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?

$ d, r: D$ D! V/ e. \1 U8 \多谢了！
+ b0 y' {% E8 Q% y0 c
3 `5 c8 n, l: n9 a6 F1 T[ 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:
" u8 q% u3 l, G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 = ?
4 K9 `5 \  i  A
: P# c' j  y0 `5 N2 q多谢了！

it is too easy:

5 x3 m: Y/ ]6 u& q  nd(a+bx)/dx = b

so
+ T  V$ a2 Z8 Q( w1 w  T! u% @" Z
b=-kc+s

7 |" J5 i. c- |4 H( Tfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
1 @7 `$ S, N$ j4 cit is too easy:

5 P9 B/ u) o5 ?$ M" h' [d(a+bx)/dx = b

so
5 C% O+ c; W: A. H" j* A* O5 `
b=-kc+s

6 F. W+ f) U) X# |; X5 q( ifinally:  c= (s-b)/k

: N# p; C4 L% {, W" Q
/ x) k6 J: e% v8 lthe 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:
+ G. N* x( N0 p* ]* p请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
- D$ `9 c6 E7 C( W1 ?/ cwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
7 y% p  q  W) d- Z% ~% Q+ W3 n, N
[ Last edited by 醉酒 ...

  [, }: D6 K; Z! c9 c( g7 gsorry, did you post another question? your statement is still not clear.
& k  c0 z, z5 s; v7 O; i* d" v
a, b, s, k are  constants? c means c(x) ?

please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
% ?; {8 p4 `) m5 f5 w- lsorry, did you post another question? your statement is still not clear.

* T/ Y! ^$ T9 u8 ma, b, s, k are  constants? c means c(x) ?

please tell me.

, r) P2 N/ N. u" [# I+ e对不起，写错了，忘了变量c,应该是
- y3 r) u2 N6 l, K% Z1 q/ Td [(a+bx) c] /dx = -kc +s
- I+ k1 _" e) B# B& J# G4 x/ W; v  _6 }
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

/ L6 I4 w# r2 s4 u  I; j, o多谢

6 Q0 ?: m. ?2 I( o7 Edo you mean it is:

, w( F/ H) |- Z( w+ }) i; ed { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?

, @* \. F( q, c' s* F- M! }& d3 Z1 Fif 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:
' P5 |7 o& o" q4 b) F9 w% s6 j
d { (a+bx) C(x)}/dx = -k C(x) + s
- A0 h4 _1 b+ a4 I
0 K6 N% F! h; K/ ?. `; \( }' nwhere 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:
4 _  @7 Y' n7 S8 b3 N8 k& M
8 R& h" G; z& w2 |C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
$ u4 o! ]2 w' s, j0 O3 G
% ]+ |/ `2 J! D* V* p  L
procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
+ V5 W& @% F( l
- Z9 j. l2 Z0 y7 F# ~7 Z) ?bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
8 o( ]. t8 y/ T' A4 V+ F$ T1 d! ei.e.
$ U9 c% v. L3 R  _
3 j2 |( S& W$ ?1 w(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:

. V- E8 l: F5 x$ F5 `0 t  W{(a+bx)/K} dY(x)/dx=Y(x)
+ X7 d7 G; s8 k9 u# o) }4 h
from here, we can get:
5 B* I( V# g1 y+ m8 T
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
% h" {+ d! u+ v7 X% p
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:
; {4 h" V3 f3 K6 Y- G6 a; _- z) p
; e5 a7 x& S' [  K- S7 ]( i-(k+b)C(x)+s = (a+bx)^(k/b+1)
1 }; E  {# l1 `# M* O# U/ w
+ P, X, E  Y, m( Rfinally:
6 M. n, z- J2 M+ o6 X7 C
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:
' g+ T: S) _5 A: S) c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！

/ W; a1 M5 H0 I+ X, P
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。
% C, t4 S- M5 `5 T) a0 ~) I- N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。
4 m4 A9 p3 k' [* A7 Q2 P% W3 I您说：

6 h5 M3 P4 e" X( H  T, s2 O: p0 dd{(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
- P3 |- ^- Z) _2 {: e! Sd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
7 Y1 I. b, D6 V9 X0 i8 H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：
/ F7 c" C! f7 }# R! X& M
2 M4 K, b' }! z8 f: s" C$ ld{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
' w2 P% m4 i  b4 ]最右边是x的平方。

6 X% e  @4 c& I% T所以您解的有问题。

" F: i. s; ~& ]1 w% e% U6 ]$ q[ 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:
7 y* T5 N# ?: S% `+ Y' B9楼的答案有点问题吧。
( _5 P+ [) q* @) W您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
! q" L' G+ q& J; q5 u( iso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
5 ]5 m. ]0 b8 j7 v% e9 e9 `6 |
) v, Y9 x* V. D% V: Y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)* ...

( Z% n$ I) r4 Z# qPlease note here:

% I! `( q; @0 I1 a% G3 t/ sd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
% p" l8 r% k8 b  X# I: nit 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
' h( u, s# B/ e+ _" \( K并不是   (a+bx)*C(x)   的导数除以   x   的导数。
$ F5 e4 ~) U9 x* R4 k  ~9 S( t8 P
+ \' E, c: ]3 L" x6 yno "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
; ]. a9 u7 M+ _" a: G0 J....
( N( P/ K2 f* Md{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

+ w$ R' v0 r: u) Z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
& }' n& r' @% M% s& Y
- c& u5 ^) |. z) `. v3 rIf 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)
4 p; T+ l9 @" {
[ 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。
+ u, [- ?/ v0 d$ c& G
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

$ _+ _9 g: ?/ [If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

8 Z, |; ?' k) H2 j这样算混淆了微分和导数的概念。
1 r5 Q0 p* C; R1 t1 Q9 |
4 t) G* |! v7 ~% @3 X8 t# a* 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

4 N  p- ^: Y. RAh, 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

2 X; h. W. x3 E& ^
4 g5 d8 p4 v0 `# h4 ]7 S1 MIf 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)
1 Q( `5 p: d- g5 g7 c# z
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
& \+ M) B/ H$ ]! Y# o* q
/ I: x# I( y8 p3 {: d% j' Wh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
0 T2 H; z( _) L/ U2 I这样算混淆了微分和导数的概念。

. }# @. a( h: k' y& Y, {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

, }) {% d# U: v' C' vAh, you made a mistake: it should be:

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

2 b" Z, w5 U: t3 U2 D% G4 dYes, there should not be " ' " after "f" and "g", it is a typo. :D
, B7 G, A) w% d2 W2 N4 I
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

- M- C' B1 O% j& {But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

! O3 F6 v0 ~1 f- mThanks, 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:

8 T0 r; E6 z8 w7 _& m+ @' `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。
- E, V8 V5 A- A# C- Q) Z0 R1 e7 T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 ...

7 @& Q. C! C9 i7 ?2 a  K+ N2 Z
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:
+ f& n( ?" V$ ^& ^! \1 T& F( H1 ?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 C5 A3 A- ]$ x1 w7 s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。