数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
$ I) e+ @: V% I# c: s6 u
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！
# ^+ o7 ]: Z3 f- g* C
0 N% V% Q7 Z+ X/ Z3 e" G[ 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:
# ]9 _# D( g: m- h2 |请教大家一道微分题，多谢！
( Z& A% x* B% e( l' l0 I
  V% X* t5 e6 ^, d1 fd [(a+bx) ] /dx = -kc +s
; a& o, _* H3 F. M2 U, q# ^where: only x and c are unknown, others are all known,  requiire c = ?
) A# u1 X! _; E8 t+ I) z
4 P3 h9 r: C" D( ^7 d; o2 M! ?! m多谢了！

& Q  S2 y9 h9 e
+ D+ z  X8 p1 g$ l* H- Z+ ~it is too easy:
7 m# ^" G1 l. C6 p1 ^+ M; N
d(a+bx)/dx = b

so
- J: b- Q* Q( z9 J+ c- ], |
b=-kc+s

% M" Z0 v8 Y. i+ r; U) Z8 V( H8 afinally:  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:

d(a+bx)/dx = b

so

b=-kc+s
: J" B  G2 W* J5 Q9 S# C# i- l
finally:  c= (s-b)/k

6 }) Z2 q3 H+ [7 I
  r, b$ b8 h  nthe 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！
+ ~( m6 h: c; M& }
  g8 |# t# N6 g. Vd [(a+bx) ] /dx = -kc +s
0 N; C7 }* b, |where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
- g6 Q6 d. }1 G, W
[ Last edited by 醉酒 ...

9 X* j! r! u! w" b
sorry, did you post another question? your statement is still not clear.
: v/ T7 S  t! F1 B; G9 S
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:
7 w# @) z& K- _' N; Z& \1 bsorry, did you post another question? your statement is still not clear.
7 k# g6 R, J$ v5 N4 F8 r( m; k
a, b, s, k are  constants? c means c(x) ?

please tell me.

$ D2 b: B! u( Z6 Y" t; ?' C% v对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

) {# `* N; `- N7 t5 W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
3 Y: `3 ?& c6 n* O( U% id [(a+bx) c] /dx = -kc +s
5 ~$ ~7 o8 g; s# X3 B
多谢

do you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s

$ y; L; X7 {1 t/ pwhere a, b, s are constants, you want to know what is C(x) ?
6 V' P' [) J3 q% f' ]* W( A% }/ F
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:

d { (a+bx) C(x)}/dx = -k C(x) + s

; |& g: o# M4 W8 W" p' F2 L& g" dwhere a, b, s are constants, you want to know what is C(x) ?
' y9 L  x* |$ h+ M6 c$ \- G' s
if so, this needs to solve the differential equation. please comfirm it fi ...

. Q/ @5 G4 R9 w- B+ }: S# M
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:
! V0 x) ~; e2 j+ a- s+ Z
8 ~! v3 [8 l* Y# M8 b6 k2 ^9 |C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


. O- v" C) a4 B! }+ X( d4 @* Nprocedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

5 W+ X' l5 V4 u9 c& R) v9 lbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
+ P& G% w2 V3 ~' l% h4 @6 F
+ r( w) t7 L1 H8 q# S0 M+ z; a
9 x, o* d" b  \1 B# v- M; a" }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:

7 b* g. p+ _/ X- o: {9 B, E{(a+bx)/K} dY(x)/dx=Y(x)

7 X( N# A1 a- R) {' U9 _from here, we can get:
6 U+ y6 g  o2 _( h0 P. B& C
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
  H: T$ x7 ^! G' K; K
so that:   ln Y(x) =( K/b) ln(a+bx)
3 R1 B9 J0 S$ D' N$ \* |& Z) u
this means:  Y(x) = (a+bx)^(K/b)
/ p+ f/ G  {. q$ u. L2 B# lby using early transform, we can have:
  B; y- o6 L5 y+ c% |6 m7 e% R) p5 O
: s' \  O$ D6 b% o-(k+b)C(x)+s = (a+bx)^(k/b+1)

- t& }2 D* ?3 ~finally:
& `3 I0 G7 w1 o/ H. d9 g
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！

0 t& y9 _# J. ~& S. q3 P: h8 n+ t
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。
% @2 j2 x* }7 P6 Y  {% B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。
; K2 S! b! u0 r8 {/ A' O% \6 L, t您说：

  f' [3 Y: ]8 W2 qd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
- [/ a4 r( {2 q6 A9 ybC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
3 r0 g/ N" y# z. U+ D$ t" P7 j4 `& u
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
$ c) Z7 p* m' p5 q  N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：

% }2 s9 l9 X6 `  Kd{(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。

" w$ k& T+ s) f! {% j6 B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。
4 r9 j% @  d# V0 b/ z您说：
& ~1 U7 o! U$ H+ V: U: v
d{(a+bx)*C(x)}/dx =-k C(x) + s
$ C* J% x: u& u$ L2 W3 G9 b, t) Fso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

2 X  I$ Z5 J/ @: |3 W也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
( l: I2 _# A6 g: o3 t% a6 h但这是不正确的
d{(a+bx)* ...

Please note here:

2 Z- I' U2 x3 k7 p4 cd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
5 G) y+ @5 u. C, R# Q# M5 Sit 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
( a5 }# r$ i9 U# K并不是   (a+bx)*C(x)   的导数除以   x   的导数。

% l+ h% z' R! @& N3 Y7 Tno "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

1 o& j1 E2 X9 N, s, d5 T
$ k- p$ \( v) C$ Z4 h这样算混淆了微分和导数的概念。
2 l  c7 `# I. n; a- Z5 \, T5 [* U" p/ 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

" I( V( y6 B5 BIf 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)

, \: m7 J' Q- z" n* ^, {3 z[ 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。
( r3 Y5 z& X! s8 i3 a
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

* G  \" B) |/ O- n; C) ?If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

  l$ T5 p* V4 @* [1 o5 t+ E
这样算混淆了微分和导数的概念。

# ]7 J! i/ r% F: C1 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

- m* M1 i! x0 v9 \Ah, you made a mistake: it should be:
. I9 a& l5 `5 T. N" N+ y
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
5 i- m( Y& J" ~3 {7 U8 _( N; v9 k

! C1 h& q2 d6 H5 Q; b! `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)

3 ?3 E, W/ e2 `+ d1 U" G+ Wfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
! Z8 ~. E6 v% F$ c* p8 t7 v
h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
; \: l, f: X% ^) f' Q4 R$ J+ a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
! q. p% F5 ^1 P9 q8 E
- D3 K1 e2 e6 r* Z; F9 c2 X8 zAh, you made a mistake: it should be:

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

; S1 f) k) c  ?1 |; I2 n
Yes, there should not be " ' " after "f" and "g", it is a typo. :D

7 |' S1 W& c( A5 T. S* s2 n1 `+ uBut 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
1 F) S. b4 `9 ~! J- |+ B# {% t# s
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

( y6 c+ `2 A9 P' M6 C% Q! A5 h! j* p
4 q) S# @* d/ k  M% I; GThanks, 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:
0 k4 b+ i/ b1 m; E/ \) V' C. t: _) s
* N" ^# d( K, u$ B3 t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 ...

- i/ _! K$ Q, T  h
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？

1 |0 I1 ?! J. d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。