数学高手请进，多谢！

醉酒当歌

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?
- T9 g& {1 g1 x- T
1 S/ m9 P5 T# ]' _2 F0 Z; l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:
( `  g; X% ^- b" e# b请教大家一道微分题，多谢！

/ u- p9 X4 w# d4 H+ S1 Qd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

# v! n* @* \$ E9 p+ Z- i多谢了！

. K  d% o/ B: g8 H! a4 mit is too easy:

d(a+bx)/dx = b

$ w% _$ s- x9 E# X3 e: {so
/ G- V9 J+ W, d( T
b=-kc+s
/ D/ r! P2 n* s5 h7 F4 E) u% g
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
5 [( ]) ^! V# q" t0 L2 R7 Qit is too easy:

d(a+bx)/dx = b
2 W) e/ g* W9 T' v9 r
so
/ K* [; d* S+ i& C( R
7 p  y  u. ?4 e: Q: g4 E0 H) P4 k  eb=-kc+s
. ]5 U, ^" [0 j6 p4 G
finally:  c= (s-b)/k

! K4 R& k( p& ]  b* H8 x$ R: y. u* i
$ h1 b% c; X* N- Sthe 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！
# K0 r  W2 s- C9 `' j  @
4 R% A* ^  g; d, F8 H, Md [(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 ...

sorry, did you post another question? your statement is still not clear.
# e& s8 A& L1 z3 Q7 S& [
% k( S6 ?7 c; q+ ya, b, s, k are  constants? c means c(x) ?

' c" v( B6 e0 [1 z6 J. x' i) L0 yplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
' U) _. ?. a) g: _sorry, did you post another question? your statement is still not clear.
6 O0 P" X2 Q  t. K  z
1 J3 S6 a( ^5 u1 Ta, b, s, k are  constants? c means c(x) ?

; \5 j, z3 C, d; P3 x6 h( Vplease tell me.

对不起，写错了，忘了变量c,应该是
, o& k' Z" r9 j  I- j0 y" l# Pd [(a+bx) c] /dx = -kc +s
, ?  |& C+ i7 e7 L
8 `  ?; c1 I' s7 i, ^- f& g多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

" M( O3 _9 T9 Y0 G& t- f多谢

) m$ }4 s. K' N& T1 v1 K; w0 q, W
do you mean it is:
6 g/ f+ D& i7 G; p; q
$ G$ K) R0 ~* h9 l7 D* Rd { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?
' O* c: j) d3 p+ k9 [
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:
/ R3 _- F$ E) Q  E. hdo you mean it is:
! u: ?$ B9 I4 O2 m9 J- `
d { (a+bx) C(x)}/dx = -k C(x) + s

! P0 ]( \3 d. a* Ywhere a, b, s are constants, you want to know what is C(x) ?
) b- G0 X0 E% k: W$ q
& m) _; N6 G& ~: G5 M7 D) eif 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_

answer:

1 R; I2 l  r# J3 k) O( `2 c& ?/ \+ J$ DC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
& Y( p8 i. G( I. f4 }
" `- ~3 ^  ?; t
procedure:
; G" r/ z* c  D& M. @4 o7 Q( _; J
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

% s1 B( M3 t; tbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
' @3 _! ~+ P; b
2 S5 ^1 ^0 n$ O5 G  [(a+bx) dC(x)/dx  = -(k+b)C(x) +s

3 W1 s. F! l: I6 F
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
! d" Z5 `! T& m) V! G, iwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
2 [0 z2 u5 a9 [+ i$ b
{(a+bx)/K} dY(x)/dx=Y(x)
0 J9 j2 [8 v& r# k7 v$ r. t1 `- ^
from here, we can get:

3 O/ [; z; T1 D) qdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
! d5 N8 n; G  `/ `: ?- H+ c+ r/ f
4 j& N  q3 q8 R$ Iso that:   ln Y(x) =( K/b) ln(a+bx)

4 I( j3 j& B# Bthis means:  Y(x) = (a+bx)^(K/b)
0 \. K7 s& V7 r& T  sby using early transform, we can have:

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

finally:
' }5 m( s$ O. g! R6 y" }
$ }+ h* O& p- y3 {1 ZC(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！

; g% @; @" _, M) K2 w  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。
2 `3 m5 ?3 v5 F/ [4 G/ B8 `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。
( [* K6 w' |3 V: R. z. D您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
' ?! P* S* A8 j! Dso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! w) n% |/ j: a" P
  c* l; `. e# W+ I. Z, M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
# S, t* @! z0 h$ j: B; Jd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
3 v) q# `* g; o8 x3 @* u1 J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。
4 w1 z7 ]- F/ h% S2 D; _因此：

0 G3 y+ v! V  ]) |) y8 J8 wd{(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。
$ ]2 x2 d) m& c7 ?+ i" ~
- e, s( y8 k$ s4 O3 b& R3 i所以您解的有问题。
8 c: k' c0 z% o6 ]8 C
[ 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:
* ^6 `5 Y$ D( X. i: l3 e" X& \; {" l9楼的答案有点问题吧。
您说：
) g8 ^* E8 K6 n  U& r
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
6 A& E) x  L! P6 ^/ o. t% O) z" O! SbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
3 P* G& G+ g; H( L! A
6 Q6 z/ P& r% m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)* ...

: {; A& T8 m/ u' h) mPlease note here:
! J" [, Q7 x5 v* I5 ?
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_

但这是不正确的
- x( h( _8 h. o6 B3 U9 n2 Pd{(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。

# W/ K3 u4 |6 `" `) J) {0 R$ Pno "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
/ b$ \- `/ t+ |5 N! s, M  \, _....
" X' k% w" B4 g% F, N9 \2 Zd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

  V$ Z) n. M( I6 R这样算混淆了微分和导数的概念。
* T9 T6 h* r  e( l
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
+ R1 S5 h8 D, o- j% B2 a
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)
2 g" R% n* N$ \; D4 e8 @
! q% {% [$ D% n2 B* t$ Z[ 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:
; g% l+ [" c6 Y, j& L( H这样算混淆了微分和导数的概念。
! o) @: ^  E( ?# ~5 a+ i) 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
0 I! Y2 W- \. o9 n2 i& H
! n3 H! V& T; X; O2 iIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

3 p4 V9 z$ m0 u7 [7 g8 j
这样算混淆了微分和导数的概念。

7 h) X0 l( ?5 m* a: [8 f& Fd(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
+ s$ k0 Q9 `: x+ h) y
: \; s3 ]1 I. _( E( w8 W$ X) WAh, you made a mistake: it should be:
/ Y- p& I" R1 O$ }' d- D
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
6 b# D4 j6 e  t( s

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)
, b6 C' @, C4 s8 x
) @4 p& Z9 G5 J  B( W( r4 A6 Ffor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

5 T) P& w' {$ a6 Mh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
# V) h  i' P5 S' N这样算混淆了微分和导数的概念。

' D/ t, s3 V9 \- j) E- N2 Dd(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
; I& `8 _) v- A- H1 N
Ah, you made a mistake: it should be:
& C( ?, Y: C8 c
4 [: f) n: K5 _; r- k- od(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_

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

: M6 T4 \+ \% V) o* x. N8 ^( gBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

, e4 J2 ~/ v4 j2 O
+ n7 ?1 P3 }# 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:
( q, y* p7 \, _
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 ...

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？

* ~& o0 D3 Y4 m# L7 B# E, ~# t3 j: w9 U
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。