数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

7 O6 t5 C; u- a  g1 T  c) J8 gd [(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?

% W8 u2 W- l; ^) W7 R. [多谢了！
& w5 a2 b6 S* w/ i- g5 B+ e8 \
[ 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！
+ \( U  E& Q$ O' j6 u& _4 q
* @& J4 _8 {3 e7 a% Z: }/ Wd [(a+bx) ] /dx = -kc +s
" p8 p  X* q& `+ d! J0 ~4 F, o! Swhere: only x and c are unknown, others are all known,  requiire c = ?
/ c+ C0 c  z& }
( X) q! B! ?$ p多谢了！

3 D8 l' G* h" F. A( U# Q
it is too easy:

$ n, h3 x- c. `( O0 ?' d* U2 Gd(a+bx)/dx = b
) G4 z: }$ P" |/ f6 N$ `: r
1 t( P  {& B  Lso
% ]5 E( q1 Y# @5 t! M$ f7 R" j
% }) U1 ^% j0 n! ib=-kc+s

" }0 |8 d) z( zfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
8 S4 I2 n) r0 v% {! {8 Nit is too easy:

d(a+bx)/dx = b
. L( a2 ^+ ^5 g1 u- }+ D$ f
+ k* u4 h; L% u( tso

b=-kc+s

: y4 t7 I& A, ]' Rfinally:  c= (s-b)/k

6 D& j" H( x( f: f/ [$ D5 }( }7 _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:
# Y) Z5 ^2 b4 N请教大家一道微分题，多谢！

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！

) A; D3 }, K% K7 ]4 Z  d[ Last edited by 醉酒 ...

: z; b2 w! \1 r. q# g2 i
  v/ L5 }( v1 i1 A3 X$ Vsorry, did you post another question? your statement is still not clear.
; C& Z. y. J; X( V6 U
; j+ T8 S- F5 G% _a, b, s, k are  constants? c means c(x) ?

& E3 J3 X. w1 uplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
. d1 g- D% i5 y9 j% _. T! }4 Fsorry, did you post another question? your statement is still not clear.
& }- C4 }3 M; P' |! z, L: T: M
# C6 W  Y1 z1 E5 d, R- Y# a0 |a, b, s, k are  constants? c means c(x) ?

3 m# x4 q- `) H! R! P. T; o2 dplease tell me.

对不起，写错了，忘了变量c,应该是
, |8 c0 d7 c3 j4 M% K# O; wd [(a+bx) c] /dx = -kc +s
. ~$ j- L# r# k  ?$ ~' n! @
! F( Y  Q- b$ A, D0 r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
* F- l' {! y- m+ q, N5 _d [(a+bx) c] /dx = -kc +s
7 p, [- `$ z1 J; j
多谢

$ q% i+ t5 b" I5 Qdo you mean it is:
  Q- F  m& [0 U6 F/ Q( ?, O
6 j( m6 a" b6 q4 Bd { (a+bx) C(x)}/dx = -k C(x) + s

2 M( c# R2 n. ^* \8 J8 Y3 Lwhere 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:
' S+ _" l" b+ ?' f5 j2 fdo you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
0 q- R3 t$ Q7 {6 s
7 H( j6 a, g8 o3 Nwhere a, b, s are constants, you want to know what is C(x) ?
+ m! \: E! K& D- ^. @( x
if so, this needs to solve the differential equation. please comfirm it fi ...

