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:
2 ~7 {0 o+ p" l1 V, l请教大家一道微分题，多谢！
7 }& ^ {) h( T2 l& r: O5 V
) N% h5 b, e8 P7 t1 Wd [(a+bx) ] /dx = -kc +s
% @: [2 C; V9 p. d- ]where: only x and c are unknown, others are all known, requiire c = ?
2 ?7 l6 |) G8 i/ K1 s/ X: @
3 j/ i! x' \' X: ]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:
# n) Z3 a( K" _; _0 W9 |1 Zit is too easy:0 T, h# U' X) I" @6 i; S. s/ E
& |% `( H7 s$ z0 ]9 ~5 u" U, s4 }4 W5 I
d(a+bx)/dx = b6 e2 v! P8 A5 T& N

3 R. ?) R i, G8 G0 _+ Bso
" ~" s. ]9 M. @0 @& a
/ B9 r. F4 L5 A' T2 i6 ?' wb=-kc+s, Y0 d: V' ?2 T
9 n) l% [6 ]& i0 `4 j* B5 c" {
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:
# D8 A) L7 l+ t( k请教大家一道微分题，多谢！8 w( J# v2 y' \# X2 w0 [9 j" T* b/ p. X
; J7 k' K" P" F. F5 ?# b9 h  x1 A" Z! P
d [(a+bx) ] /dx = -kc +s
3 V- q( d8 P* M! ]+ j1 Zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
  P/ Q; F* D" U3 c2 i/ }& }$ h5 H
( v, L% g* y7 w9 ]* X多谢了！
, m) F, ~6 r" [9 C+ n0 q8 ^+ l; g$ b- Z* m% E" y- @2 Q
[ 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:" _; c6 o, b- [, E
sorry, did you post another question? your statement is still not clear.5 p G( p/ a! Q& Y
- S9 P7 }1 Y4 n8 F$ n; f
a, b, s, k are constants? c means c(x) ?6 x# }/ r+ p1 D7 P
0 S( N# @2 N2 D/ x5 X. \/ P
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 x* q* O2 B1 z4 l
对不起，写错了，忘了变量c,应该是
4 H( n" A; J5 T5 g- |& a( qd [(a+bx) c] /dx = -kc +s * t5 F' }' y; T$ h! |3 k8 k

, Y3 i3 J9 u" D( \) `1 v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:8 J+ Z# h& Q6 e+ R" n* Y
do you mean it is:2 m; E# s, t- \
& W h: h$ T2 L1 y D! ?. v; I
d { (a+bx) C(x)}/dx = -k C(x) + s
: A/ Y! {+ N: D/ S# o4 R0 G1 a, b% B- |- d2 }& h- ]8 K4 P: H
where a, b, s are constants, you want to know what is C(x) ?
4 c4 ^+ U, S) i3 N% o, g, Y* a. S' [# Q
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:
9 S8 l+ _! \+ }. r; K( J0 `4 K* U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:
; n# ^# {8 w7 k- l1 Y3 U! R: B3 O9楼的答案有点问题吧。
6 I4 U# V- d" c$ v2 X3 c您说：) S/ {  r1 c/ V+ f$ h4 h' Z: `
7 U8 R" x1 L+ t7 c9 [9 J
d{(a+bx)*C(x)}/dx =-k C(x) + s$ c: S( Z5 ]! ^2 H& f% {* M
so:/ R, f3 c9 U) c; P7 K
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
: V) M/ r. i  F+ S' ]( D) J2 S! k
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
$ D# C; r6 Q4 F4 B) g但这是不正确的2 P/ a; t- H/ F. [) o0 @" ?: 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
1 q, ?3 e0 [ U: i....
1 o( Q3 W5 x. f+ nd{(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:
' J! B* u7 B2 v6 g这样算混淆了微分和导数的概念。3 v. Q/ b: s; Z6 u% S0 \( ]

0 b1 W8 u; Q# [5 F' td(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 V& K" G Z0 c/ U) Y: A. i6 h

: ~$ p0 k% l) g7 N9 UIf 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:; t* i$ l7 ^% w, E, t
这样算混淆了微分和导数的概念。
" w: }1 A! N) ?% u4 L# J. p; {* @/ Y9 y$ 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
- e0 x/ J( D- z/ E/ G; r2 V1 M! k' }9 F* z) E
Ah, you made a mistake: it should be:
. ?1 m- E$ s0 g6 \! U' u% m7 t
5 b8 \" j8 b8 l$ F9 y4 Ed(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:
8 b6 M1 ]" a5 Z- u1 n' JYes, there should not be " ' " after "f" and "g", it is a typo. :D
% {: o- t2 ?* w. D" C! m+ W( g
3 R8 J. G1 }& O9 y. X9 i; mBut 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:* _" ~' N% O d E3 M9 { U2 Y/ s

* e A. L" [& [/ {% o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。1 e' n$ K7 I+ ~6 P" y( {
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 J# c0 h* R( f x; N1 k1 ^* a+ n
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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