数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

8 a- \, ?, z  {& O& L1 sd [(a+bx) ] /dx = -kc +s
- k; v  s0 O: R6 X% j' Xwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

, r: h2 \  d5 F5 _3 u多谢了！
5 D, F; Z; i2 o0 I6 [
[ 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:
请教大家一道微分题，多谢！
& N7 R- [$ x# N/ J, v& H
d [(a+bx) ] /dx = -kc +s
7 j2 B) ?: u7 y7 R3 o5 awhere: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

: B5 j  d) A9 xit is too easy:
+ W" K) ^& O: C  \7 Y: Y9 Z
d(a+bx)/dx = b
% e$ Y, z( S: y& r& b: a
5 Z8 i5 Z5 [1 x1 D+ p' h  o7 {& jso

b=-kc+s
1 U7 ]- m5 k- H( J' e' Y' L0 m
) `4 B; e! Z. ^+ @) [% x" e: f) p: t: yfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
  C# y- p  ^0 g3 T! Ait is too easy:

d(a+bx)/dx = b

so
% \6 D* y* z) |
0 F, W$ F$ w: l/ Mb=-kc+s
" I4 W& U' O" I! e; l) M
finally:  c= (s-b)/k

# b' y5 L& m# d5 i) A0 k2 ?
0 L4 P2 D' Z' d( wthe 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:
( b, p8 B  J) `* t3 B6 R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！
# X6 u$ a( a# w
[ 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) ?
$ _% t. N5 p7 X5 o1 ]
/ m% h/ k8 X& W9 T) C) V; b4 gplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
sorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
% ]5 g% u; c& D3 ~$ z
please tell me.

; ]2 P0 H3 H$ j, |+ H对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
6 @$ B# ~. c0 n8 m; X
多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
! e$ n1 r6 L. Z4 t* \对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

0 B+ k% r: t. F3 U; U4 M6 V# m
do you mean it is:
# Q( p8 p( m% J! g
d { (a+bx) C(x)}/dx = -k C(x) + s
' M) h1 r' b$ u$ E6 P4 `5 Y, }( [, n
% |( |8 D3 X( t" Y3 n+ Kwhere a, b, s are constants, you want to know what is C(x) ?
: j. T2 Q5 o9 `: z" i' {
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:
- q8 L+ L/ }4 w1 u  M) Z: x' B' ldo you mean it is:
; @9 M6 F! ]+ N# q( g
d { (a+bx) C(x)}/dx = -k C(x) + s
6 |$ s: K) `7 V
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 ...

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:
- w" m- J$ n  V) t. j* m& c. a
; N' U  ?& d8 u: uC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
, \' o2 f9 k' @* F, [, a) g  ?4 {. |

0 R/ I. ]. `1 I* H, y! Hprocedure:

7 I( _- H% J2 q* R7 NFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
9 i$ @% S+ D! Gso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
$ g$ Q# N) L/ j! R; k% gi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


& t+ q) K4 t+ n) `/ Eintroduce 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
4 C5 q0 r8 u6 f5 c9 ~% etherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
7 B- s- ?6 w% t' D, s+ a
! k2 v/ H+ @, Zfrom 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)
+ `) r9 l* S, q9 T. h
so that:   ln Y(x) =( K/b) ln(a+bx)
* A# A: g" a8 g( Y
this means:  Y(x) = (a+bx)^(K/b)
7 J. Q! ~' w) Y/ s2 _) ?by using early transform, we can have:

7 U  r4 S1 p8 b7 Q-(k+b)C(x)+s = (a+bx)^(k/b+1)

8 T! G9 I) W4 s  d8 x5 {' N( N. d  bfinally:
  D: z# t6 C& [: a% d) w9 |, a
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:
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$ b$ L$ T1 K2 U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。
  {9 k$ v1 j. d6 n. S! I您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
4 e7 w. C  P0 o2 d
% ^, A6 ]6 O" n+ l( s, E; J也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
% l3 O( W2 Z9 K1 _5 v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
/ q; ~4 u3 n6 l& Y' T1 g7 ux' 意思是x的导数，y'意思是y的导数。分母是y的平方。
$ \8 y8 R9 \& P, Y! S因此：
6 E. A! n8 t4 e: l
' c2 D* p8 ?9 X% `# od{(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。

, T$ v& E% v: ~2 W6 N所以您解的有问题。

/ D- v3 d5 O. Q. u[ 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:
/ ~8 w) I: H6 t# B3 h( N$ y9楼的答案有点问题吧。
9 J: ^& q8 q) M/ O3 @8 d6 {, q您说：

8 L% _  c% h! _7 O* o. ^d{(a+bx)*C(x)}/dx =-k C(x) + s
# N. l# _, W& ]" Wso:
7 X: }! P3 Q$ mbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
0 y9 ?& u* e" e8 Q( I! U
% g$ z. Q3 o, h也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
8 c9 \+ [( |3 x; ~但这是不正确的
+ o6 I6 D7 [% R; v' p$ r1 Cd{(a+bx)* ...

% `5 Y& e6 n! v& O% ?" bPlease 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_

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。
, i% i3 l! @- o1 W3 u) L4 A
4 t, t( F/ W, k( @% @# l- ^9 Hno "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
8 |- s/ r, e! _8 Q0 v4 F9 c....
- x6 J8 c1 n4 P' {d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

3 @) ]/ a9 F, |
这样算混淆了微分和导数的概念。

2 s. o# `' |+ x  c6 @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
% O4 b/ c. ^1 q' J
3 H0 U5 J* B9 z3 a" V6 s$ {# X( aIf 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)

+ G, s. k) l$ y; M8 p) u' [8 p  M[ 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:
: X) a+ M% V. Z. b6 B这样算混淆了微分和导数的概念。
1 q$ r( w8 G% ?" p' t3 z
& J; F) z' |& K" ed(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 ...

& F0 k+ J$ @; D* r
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
4 S7 I( {; R. u1 V5 K5 j3 N
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
- G6 A0 u* O/ e, L0 n6 D, L$ }

3 I% e0 n& x& s+ x9 |3 p! C2 Y# gIf 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:
. [0 M* W* C2 o+ R
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。

- j1 r; w* S5 e; P! Xd(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 y" O. Z/ J, C( u/ F8 _
, U7 a5 {. [" k- MAh, you made a mistake: it should be:
+ r! v* F2 w  e) D* K8 C& N6 i+ ~7 ~) w
d(f(x)g(x)) ...

2 `. ~# _+ k! ~; n$ c( }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:
* C+ C3 i' G4 }+ ^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.

. P2 C; `1 U3 P& n  z3 ~
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:
5 N) A1 E4 ]3 `$ o3 C2 i! B3 B
/ @& K0 }' A) U4 c! S1 O( Y, l* f6 k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。
+ ~9 N6 l! H$ O# E6 r' 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 ...

6 B1 g& `0 p' _$ P+ 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？

8 A3 q/ ^2 w  l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。