数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
  Q" x. J+ B: }" _* K
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！
3 v# i$ c6 g. m; u* X% M" A
[ 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:
4 d  \. R& `/ e& L$ A9 i6 x/ J+ ~请教大家一道微分题，多谢！

+ x$ P$ O6 g& Z' E4 o. Q' Pd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
; c- h. X* C7 ~* a$ v9 Q
多谢了！

! W& K- T* m' c5 @. k
: Y* w* q. ^3 D8 R$ _3 ]it is too easy:

d(a+bx)/dx = b
. R- |( A  i3 O0 O# A
so
3 z8 k. f+ i* B0 l% o6 Y$ ~
b=-kc+s

1 {) J5 p% X* D8 _3 f: ^finally:  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:
2 B. T/ Y7 i% ?* @4 |
$ |) F! H2 U+ b7 Y7 od(a+bx)/dx = b

so

b=-kc+s
/ _; B5 s& v' K' v6 P. T
6 {" s8 {. ]5 z+ m4 K; Jfinally:  c= (s-b)/k

' y# v; a' Y) l1 `/ Tthe 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！
* l; Z7 S" a  {
* W0 i" u5 U% }# ^* Yd [(a+bx) ] /dx = -kc +s
8 |) F% }+ ^6 E) d/ Y7 pwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
4 a. u2 e1 o/ I) _: Z  t/ ^% r0 e
多谢了！

[ Last edited by 醉酒 ...

5 y: K# f5 ]6 C& qsorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
' j- O( {2 z& S4 V+ l. M7 X
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.
" {. q2 x# j. k  Q* `8 h2 k
a, b, s, k are  constants? c means c(x) ?

please tell me.

4 w2 V+ \& z( _8 ~5 n对不起，写错了，忘了变量c,应该是
) n5 ~/ S% c+ n+ F8 Ad [(a+bx) c] /dx = -kc +s
* Y4 {9 N/ H# F  H, }. H
多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
( E% ?: T/ h* G6 i. [对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
9 [3 n) f4 v( \0 L2 z
多谢

do you mean it is:
; N& @/ V: G, O( T8 F$ Y
d { (a+bx) C(x)}/dx = -k C(x) + s

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 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:

, p- j, K+ Q. B) Kd { (a+bx) C(x)}/dx = -k C(x) + s

7 t7 w- G1 N9 [. K% zwhere a, b, s are constants, you want to know what is C(x) ?
( {5 N' p- }) E5 ~
4 X% W7 Q9 d* I( Pif 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:
( a" K$ a* y9 ^% R/ K
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

. N9 P4 |( K% K/ g1 A% L7 X2 e
1 G$ F' ~  O0 tprocedure:
3 q: H% X) e3 F
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
# x2 I& n: Y" ~( R1 Hso:
( T. C1 G6 v. ]! ]
" t( e& O: B% r  M" HbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

4 I8 \7 ^& h, s. S4 @$ g3 r) g
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
3 x2 E3 }/ q; o) Xwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
# K& x- ~6 y  S& |, `+ _therefore:
* d0 |8 D' ~- b
" D- |7 |/ [6 B! P0 |6 n7 j{(a+bx)/K} dY(x)/dx=Y(x)

$ i# ^# Z; F8 F0 x# gfrom here, we can get:
; w- y+ J9 o. g/ z/ v
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

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:

8 l0 e) I- m4 i' w-(k+b)C(x)+s = (a+bx)^(k/b+1)
9 n  X5 J' I2 f, U% u5 x' o
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:
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！

; k" y7 k2 B4 a: H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。
& j5 q  }4 W0 }) H4 ~& u; K您说：
4 K% ^+ v$ h( ^# ~1 h7 E. Q) O" W
4 ^" r3 P# y- N- \d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
$ q& {- l# z  G4 W+ {  z8 H
- E, ?% }5 s5 n  N  x9 s0 d也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
+ o, B7 ]8 ~( p但这是不正确的
+ g! g) d9 i$ W. r2 |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：

" S% [4 z, D1 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。
5 T4 _  ^: `" @
2 d+ j5 J. C1 `3 X0 J  S所以您解的有问题。
* J. M1 t4 i- H/ a& u. I) x: ]
[ 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:
! L: L/ k& \% O( e9楼的答案有点问题吧。
您说：

7 u7 j. P; M$ C& F, v  Ud{(a+bx)*C(x)}/dx =-k C(x) + s
4 E+ N% A# I, }7 z# f/ f6 q! z) ?( W/ g! \so:
/ H; w, _) b7 P) i1 qbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

* C( ]$ k& x+ k( n也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
  R6 e7 }/ k" t. l6 f7 G但这是不正确的
% n. ]& L# }0 R4 ^+ d- Jd{(a+bx)* ...

9 B7 s. C! b8 k& D1 F' @& uPlease note here:
5 W" Q$ ]% z4 w4 q/ c5 m2 O# N$ l
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
. g! \; ?& B- yd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
8 d: v. m' U7 s. C并不是   (a+bx)*C(x)   的导数除以   x   的导数。
2 Y1 m1 b1 d; j# H# E7 [. n+ w' 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
....
: k8 ^0 g9 g2 t4 F- o9 hd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

这样算混淆了微分和导数的概念。
$ E/ P0 r! x9 P3 Y4 E; H
% Z4 b$ r+ Z7 d7 E$ ]1 P, 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

/ x  y% t+ B5 y9 B: `4 m+ p, EIf 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" H" U' I$ i; _0 F4 A[ 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:
) U7 S- T5 l1 e8 @" u; E% w4 S" v这样算混淆了微分和导数的概念。
( y1 b' z$ h3 o+ t2 b/ a, W5 r
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

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

# c- q6 c5 K7 H0 N% W$ o
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
: w1 X: _1 x1 X; v! s# g+ c6 |
! V9 o9 W$ r! A2 q( eAh, you made a mistake: it should be:

0 R- O/ B: h& X4 r* Fd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
) j+ {- X9 J7 R" O- z
# _# j4 y2 z8 Z: t- h  D8 u
# W2 ]3 @6 y& t5 a. o- U$ 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)
! q8 ~; o9 f* f1 O' Q- S/ y
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

9 ~2 l) Q5 S( N2 F) G& F. Kh(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。

/ A0 z% @7 [9 G3 Q4 }* b2 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

- v6 a( K3 w' J1 n; b0 x9 X, kAh, you made a mistake: it should be:

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

4 c+ L$ ?9 X; e; p0 m) L# s' C
- t0 k! I  _; N0 t2 f/ L; `4 R! nYes, there should not be " ' " after "f" and "g", it is a typo. :D
3 U: p: u! V! [8 \1 Z7 `% o. }  N( R
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:
  ?  ~: }5 o( I" I3 j% tYes, there should not be " ' " after "f" and "g", it is a typo. :D
5 z; O# c' H8 G, w0 Y! o+ V
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:

8 T7 l& W8 L6 l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 ...

# O9 ^, B9 V% O  L! X5 ^
3 C9 ~1 N/ q6 x  _5 q" E- u1 E" 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:
' s# K0 }9 b0 o8 [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？

; m1 y3 h, d) E) z% y  `
! v) k" a/ O3 S( f, m' r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。