" k' }8 U) ~$ @0 _# I% n4 ~( J' i
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:
: h9 }! K  T" D" b
0 J. n/ M& E/ B- r/ J9 L; BC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

3 H3 g1 v9 u9 L5 w' A( \, ]
procedure:
) i1 z: N4 c( n9 y
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

# J7 L% q. _0 t# V7 U5 SbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
6 B& Y: ~% D, h- O1 c" j' M" L' H2 {+ T
* E# y! w5 V. u' F3 D(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
& f2 H5 W8 I8 j0 ^6 p& v! j0 Q% Hwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
1 \& Y8 d1 U+ I0 }, D/ o
5 z+ u, v' m! K' }& f2 X+ Ufrom here, we can get:
/ r0 T: g* s7 N, F( h
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
4 z$ [4 _2 {  t* k+ M
so that:   ln Y(x) =( K/b) ln(a+bx)

0 G9 p3 \1 n3 `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)

  ]3 ]" g  c  X; k9 D# n: X) M( wfinally:

  J0 D; Z0 _( h7 SC(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！

! y$ |/ n5 X6 P6 d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。
$ ?( u% {! S6 l6 {& Z( a& W; y+ t4 e您说：
0 P( S, ?1 W: U) i' k( l
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
7 `( o5 u' S9 d9 D6 Q7 o. kd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
# K, a6 v/ m5 w: Q2 o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：
! C; u. T, N* u: J
. s) P/ ^7 a8 K; Y- [+ f# `d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
5 ^4 m0 W$ p0 U$ J7 }0 z3 _最右边是x的平方。

所以您解的有问题。
! B0 N9 b4 }1 D7 X! w( h( B
" H& i( U0 N- j" l/ K7 G" _4 f) D3 [[ 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。
- ~: A7 \8 l( n6 I* |: t+ M您说：
7 ^  x5 w/ Y2 {3 P8 y6 i: h
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 N% k: p$ c+ G" ]# S- J: Y
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
' i; P4 w; g  _但这是不正确的
- {  X  i7 X6 I: y5 R& I8 c5 Z# }d{(a+bx)* ...

Please note here:
- H' c8 \, W+ y6 v2 v
2 ^3 x& w+ Q' \9 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
d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
2 k8 k, C  ^0 _! s/ D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
....
" e# o" X4 x4 _3 x$ d! dd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

1 \; w6 t1 s* M! J. `  a- m
' w6 U) U8 D( o' A* D$ x这样算混淆了微分和导数的概念。
2 v& e, }: M# ]: `5 D
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
3 `3 S2 k' Z( b+ Z8 Q& A, U
- T) o! z/ A$ kIf 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:
这样算混淆了微分和导数的概念。

- w1 k! a1 O, d6 t5 xd(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
9 u/ I- R% i: K) v" l+ k. I* p( A% Z
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

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

% i$ j/ X% b4 g* [/ ?0 L5 l* j$ xAh, you made a mistake: it should be:
" Z- I" n* z6 r9 F4 U4 `! N9 f$ @' T
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
2 C% {$ B1 x: S

# q$ x7 c, ^# o5 \2 z' U. l( X* dIf 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)

) i5 A: J5 k! [. H& o. d: |* Bfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
. J: d% @8 N/ @4 k: p5 ]. I# u3 ^
h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
6 ^+ E2 n& o9 @3 Q9 c: z4 c! Z! l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
$ _9 _! g3 a) x( V% B1 j9 F1 |
Ah, you made a mistake: it should be:
( S' c  j1 S) S6 n0 _
d(f(x)g(x)) ...

$ d2 |- P. p* N! o
6 q8 ~/ L, [$ ~0 }4 t5 z) q* oYes, there should not be " ' " after "f" and "g", it is a typo. :D
& y5 t" l' R) t5 s! t3 A) G- U
$ Z6 l7 o4 ]' z2 O$ MBut 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:
, S/ C' a' G6 D+ w( ~7 b2 MYes, there should not be " ' " after "f" and "g", it is a typo. :D
! c3 h( l- u) t8 u, e  y
; G8 @- A  ?4 K2 X" q- rBut 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:
0 y0 H5 [, F# G( a2 d9 Y6 s2 p* S& `
  V+ E/ C* B1 ^$ g, _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 ...

2 `  m; x# m8 P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？

6 A9 d, e7 U* m" 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。