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 20:41:9 @9 C+ z& T, W4 |6 ~" z* ], c
请教大家一道微分题，多谢！
+ ?/ v" r4 k1 r4 R- b m) F P) z: Z. E
d [(a+bx) ] /dx = -kc +s
" R7 ~# C9 K' a+ F* }7 z/ M" `where: only x and c are unknown, others are all known, requiire c = ?
# V) U, q+ M/ o$ v4 n
! c5 A; d. U& f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:! m9 |: P1 Y% N; C9 H. t
it is too easy:' \) x$ n5 {2 R# ^. }) j. i4 p
) U4 Z6 I  F. ], }
d(a+bx)/dx = b9 s* I, w: x" ?5 X6 C. n' C/ ?

6 q7 V( W" m! d, O+ Oso* Q7 S1 b, e. R& L) m$ l; |7 P
9 _7 t4 x! J- Z1 w4 F9 d/ m# z! t  O' W
b=-kc+s4 ]0 t6 W& {: x
6 M8 v, t! r; E* {2 Y5 K" W4 R, M
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:
3 C% M' Q/ U0 I' _请教大家一道微分题，多谢！) I, n+ `- |' Q1 N# a& ^1 n7 t  A* F
- q' ]' ?1 W: I+ ]1 M
d [(a+bx) ] /dx = -kc +s 7 \+ G& r  ^6 O- o
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?# M  d7 A0 l+ ?1 t4 j. ~' x5 ?
" D' Z0 z& @6 p' k$ V) m
多谢了！& H0 @3 O" S, f" E2 w5 {

$ B" ]. `9 d7 S" }0 N6 i7 R* ]; F[ 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:
: }# N- S5 p& A6 wsorry, did you post another question? your statement is still not clear.
: I8 t! ?5 n5 A8 ^
$ D* T: e- {6 Z3 la, b, s, k are constants? c means c(x) ?! P2 C* Y% A/ k O) Z, R

1 E4 C" |) V3 N: Mplease 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:
9 q6 l; U3 T! \7 G n+ \2 L& D对不起，写错了，忘了变量c,应该是0 {; k. M& B# l f; n
d [(a+bx) c] /dx = -kc +s $ b+ K* k" Y p- K4 a& y
( n# ^$ B+ s8 [3 Z- `
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:
0 \; Y& f4 `" d# `& Ddo you mean it is:7 A6 U* Z& N. o* N. _6 n5 ~

4 s% N0 _7 Z. xd { (a+bx) C(x)}/dx = -k C(x) + s# C: n& F. p- s4 \/ C4 J
/ N/ h' M% x7 r& r3 C1 s& I
where a, b, s are constants, you want to know what is C(x) ?# Q) Y1 ~& [5 f, k0 `; E3 v6 p# K: s
& }) U" M' G6 U' a9 g) `1 U# Y- V
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:
; o9 @! G9 L) I( F) l( D5 I* B& \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:
- }( f  A' m* L  G. }( l9楼的答案有点问题吧。
: c" U0 s; G7 G$ L* w您说：! a9 P, X/ C3 i& l( b; q( s4 N% G

  o! P+ V' H  _5 Z- ~d{(a+bx)*C(x)}/dx =-k C(x) + s
8 W* C# i4 T* ]$ I4 Y) x* {: A! X( oso:
) l; S- ]/ g0 L4 W0 N& KbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s. o4 P$ O( H! p9 v4 S: Q3 |" P; Z
" \0 t1 I2 l' `& Z8 B' ^
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx7 `4 V! B+ ]7 J
但这是不正确的% I: @0 X# Y( `9 m: J
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* F" C# p+ j, B) h+ Z
....) w- e4 ?& p, L1 D3 j
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: v9 J% s3 v! ~" Q0 V: |
这样算混淆了微分和导数的概念。
& r; B2 D2 ?/ E( }" H' x+ D$ U- y0 [+ v! R- l/ [' u. s
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
, t7 B6 R$ M B! M+ A
$ [/ k) n6 N, V; W& x7 P1 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:
) R& a; @" a5 `这样算混淆了微分和导数的概念。- W( _8 G" Y$ \' I$ I

! D% L) k) H0 M( t1 s5 ud(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 q2 k7 M& a5 F: k. T
4 h4 I- [- r' G% ~$ G1 O e8 Y0 g5 gAh, you made a mistake: it should be:- t+ L! B" f) P8 D/ d1 ~8 E

9 T8 @" ~' j" p/ `# o4 C' R: Z3 xd(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 20:51:3 b7 ]1 E2 s8 k! s
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
# R% n/ p3 s) k4 s5 [
. T! \4 e3 G1 ^1 rBut 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:; U R* {* T& h& J0 y

$ U$ x: I2 y9 G |+ v6 {4 ?3 r% Y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。# O! ?# R& |$ k' [% S& ^) X/ m& E
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:
, B) _$ g5 J' Z* }3 t' j5 c+ k7 h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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