数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
* R- s  N. }8 R
d [(a+bx) ] /dx = -kc +s
9 N* B. W' r; Cwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
; \" H/ \& T  g6 F- R4 P' r/ h  `
& q5 P4 `: d& L, d# }6 _2 Z4 R多谢了！
$ Y+ ]- }$ ]7 }" N1 d% C
0 t! v! C) p# P2 y[ 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！

. H7 t& v4 m& r0 i; k6 q9 g* ^d [(a+bx) ] /dx = -kc +s
! J9 w" p* s6 ^& \/ }where: only x and c are unknown, others are all known,  requiire c = ?
! N0 q: K8 }5 G/ x& i3 {+ s
" w1 }9 c4 u$ L多谢了！

+ @, U( G& w" ~% E
it is too easy:

d(a+bx)/dx = b
( x0 k6 X9 l9 c
so

b=-kc+s

finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
* y9 [' E3 ^% `& S# p- P. `- Zit is too easy:

5 t, ~: f' n2 d' J/ xd(a+bx)/dx = b

so

6 I7 D( o' u9 D5 R! ]b=-kc+s

finally:  c= (s-b)/k

4 ]5 H- m: V/ x6 ~  V( @7 q
+ H) N' q0 e. l# T0 Rthe 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！
4 b. b6 k) g: t
8 g/ |8 D: r9 ], y( jd [(a+bx) ] /dx = -kc +s
% A) I& T' W& _% \8 lwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

$ _  `; {- w6 E0 \% y多谢了！
1 ~! ]9 \4 M- H* H3 U
[ Last edited by 醉酒 ...

3 k. o# R/ _8 @- w% M4 Nsorry, did you post another question? your statement is still not clear.

/ z& v# j: M0 R' w9 Q* X9 Ha, b, s, k are  constants? c means c(x) ?
* A3 e( U/ O! p5 t& _+ i5 B" v
8 a2 N% \' ?; F6 V; j: t8 M: ~please 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.
# h9 \7 b4 G* ^9 ~8 n! k+ G  ^
a, b, s, k are  constants? c means c(x) ?
. j; Q  C- C! \( r
please tell me.

对不起，写错了，忘了变量c,应该是
" N& d, w- h+ |1 d% h) p$ G! Dd [(a+bx) c] /dx = -kc +s
7 f$ k& \/ G5 q
多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
- t5 N7 Q8 Y# F" {对不起，写错了，忘了变量c,应该是
; J9 S( c! |7 Md [(a+bx) c] /dx = -kc +s

多谢

do you mean it is:
. [/ X$ s9 S* M% j4 ~3 X& v6 i
. E& P5 B% C: d* cd { (a+bx) C(x)}/dx = -k C(x) + s
" l$ X2 Q+ T6 q: A3 E$ A! Z* |
where a, b, s are constants, you want to know what is C(x) ?
$ c( h+ r! c: 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:
do you mean it is:
9 f+ X% T2 D8 m6 i0 `
0 a. a( T% X% ^8 c( }d { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?
. u+ ~1 [( \  C+ E$ ~, z
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:
% h# ^0 L$ e1 a; L0 J
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
8 u9 E1 ~- p( n
7 `0 U" t: p% Z+ @8 a! N
9 b, ^) Y# M& Z! h: P$ ], Gprocedure:

. V8 [5 W3 \9 @  V* m$ u5 \9 G5 r+ lFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
* I. s8 r+ m3 Q( }' B
- U6 @( O9 Z4 m( b0 n- T* ]bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
1 e4 m& p% F7 l" M) Z7 c% n
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
% }0 j5 I! s4 ]* G1 |

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
- Y/ p  Y; r6 kwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
- ^- F5 o' X$ f# I1 H) ?% P9 ctherefore:

6 g+ ?) A/ c5 U{(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)
* D# N% T- ?6 a
so that:   ln Y(x) =( K/b) ln(a+bx)
1 a' D. @" O0 t9 L* s
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

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

finally:

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:
! D6 c, y% \9 Z, A; T  D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！

! V$ R8 D& ^* D/ A5 O; `6 L( Y0 W2 j
2 a! w$ R9 \3 Q: z, O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）。这种题真的不是难题。

龙舞九天

9楼的答案有点问题吧。
. W- O6 J1 B* r5 u6 z! U您说：
, M' F9 [  D; c
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
0 s  `0 W) J6 A2 B) JbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

9 E# z+ O! P) B0 s6 W; a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：
6 ]0 s& T, x8 d5 h
  T: b. c; @5 \& z) u( R" E/ Yd{(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。
) ]4 p* V3 \4 C' P+ ^
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。
您说：
" ]# C4 ]" S' p0 x) [0 R: p! v9 e& l4 ]
5 s! X3 h: B' l* v" ~d{(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
9 T) ^! A3 `( p8 F但这是不正确的
. Z- e8 ]6 W" t; [' @! C/ ]d{(a+bx)* ...

8 t' o% r* |) I4 n5 K  n. L% d. ^1 k
Please note here:

, G. s* Y) D5 E% e5 _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
: M; X5 c' u: K% [, G! p! h并不是   (a+bx)*C(x)   的导数除以   x   的导数。
3 E( }% B$ n9 Y9 S. d( z
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

6 i( _7 x1 v7 i6 K* m
3 Z3 w3 _8 E  S# D: ?* @4 }5 N这样算混淆了微分和导数的概念。

3 d7 r0 ?1 o8 {1 M- Z" D7 B8 Vd(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

) `1 x4 G  N- F; Z7 \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)

[ 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:
* q3 O& ?  Z! \& G这样算混淆了微分和导数的概念。

  \9 e: `  X9 K. d% x* k5 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

6 L* @. _. V9 ~7 c* \If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

' T. g: E& E, M$ a: r
" |. |! h8 p2 w# G" m这样算混淆了微分和导数的概念。
$ |$ ?8 {# Z! y
6 F6 ?. ~0 M: a  I0 Nd(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
5 S1 l" a* a5 g; z- T: i
* z* E5 I) K) dAh, 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


) }1 m3 q- Y. XIf 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 h, I5 p) h7 @! p+ j
3 R" ?8 i# x2 w' Q, Y( f  P9 {2 ph(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
/ k, A! f& Q! g这样算混淆了微分和导数的概念。

2 C- l; C  \5 Y5 }2 U$ Q' Cd(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
; X( J  r# v  {
( O+ r* P9 X& _# k) zAh, you made a mistake: it should be:
" F5 [7 Z% K1 j# |) N
d(f(x)g(x)) ...

! Z8 |5 w) D3 xYes, there should not be " ' " after "f" and "g", it is a typo. :D
# f0 m7 O/ B; \( k9 I# ]+ L
+ G% g6 X1 z$ L$ r* YBut 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
. v5 v  K* ~  m( N3 r6 r
2 n5 r5 Q; g5 aBut 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:

! R+ e# y6 Y( K) ~/ e+ 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。
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:
9 A. x, u- q2 \# A( a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？

& p+ e: e6 F# ?3 F
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。