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:
8 M" u8 g: N; U% F6 z( |请教大家一道微分题，多谢！8 t0 J; a5 R+ o( l7 ]
4 r: v' V. ^4 `/ K8 U2 U
d [(a+bx) ] /dx = -kc +s
$ M1 ?% Z* c* A4 M `9 Cwhere: only x and c are unknown, others are all known, requiire c = ?5 V; h& _* F% d' R; w

3 z/ R) L) w" P/ n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:
' a& e' Z% U, |" U0 @it is too easy:+ z& [6 g. u% X, `

, w, S- g. M: l' [d(a+bx)/dx = b
& x7 n# b, w5 ?4 q' O( l1 M- O# R, M$ G; ~, i& c- @% y
so
# y" F: E1 j2 N- h
$ N3 Q" P' {) C  \/ Kb=-kc+s
& K* S  G4 {/ t( o" B' i, D
: e0 D2 Q- S* a( S1 Q/ efinally:  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:
! Y5 W$ Y4 g! P- r& d请教大家一道微分题，多谢！
' h4 |% [$ N# \, q. C4 `4 l0 _; J0 D5 q/ E; e
d [(a+bx) ] /dx = -kc +s
. H' v; e! T% i9 x3 T/ W$ zwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?& H' S2 I$ w+ U: x: k/ B9 b6 D& e

" u2 u$ j6 u5 r, D. |) J多谢了！! r5 i$ C/ y+ p6 n* p4 P% s' N" W

9 i( _& J9 ?" p4 v[ 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:- q" F% l( W% a4 q2 L8 }
sorry, did you post another question? your statement is still not clear.
$ g1 a5 H4 w- t( b
1 T- w; D! k/ M- H3 O3 D. ka, b, s, k are constants? c means c(x) ?8 D( }! M8 A$ c$ ~6 A
% u) E6 [+ ?& J, g' A$ u4 {! Z2 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:; ^0 j1 d. H# v: I1 C
对不起，写错了，忘了变量c,应该是& Y" k, S | k
d [(a+bx) c] /dx = -kc +s 7 {3 ?: f, T/ o
/ I. U' v' G, P& {1 U6 k
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:! T+ s5 O# P2 S; \5 R
do you mean it is:0 y8 z$ X1 \. U

$ A2 q1 M6 m& Y9 C3 O9 z0 d( ]d { (a+bx) C(x)}/dx = -k C(x) + s2 o q4 S* F* N+ G7 X, i

8 o* g6 k# N! L& v) ?where a, b, s are constants, you want to know what is C(x) ?4 j' X+ F. N$ l# N/ N ^/ A

& J* X. `+ Z1 H) u4 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:
; w [/ U8 ]" A. V* r7 h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:% Y6 L' w, j: s7 v  z' K
9楼的答案有点问题吧。) G- ]8 A2 l$ y3 l
您说：- w, L1 |' y; H9 r1 \' _. Q

( F- t) a3 A: Y% p0 ]0 Wd{(a+bx)*C(x)}/dx =-k C(x) + s
" t6 W; u# z+ T, M! nso:
1 }4 p3 Q* a1 M2 bbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
$ B, L4 Q. A5 p1 [
# m; a5 N2 |  p6 r9 V: W也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx. |( U; K( ~' `7 R# p  V. {' E
但这是不正确的" f5 \" s0 a: j7 M! H( Q6 e
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
, a: ~3 g/ F1 S5 y1 Y- H3 }: L9 }, o....5 H! _& S- f$ B
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:
1 U5 u7 k; {7 q3 z7 d这样算混淆了微分和导数的概念。
6 v3 W7 ~2 j f/ v8 W! I3 \' V9 k c- Q! B
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$ x, A' R" r9 w/ p; x9 {/ v

" u9 ?) F$ q) hIf 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" a) E2 d0 v6 b/ o: I* F; e
这样算混淆了微分和导数的概念。! ?0 ]" ~  ]& ?3 x3 h

, t: M7 W- n! p6 [! h0 w3 }6 ad(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7 o& V$ J% a; P+ N  Y& `
9 l4 o8 N2 {5 k( X7 ~* r2 a) `
Ah, you made a mistake: it should be:* s0 b) B4 W! ~' z$ C

) E  o7 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:* h1 c* l# ^7 {+ O
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
( w2 ?( p! a' p: p' E+ a
+ m1 c3 n1 z# J6 p7 e1 B( X9 `9 [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:
! I/ y' g! N, i- v5 @0 N1 w
7 K, K1 z) d8 d$ U' h: d( f/ i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。
* W# ~0 ^1 S7 B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:
6 ]6 _" r) q, u/ A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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