数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
8 {2 [1 U' f3 \6 ?1 U
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?

0 P$ g- c3 J" W. b: M; w9 U1 y9 X多谢了！

[ Last edited by 醉酒当歌 on 2005-4-3 at 07:56 PM ]

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
( e' U/ A1 O1 v, c; K请教大家一道微分题，多谢！

* T, H8 o! u+ `: U# d+ \+ u* [d [(a+bx) ] /dx = -kc +s
/ v) U* W  s1 A, u8 b4 swhere: only x and c are unknown, others are all known,  requiire c = ?
6 E& r  ]% n' S( Z: w
- K% v1 ^) i$ w多谢了！

5 a9 F; x/ _: b  v3 M0 f- g: [
it is too easy:

d(a+bx)/dx = b
! @6 {7 A8 I  \/ s1 K
$ N3 q# n: p$ {/ b0 tso
8 w, ^/ L7 i+ u* x: l+ M1 i
b=-kc+s
4 e5 V) @. b6 F- h" v9 @1 r6 Z
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
it is too easy:
1 |# n) y3 L+ n) K( ]7 ~  O$ i
d(a+bx)/dx = b
1 x) }' `  I9 [2 q+ i1 J2 l
0 P. ?" I* h6 wso

b=-kc+s

; E" Y% V: ?8 s  S- Mfinally:  c= (s-b)/k

% I6 [: w/ W& l0 h7 I9 Gthe right side cannnot contain c, but b contains c, so it is still not solved.

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
0 s* h0 {$ F" C; p& T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！

# B- r+ H- w5 a; }* Q. H# j( @[ Last edited by 醉酒 ...

) G) v6 N" W8 M& l/ D" a# s( w7 s/ x
/ ]3 ~, R: V1 d* _# |sorry, did you post another question? your statement is still not clear.
& e7 c% I& [5 }7 Z. b
a, b, s, k are  constants? c means c(x) ?

1 _( r) ^$ }" N: B+ }7 Aplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
) s, N( O! ?* hsorry, did you post another question? your statement is still not clear.

4 ]+ j; [; a; ^a, b, s, k are  constants? c means c(x) ?
5 l! A. u6 `7 k- w8 Y" O
please tell me.

对不起，写错了，忘了变量c,应该是
% o0 E2 ?% Z$ N7 I0 I; N& Sd [(a+bx) c] /dx = -kc +s

$ L8 ?* e& \8 d& n多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
% x$ M2 b! Y' }) M5 m对不起，写错了，忘了变量c,应该是
8 ~; ~+ ]7 U0 \* [, U/ n! cd [(a+bx) c] /dx = -kc +s
- m" A2 p; \& f1 A$ f
多谢

7 \) x( ]: `0 d1 a$ f3 N
do you mean it is:
( L& J: i+ a/ a9 z7 r6 u) }
d { (a+bx) C(x)}/dx = -k C(x) + s

/ N  l7 Q& a4 l( z8 }9 \! m" Cwhere 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

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:
( W2 O* x4 b; w6 x) J$ k6 {" ?do you mean it is:

) [5 b( M6 v$ hd { (a+bx) C(x)}/dx = -k C(x) + s
1 l* M, I5 {+ P) I3 E
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 fi ...

) b3 L& F3 Q' Z9 Q$ x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_

answer:

C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
4 k9 R6 s  b% l6 T1 ]2 K
! ~' K0 s$ E' U4 p/ d# U5 G* P* s# J
procedure:

: @& `; m8 o' {/ o! qFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

4 @6 Z* l5 T, ibC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
  |/ H2 Q8 B& t* p: `

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
2 y0 d; e$ G3 G; U! owhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
, @+ y# g/ `9 r. G7 Y1 G
( L( F# |9 {/ r{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
" K+ p/ ?6 U5 R# G1 I/ o
" C+ l% l, \1 I- _: j6 w* i) ?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)

5 c2 j% k" y8 }this means:  Y(x) = (a+bx)^(K/b)
; i4 O* a$ t* d% `' [; S# }! \: Rby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 G+ Z5 G6 G! [finally:
  d5 e& N; J0 d
( V& Q  o5 b8 y& I: C5 f+ rC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)

醉酒当歌

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_

Originally posted by 醉酒当歌 at 2005-4-5 14:33:
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！

1 I- U+ f' ^+ k( x
. g4 F8 G% Y( O/ v; Q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。
$ i6 t% _: I; F( j; K  I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

9楼的答案有点问题吧。
0 L( r  z; l2 {& r/ I您说：
7 d, t; c6 v) M! F
2 `) F/ B; |- {# W# ^' [/ |5 \d{(a+bx)*C(x)}/dx =-k C(x) + s
; Z, c3 S* T& p# d7 C% J: ~so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
1 W$ S* o" N/ @) i/ e9 v
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
: {+ X8 R4 W& ?d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
: Z2 K* c; |" S# U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
) C/ H( `. `2 C7 f. [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：
8 }; k* y% E' w6 s% F8 y
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_

Originally posted by 龙舞九天 at 2005-4-6 19:34:
9楼的答案有点问题吧。
2 `3 `9 b' m& @) \2 y您说：

# A- s' t! M3 Z: `) h1 S9 ld{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

  \: v& Z  M1 j' r' ^! D也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
- l  _6 t' [" F, `但这是不正确的
! H* i, k. D0 A5 ad{(a+bx)* ...

1 c: x% I7 H  ~! j3 j  _7 c2 J- ~
  L% ]9 y- R/ {) G* ?( g+ [Please note here:
! Q/ L$ N  [1 X3 I6 M: a
" |" n  }7 h3 H5 e+ O8 o1 X' yd{(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_

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

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 _: U, H# B, i% Fd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

) W4 ?! d5 P, b. t: m  d1 P
: s7 k+ E* z% E" @1 ?) U, 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
6 ]% ^5 @/ G3 T" Q: [$ a
, z: h. _+ r2 s; l9 yIf 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)

* i# L1 `4 ~/ `[ Last edited by gnome-kde on 2005-4-6 at 08:31 PM ]

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:30:
这样算混淆了微分和导数的概念。
( X# C2 g# g0 y7 N3 _1 u" k
/ ]+ {0 k) W0 m4 N) H- \1 r0 `6 f1 K$ md(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 ...

0 [  k0 q0 \) s4 c7 F  X& o. p0 e) l
这样算混淆了微分和导数的概念。

3 q  u3 X2 g) N4 t( @- Rd(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:
4 o0 J8 Y9 }: y$ }: w3 k- }
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

8 s0 E" Y( x( F7 F1 C/ g/ e! U
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)

& M1 k( g: v3 T8 \4 Kfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
8 n& v$ s7 k3 D/ H6 ~
" v3 a$ ~# j& Z! k# |0 [h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
这样算混淆了微分和导数的概念。

$ l* W/ G- M5 C! [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
2 l! M+ S# g- j0 A+ {7 d
3 E1 e% a+ P2 q+ nAh, you made a mistake: it should be:

d(f(x)g(x)) ...

: ]7 o9 A  z$ [, X( T
; v% j" [* _) b( p% IYes, there should not be " ' " after "f" and "g", it is a typo. :D
. E+ o+ |( d6 r
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_

Originally posted by gnome-kde at 2005-4-6 20:51:
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.

6 T$ [0 x6 v! t/ O
% A, x: \: ?  B/ |' |6 {  T% Q! SThanks, I am happy here to discuss math.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:

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 ...

. w, Z" X/ E' g  c9 ~- P5 o# @+ y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_

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
# ~2 H) s. s( B* @* V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？

4 j  _# T  i% E. ^4 \8 G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。