数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

; A, R5 Z$ l1 pd [(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?

3 {/ C$ X* w- m1 S, v! g多谢了！

( G6 [7 |4 c3 ]  @/ R. d[ 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:
" \) q- o9 ?3 a4 N请教大家一道微分题，多谢！
% [' ^& @* D6 r3 T! S' y0 Q% _' Z
1 e$ ]9 }/ @# M/ ]/ fd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

$ D5 V! R- Y1 e; V$ h- {多谢了！

* ]# K+ l0 X2 [* y: v% o
it is too easy:
6 y4 ^& q" }  o, f5 E
d(a+bx)/dx = b

8 m9 v1 D1 b+ G1 z% _; ^so

% [2 K. e+ a+ |b=-kc+s
, D* J. n, H0 Q7 L
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
" g6 T2 |  `/ o  sit is too easy:
- c4 @5 c5 i8 u+ |3 `! d
d(a+bx)/dx = b

so
) b' ^9 N. H$ y
) `) o3 e8 l6 N7 y7 A* h0 Jb=-kc+s
9 C6 a4 O' b6 r' U7 t# G
finally:  c= (s-b)/k

) C" R9 U3 h1 T. V) a. Q2 O7 k6 Y
the 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:
6 y9 t8 |+ x2 i- S8 F4 _请教大家一道微分题，多谢！
" u# E# c; W) s2 r+ ]+ Q, a
  t  ]5 S: Q4 M( d" Qd [(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?
# h. I4 ]8 h) @  Q& s9 a
多谢了！
  V* y6 [& D3 Y# P% C6 t7 u$ ]
1 f, `5 `* f1 C* t9 X[ Last edited by 醉酒 ...

$ R# B$ q5 J% B$ b) |; \9 G" ?sorry, did you post another question? your statement is still not clear.
+ O! |( t: A3 Z. z) z0 r9 g3 U
a, b, s, k are  constants? c means c(x) ?

0 r/ ~! \/ r# z' \# W& k" Oplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
* L  y* ^; m+ w/ R  Lsorry, did you post another question? your statement is still not clear.
5 f9 x; M) f. W) I2 {
a, b, s, k are  constants? c means c(x) ?

please tell me.

9 i, j- X, H; N对不起，写错了，忘了变量c,应该是
7 ^. O6 \1 ~9 t4 `7 ?2 `7 W* [( @. T5 T/ {d [(a+bx) c] /dx = -kc +s

多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
/ t4 K0 e" W" z+ F! T/ Y对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

do you mean it is:

/ L# a* {2 j. _7 z2 l& rd { (a+bx) C(x)}/dx = -k C(x) + s
7 L1 U: x+ b( v3 L  M- T( y
3 e! k  I$ K8 W3 W: iwhere a, b, s are constants, you want to know what is C(x) ?
2 M# l: x8 U8 n* d7 q; ]+ m
. W" Q' e" d3 @7 zif 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:
  {9 F  w5 Y, j8 pdo you mean it is:
9 V& _+ P9 j3 q' f: U) }
/ {' [* D/ ~9 ?& K0 |d { (a+bx) C(x)}/dx = -k C(x) + s

/ \! L  N9 W  iwhere a, b, s are constants, you want to know what is C(x) ?
& K+ `% B+ }1 W5 A
% Z" j% q: N7 M8 a1 {* D. jif so, this needs to solve the differential equation. please comfirm it fi ...

6 T$ j. S( Z* E, @) ~
5 y2 N" {- g+ S2 H2 C8 g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:

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

十年移民路_

this answer is the good one.

* C& ]' P( r. m* j
procedure:

& f, g8 k+ O! g; u5 s6 gFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
6 b6 ?% B1 ?  M3 y* S" _8 q7 e1 q9 }  Cso:
# ^( x+ U: ?# U( w  x: y6 U/ g
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
) F+ O# T" U) z) _+ Ci.e.
' }# R- s8 C: ?5 B4 n
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
0 ~2 e6 B' l/ _
0 f6 A3 k0 @. N0 m: q0 i% l) f
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:

{(a+bx)/K} dY(x)/dx=Y(x)

$ }( r! V) u. _5 ~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)
% N! K, l3 @* @, ^2 z2 m
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:
* a9 t3 B; ~3 @, ]2 a7 N7 B# B
7 ~' }% S$ F, E+ O; D" F-(k+b)C(x)+s = (a+bx)^(k/b+1)

4 i, _5 H3 r- d* Q/ f( q1 \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！

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。
您说：

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
但这是不正确的
3 R+ S4 s1 K3 vd{(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* d4 {9 A9 n: f% q1 R  |8 @8 T' P2 J0 d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。
: O3 M. p) ~/ r+ e# g6 q3 J- o' p因此：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
& h$ I1 W9 ~' ^" C* S$ }最右边是x的平方。
6 o( O" g2 k; A* [: v$ M
所以您解的有问题。
/ L9 i( w( L& C1 f
" y  r0 f- |. k/ z9 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:
9楼的答案有点问题吧。
  T" I8 G8 p  S' ]  O+ f3 T+ P/ m您说：

2 v, j% C7 p2 ]: y: n1 @d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

, ]2 C/ u$ H; l  J, h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
* C) |5 D. L% K# u% J% Gd{(a+bx)* ...

Please note here:
9 P4 V3 J2 m; @' \  V5 l
& ^# {9 z" R5 e9 G/ N% v6 Hd{(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
- I3 p1 Q7 t2 jd{(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。
' Y' V0 H0 X# j7 A
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
....
6 q( \$ P  W: g: O1 k9 fd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

  n5 C% n7 O( @1 E; q- `6 {
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

) l. F- B5 D* k" d6 g5 W3 {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)
- y/ ?6 {+ w; W" K
[ 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:
2 K( c# B% v6 ^* `: w+ @这样算混淆了微分和导数的概念。
8 Y0 ?. H  B4 f: 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
" J& G/ `9 E. B0 ~: I
+ B' B/ I& I, [/ j9 B; ?) a; QIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

: e3 a0 a, p* k+ r2 s这样算混淆了微分和导数的概念。

4 M' j  B" ^2 ~8 u0 id(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

Ah, you made a mistake: it should be:
- E. V0 }$ }/ M. l1 h" d6 n) U
0 r0 V! p1 }8 Hd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
5 U; [% Q( v% i# h- d& h
+ R2 x( N/ n* h1 J( H" v/ _
, n, _% q# y9 C- TIf 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)
  S& \, b; d+ w, N1 K- [* q
# }* N* T0 r" d3 f. {8 I# wfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
: j9 ?2 _& D, [+ g" t" Z0 |% m# o8 K
3 F3 W& a9 p! Z$ O) b( c) Q  ~h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
9 S% n/ z0 ?" |7 Q) h1 a% }这样算混淆了微分和导数的概念。
( x3 z( M0 X" }7 e
/ P, g# @# |# Wd(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

Ah, you made a mistake: it should be:

* w. z& L3 o: r6 \! r; ?( Ad(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:
. M, z/ v, t; HYes, 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.

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:
  j# L4 n0 z* X, B$ ]
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 ...

9 e; @% F- @4 c7 {  e
# l  e9 G  V! U: S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？

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。