数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

' R6 ^: W& K+ F, Fd [(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?
& s% E2 Q2 f( S- p/ i
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:
8 h1 M' }9 a" t3 O7 n请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
& s2 V2 i& p- s- \5 P1 C. L- H, J" Fwhere: only x and c are unknown, others are all known,  requiire c = ?
. v' n% F! B: P0 v/ C3 ~1 X
多谢了！

it is too easy:
5 _" n+ H& J; w$ |
/ s' p0 @9 }2 qd(a+bx)/dx = b
9 L9 @- N2 X! F: W+ h
7 L% m. X& F8 {) _- nso
5 M/ R4 O; y0 @5 R; F2 z& m% l
b=-kc+s
1 j: t8 W/ z4 D: x1 s& _! t
  q$ I  P! }; e) ?; s1 |. w* nfinally:  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:
5 ]1 c) P) x& A9 Z5 O3 M' Z
% `( b8 {. D8 r3 kd(a+bx)/dx = b
" z$ I* J% G; s* @
so
0 h) N0 {9 L: w( r* d' [2 l- o
b=-kc+s

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_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
请教大家一道微分题，多谢！

0 L% ~0 x" E) l4 td [(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?
! F9 P: n3 _  b- w
4 ?+ `. O% b  g( x多谢了！

! K1 k8 i: ]8 a7 b- d[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.
! i+ E8 ^5 K! y, w
a, b, s, k are  constants? c means c(x) ?

( C/ u+ z. o1 m: R) N+ \! Z7 e& H: pplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
9 J+ M: b' O1 R# F; A6 fsorry, did you post another question? your statement is still not clear.
+ S# O% T: l9 R
+ H# U+ e1 D3 v3 g. I. H! \/ ^a, b, s, k are  constants? c means c(x) ?

please tell me.

% S; e5 {: Z  M: f对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
& o8 K* {( x& f& s  U+ d4 h( w
多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
; S% s9 N( }5 _  P对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
2 }1 p3 S/ O- ^4 n+ f& t" Y
0 b2 p0 H( I! q0 E: X多谢

) `# `- W1 M6 i6 u8 Y  J, {. l
do you mean it is:

6 I2 j' ^% U0 ?1 }3 K7 C+ X+ ed { (a+bx) C(x)}/dx = -k C(x) + s
3 F1 L& e5 f9 B; w8 ^+ |* X
1 @5 U) V6 m% P& I# Ywhere 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:
do you mean it is:

  e% D) r: i) I' [9 d" U: cd { (a+bx) C(x)}/dx = -k C(x) + s
  |. g# k& _/ F: x
2 V2 |, ^6 }- {6 Mwhere a, b, s are constants, you want to know what is C(x) ?
0 o  h2 ?: d/ z1 Y, [" \
) T5 u! `" O; ~6 aif so, this needs to solve the differential equation. please comfirm it fi ...

2 @# y0 U0 [4 z9 c1 u
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:
' O- Y# w- D) l% r- b
8 l  S% @; Y& P6 {- GC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
6 u" |' K& F- Z1 g
( t% [9 W# g& ~
procedure:
! D. h" \0 n3 n
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
. S# l- i0 h; m& U7 y3 h. ^! nso:

: X" g. l( O, ~' e5 v1 |bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
' R, `8 j. u; N* V+ Y
" h8 b3 @% W, o) T5 j* n; n# L
" m# m3 U+ p7 |: l, ?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
, P4 f* w# [; R0 H' ctherefore:
  L) Q9 T* a/ d4 b( G
{(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)
* I6 e5 Y% i& L% E5 f1 }
! ~5 y  g: [+ Y( B# u8 cso that:   ln Y(x) =( K/b) ln(a+bx)

) ?4 l$ }+ P& {7 ?* J, Gthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
- s3 Z% f. Q  M, u4 O! V
-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 B. `8 E# W. L5 Z$ G8 bfinally:

C(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:
) N& u6 ]- f1 u  L- ~% ]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。
) h: ^6 r0 ]' I) P& o" `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。
您说：
+ m8 ^% k4 I7 O$ v3 x1 J
  W' {; ?2 Q% [5 @; I) U2 _d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
4 \5 b3 b5 O8 E9 m7 lbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

1 K4 U2 k" |# |+ w# ~1 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
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：
* E- i- R) q1 {/ _0 W
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
# _; l7 Y# c1 ~' S最右边是x的平方。
3 Y+ m5 D# |3 O+ S1 O3 `
所以您解的有问题。
0 e, n: Z3 Z2 {% ?9 P" k
) l- z( w4 U  S% k9 G[ 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:
; S& f0 P4 |( O0 g5 Q$ g9楼的答案有点问题吧。
您说：
  [6 k- h+ g  U+ a
d{(a+bx)*C(x)}/dx =-k C(x) + s
- r6 k6 Z; J+ X6 k5 c. n2 D( gso:
, O+ [. ^0 y! j2 R( G# w& U3 obC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

5 x, o) W) g0 A$ C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)* ...

; v5 T( V, `# j* H. ^1 V) EPlease note here:

d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
* X. O$ X& s3 Jit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
1 {" A+ p5 `9 z5 V' _1 n  _d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
; j  ^6 z  J) E* b! {! N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
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

3 k/ o8 [& S) S% p& L- m
9 `3 R+ v6 }, m这样算混淆了微分和导数的概念。

8 {8 o' ~8 m/ \& 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

/ o1 }5 N" h7 d/ ^# ~7 NIf 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_

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。
/ n: y. O5 ~( R0 e+ x  P# i% Y3 t
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
8 l8 n9 E, G9 n
# Y9 D; d* ^' R; _- i* x- SIf 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。
1 @& R% z/ o; T. ^& d3 n& ~1 j
0 S3 _4 S$ N. m2 C+ J( Sd(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

/ ^! w4 u8 V; @' m; t4 gAh, you made a mistake: it should be:
, [- o9 |, x' ~4 j# w
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

/ ^' I. k; U8 ?5 m
' P* L: ^/ c; t3 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)
( w6 ?* ~4 i' Y" j' a
3 _" C+ F4 A. I# hfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
1 f$ G4 c' ^1 `+ y* A5 U2 O6 d
8 z" Z' G0 E! e: b3 ~h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
( [( m# r8 g# ?0 @& B( h$ l这样算混淆了微分和导数的概念。
/ w% X9 ]5 l. v
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:

: S  G- B" V! x3 a) l. pd(f(x)g(x)) ...

. ]5 }) ?$ q1 Z, x/ P# cYes, there should not be " ' " after "f" and "g", it is a typo. :D
5 z7 x# a9 G3 v
" _% X1 _/ N4 r1 v) \; ~" ABut 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.

Thanks, 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:
* M- V' M5 l1 D8 S) X6 a
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 ...

% J7 S# c5 X6 {/ j! H% @/ I
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:
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。