数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
  i, `+ d& {+ h& e) F/ H
d [(a+bx) ] /dx = -kc +s
. g6 _* N, l. s5 zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

+ I+ J- }6 g+ P0 S3 p; r多谢了！

' H, I! j; j5 r% x[ 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！
6 \; b% B7 Q0 `7 s! z1 H
d [(a+bx) ] /dx = -kc +s
4 w  L4 a! w1 E- U8 |# ?  fwhere: only x and c are unknown, others are all known,  requiire c = ?
5 D6 ?$ f+ ?+ T1 f9 ]
多谢了！

) w# Z) c: [2 U( \! U
it is too easy:
6 z! D* j; o! r% Z
d(a+bx)/dx = b

so

' A, r0 c" {3 {6 r5 V: j9 Y8 ~. Sb=-kc+s

( b9 m8 o6 O, Z* \' o' l9 ]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:

: b0 B- m- Y1 Kd(a+bx)/dx = b
5 O5 }0 h  m- L8 @5 K- n6 z
so

" C* o2 Z/ D* J0 u5 z3 s- z$ Ub=-kc+s
$ P4 t2 q8 J$ f* z2 ]* H3 r
finally:  c= (s-b)/k

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:
+ x/ ]6 J  F$ p! l( z+ |3 M请教大家一道微分题，多谢！
* c0 s0 L4 r. x! A5 C" _
" m( O- Y+ l7 D/ I1 K, h8 D5 Td [(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?
: t. a% t8 f+ C/ o- v1 R
多谢了！

[ Last edited by 醉酒 ...

! p" N# ~8 F: k  o
2 d  m  {) B  r! Bsorry, did you post another question? your statement is still not clear.
- F8 E# r; E: L0 L" T$ F* P0 e
  e8 E7 j6 r, c0 ha, b, s, k are  constants? c means c(x) ?

9 i8 d2 ^" F0 C% _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.
6 D. Y* \5 p6 x0 }# W: V/ ]
' C) D" x+ ^3 ma, b, s, k are  constants? c means c(x) ?

8 \7 P! K9 ?: k! i1 M  a& P% Kplease tell me.

对不起，写错了，忘了变量c,应该是
  O% W2 t, H2 U0 g& B, E5 a5 u4 Bd [(a+bx) c] /dx = -kc +s
- m0 l/ D/ {) q5 T* \% ]
2 R, t$ [% C/ ^; d+ K) v5 N: F3 ^多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
3 I/ D4 q; Z  K( y% @对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
& h# h1 D8 d1 Y9 J" g, |
多谢

do you mean it is:

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:
: j% q$ s" M0 n# vdo you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
+ w5 g* w# f" z: [
, T( ]7 y% s% Z6 i# q: l) Qwhere 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 ...

4 F$ j6 T2 [+ E  U
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:
+ t. [8 o) D9 H: M
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
* ^- b- o; |. p( Z. L

procedure:
$ ^- A" f/ D9 Q& F
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
4 Q1 }7 }' q6 i' Y# T
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
5 D3 r; v2 n4 B" U: c. I( K1 V$ E- ?i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


% Q8 ^' l3 M: `introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
& \( }' v1 c# U- a* x, jwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
' A2 v3 h5 c7 w9 K, Ztherefore:

) f8 i, J& n; V7 u/ H{(a+bx)/K} dY(x)/dx=Y(x)
& @) p$ d+ O- z1 `  n
; ^/ n1 n) s8 K. g2 Dfrom here, we can get:
+ s! i4 Y6 C: N9 P
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

/ H$ {0 u* t: x; H7 Qso that:   ln Y(x) =( K/b) ln(a+bx)
( ?# g3 F! ]9 V9 |  \$ P
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
& a- l) x% i0 V9 n9 p/ O
! X& z) I+ A. q8 D8 i-(k+b)C(x)+s = (a+bx)^(k/b+1)
* f* u/ x; S( O- I9 ^
/ s+ K5 i7 g0 s& \( u& v; L1 ?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:
! G& d, H  `/ q# f3 j. 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！

) Z8 \; q: [0 ]$ I6 V9 [
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：
* ^6 V8 G0 q+ Z* n" T2 B
9 W7 F, ?( A5 Z) ~0 h9 ?. Ed{(a+bx)*C(x)}/dx =-k C(x) + s
so:
, s8 s% Z" S- t9 h7 d+ KbC(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
/ `# j5 A7 V0 y但这是不正确的
  Y- i+ a  U/ d6 t0 q) s7 W1 u6 ld{(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 |& P9 O, n  W- q5 @, z( J) H; I8 ~
d{(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。

所以您解的有问题。
" R2 Z( r8 W6 S2 a
. s; b# }4 z+ Q0 K- Z1 K[ 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:
$ E( S$ @4 N1 \; r1 T: P3 R/ F9楼的答案有点问题吧。
您说：

1 @( h; n# Y0 \5 `- p/ Ld{(a+bx)*C(x)}/dx =-k C(x) + s
( a# p: P- b& eso:
7 l. ?2 ?, ~& a' R/ f" f7 abC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

% ?& I& {2 B% T! G也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
  x. r/ r! G& U& }, e( u但这是不正确的
  C) \0 n3 o; Z6 P: u3 D+ G+ P% F6 `d{(a+bx)* ...

Please note here:
6 c3 r/ y! l4 M+ t2 b
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
* f2 W9 L# `3 F" \3 V$ J: Z; {% q/ Hd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
& _# Y5 n: k5 A% ~并不是   (a+bx)*C(x)   的导数除以   x   的导数。

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

3 Q5 j/ s) a* m2 P
这样算混淆了微分和导数的概念。
2 V7 Q. `% `9 ]2 e% x4 f; @" G6 T0 y
. `3 D: h: `4 ]. x1 hd(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
' ^  V, T; e" H
7 Z! c4 C7 Q' SIf 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)
8 |& p! V4 |% a% D, |
  }- G) G% U  M' m% q2 e[ 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:
9 c; R$ u3 H* @% N( T- \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 T: [0 ]! b6 F7 B0 X& \! q7 ]If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

. p$ R( ?7 X% d( ~
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

Ah, you made a mistake: it should be:
+ z: ^9 B4 q8 j! t- U' Z
1 @2 m8 S" ?7 ~7 n4 od(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
% G' P# G% O9 O( u& f1 C8 Q" I1 C
+ C$ ?* r+ h1 A$ e
# V% Y9 Q, f* d5 U* x3 nIf 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  H4 g( S& H1 a6 K# a# a3 t: Bfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
, q: F- Y, f( _/ Z% B! S
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。

( t& p/ `; Y! B9 Y1 d1 Zd(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 Q+ v% ?: @( q% ?- \. W) z
Ah, you made a mistake: it should be:
' U  z, E, `  ]! C: D( ]; e' E/ B
d(f(x)g(x)) ...

Yes, there should not be " ' " after "f" and "g", it is a typo. :D
" Y& L9 B2 N/ m6 F9 F
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:
8 x- @0 }7 n- B4 uYes, 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.

) `9 P/ A( J: B- Y& b7 @
- r, [% B2 W1 n: {! SThanks, 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:

4 z" d7 ^' m0 w9 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。
3 A, Q, W2 k! ]* c: @! ]  j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 ...

  f  ~. P% C% B
( L5 T  {/ H. z+ {  ]- ~: S4 Z. c4 m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？

* Y7 |; |) y, ^$ N; t. s; q6 E. ^
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。