6 |5 T8 I) p0 N5 p6 {+ [: x0 C
do you mean it is:% R$ M0 r& K7 ?0 x( i% I2 A& e

3 x- }. f, y/ Y, U& E5 md { (a+bx) C(x)}/dx = -k C(x) + s
! S* K0 K! H, _: W: v4 e
' }0 L4 u* |6 y: |' Y% ?: Uwhere a, b, s are constants, you want to know what is C(x) ?1 x- y/ c' Z0 Z! j
1 \1 z" ^+ B, [4 ]) i, 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:
5 R% {: `1 g, T1 c( D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:
# D( `7 U7 k5 t2 n- W, a0 l9楼的答案有点问题吧。4 m2 G. o  ]3 ]: u6 S4 t0 z7 ?
您说：( p" r% B0 A3 A6 R) p6 e  k, p( K
) t/ X/ k  w# ^: ^! U" \9 s
d{(a+bx)*C(x)}/dx =-k C(x) + s
# A: X0 @& L3 [4 f" i7 lso:5 ?. K7 U$ B# q; N9 I0 F- Y
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s8 P( d) f& U4 s7 ~% H
1 I- s! ?5 ]; F' V/ ]
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
* Z: C, f! c& O. p  E9 y但这是不正确的; ]$ s! @0 ^3 @. U9 A6 H( T9 i+ o2 D
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
3 ?& ^ S% T+ g( F0 S$ J8 w2 {....
7 p/ K6 Z8 O0 k/ Vd{(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$ I* j6 |: J+ m
这样算混淆了微分和导数的概念。; Z! C) \5 q) S+ h' p s. p
& m& u: V+ S( p( F( N1 J
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
+ P' i. A3 k8 v! Z* V" J( H" \$ p6 n; t/ ]
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:  U" i9 e4 s! K/ _- T- h
这样算混淆了微分和导数的概念。5 c" X0 D( n5 h
  R% r0 p3 f) Y. E
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8 M. }4 X" _  f, m7 c* W/ m

3 p$ k6 d2 a- q# bAh, you made a mistake: it should be:% L7 |$ I1 ^% Q1 ^
0 A0 `7 ]5 c$ \9 ~$ {
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:
+ E9 d' H) l+ |0 u/ nYes, there should not be " ' " after "f" and "g", it is a typo. :D
0 P5 e, |/ q2 S! e' g+ o* P3 A9 @/ y, _* `( o% p0 I
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:
( z5 X( p6 f4 ]9 [+ i6 M9 r2 }0 S7 F$ z7 C6 v
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 P1 z+ T; L3 e+ R4 v
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:
% H2 i$ W& l) J) E$ J3 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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