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 20:41:3 J d: C/ l* c8 R' x6 r
请教大家一道微分题，多谢！
$ w- T/ i9 S: `; _* C
" D; ^, m9 n% n& q" yd [(a+bx) ] /dx = -kc +s
- Y$ u% L/ `. ]% a! ], N9 @5 W+ ^6 [* Pwhere: only x and c are unknown, others are all known, requiire c = ?) r A% Y- C8 Q4 K0 d5 _% G- e

0 l( N' w2 l% i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:
& o2 g3 W( M  i! e+ Dit is too easy:
7 @) D( [) \6 ^% \: J( E2 L7 P/ ?
  j6 u0 k1 y; k& f% ~' a$ Xd(a+bx)/dx = b
4 e7 G* s, K3 G+ {6 t' a, ]% X( H* s3 f5 U
so
( F& i7 |8 c, G& W9 T9 a
" A  P/ V2 Y* `$ Ab=-kc+s7 ]9 s; X* `& `: A. Q/ y) A

, \. e' _( n$ pfinally:  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:! t9 N# F: c/ [
请教大家一道微分题，多谢！
& R( N( E" H8 R
! N+ K) `3 a) Q* Vd [(a+bx) ] /dx = -kc +s - p" R/ i+ G7 n
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
: b. A8 Q# O' b8 y, [3 ]. A8 O) F5 M) b; w1 _) _3 y) K" O
多谢了！& U3 a, O6 f" E9 K/ Q& e
+ x+ L: O4 S1 y) z4 `# s
[ 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:$ E1 {$ [# ?$ d. X  @) j
sorry, did you post another question? your statement is still not clear./ t, V+ |' K9 _" G2 z" v

6 ]# v) s; R* o! l- G' j5 g/ T/ La, b, s, k are  constants? c means c(x) ?
+ S- O+ q  A# r8 U/ X1 e' f+ D2 f- {5 H. F' R
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:, g: W, A: I C. j) L+ e9 ?
对不起，写错了，忘了变量c,应该是' w1 \4 ^' F1 t* q- @
d [(a+bx) c] /dx = -kc +s
. X8 }( e! X! U2 j' N; J1 F( i6 I" C/ n
, v) a; p% k+ 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:
! q0 a$ t  e9 q& g3 ~& r' v  fdo you mean it is:
# |6 J" f0 v9 J2 _: ]$ U" P1 O' y+ N4 J: g# l  P% U6 j. r
d { (a+bx) C(x)}/dx = -k C(x) + s
+ \" u& t0 H" D* c5 j/ S1 f* G- _3 ^
! c5 [  W/ u, `& K; [. s; S( ^where a, b, s are constants, you want to know what is C(x) ?
5 |. Z* k, N! |3 t0 U( B8 i! n- r$ n
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:+ @1 h( m. S. y6 q3 R/ M
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:
- |! w; Y6 j9 Y/ Y: S) q8 H9楼的答案有点问题吧。
& l) j4 J% g; O/ B, _您说：
" v9 w, t8 n" g8 I* _& h% L
, s# p+ S7 v! Dd{(a+bx)*C(x)}/dx =-k C(x) + s0 M0 A+ `$ s. T1 L
so:
' x8 n* u. w1 c' F7 _3 hbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 o& g) x# c- h) X3 e# K
: @& q& g6 Q' ?! \' O) Y9 m# `也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx7 }, S' l/ z! {2 c: ~9 w
但这是不正确的) _3 v9 O2 m% P c" l- q
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6 P! x8 ^3 ] T0 c# d4 j
....# X- c$ D! U; Y! v
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:/ }$ V2 b2 p7 x8 M( x. @3 z
这样算混淆了微分和导数的概念。! U/ ^7 n- Z0 d. Y, `

2 L: O9 G; m) V6 t( { Q9 jd(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5 N. w7 Z% l+ U* ^0 B/ O

: d5 r6 g( a' Z$ M: _ c; bIf 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:' y# U, T9 t/ i5 Z
这样算混淆了微分和导数的概念。/ L: B/ T! |0 k" L# C
5 l6 p9 \" V# z1 o
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
4 i" \8 P4 l; R8 B+ N& O# e b2 H% t) z+ v: H, x R
Ah, you made a mistake: it should be:, ]6 o9 L1 B8 I; E2 N2 u( i
( @, r% n5 f, J
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 20:51:7 G- S( e H0 A- ]
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
/ l, w9 C/ T( S I5 }" r& N, U5 s. v) L- L; X1 z
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:
5 ^$ H; i5 v9 d' t' u/ u$ t
# u$ I+ _$ ~2 M( @) v$ z$ z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。* | K4 X8 Z w6 s
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: n7 x9 q& ?/ f4 q, ^+ P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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