数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

( n6 M8 U/ K, N+ xd [(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 y1 Q# d3 O2 _! F& y9 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:
0 `- y& c$ H* w# ^6 Y请教大家一道微分题，多谢！

  @& q2 y. V$ g9 Ed [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
2 m0 j4 U5 J4 j
多谢了！

2 }' E4 c+ S* p$ Yit is too easy:
( O: z$ D: T( R7 Q2 ?$ B. t5 j7 e& A
d(a+bx)/dx = b
8 y+ \4 e0 U: V. x* z, F
so

b=-kc+s

finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
7 ~3 e1 O1 i! ]! G+ Sit is too easy:
  Z# D+ m2 V; I9 B6 D
4 z' z9 [8 b5 h" b+ [d(a+bx)/dx = b
. U# `) n& a1 W4 a2 {: c& R) X
7 ~% t5 Y) W3 O# B4 Lso

b=-kc+s
# h2 `1 K1 ]$ ?9 I5 V- X- V
finally:  c= (s-b)/k

$ Y) J( j7 l8 H5 U1 X& ?8 jthe 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！
+ p- F0 `( O7 V( t, _0 P
" h2 s4 K5 b, Gd [(a+bx) ] /dx = -kc +s
  k0 x* X4 C# t5 A1 B( kwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

9 R$ [. {( z9 ]! p[ Last edited by 醉酒 ...

4 a  T, |- d% d) Hsorry, did you post another question? your statement is still not clear.

: ?  t0 p: p  C7 R5 B$ Pa, 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:
- i" P0 i" M) Isorry, did you post another question? your statement is still not clear.
: t! t; Y) U! X1 n8 T' e( W% @/ B
+ v9 U7 O% G* `6 t. u1 z/ ra, b, s, k are  constants? c means c(x) ?

* s: b' T  {; Q& dplease tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
1 d* l9 q( ]. Y
+ m0 D( h0 _1 P. ^& h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
d [(a+bx) c] /dx = -kc +s

多谢

9 c3 I( C+ g% h& ]( y- [do you mean it is:

/ V" X8 P* E. a" Bd { (a+bx) C(x)}/dx = -k C(x) + s

" z/ v! \( x8 \% n/ m( A% \where a, b, s are constants, you want to know what is C(x) ?

2 g2 l6 U3 B# r! z5 k7 u+ K" E5 pif 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:
+ U0 g) _" A6 w, O1 G
d { (a+bx) C(x)}/dx = -k C(x) + s
3 N9 j+ ~, y; e3 C" Z: j1 c3 J/ [6 z
where a, b, s are constants, you want to know what is C(x) ?
; ~2 s; c. o& ]4 B5 o1 P8 m6 U
$ g! z' h) {/ x+ o; aif so, this needs to solve the differential equation. please comfirm it fi ...

' C8 `  m) c8 `6 f( f% `
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:
9 B: }' _4 _: _3 K+ ^  k
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
8 Y' F) M$ D+ A. m
  }2 J/ X5 G- q( @# p
procedure:

9 d4 F, a6 y* QFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 j; \% {1 d' H) F4 u8 Q' Pso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
% n3 }3 Y  s! w/ i3 @+ Y' \4 Y  i4 ji.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 g9 D+ l0 M) c5 o1 M5 l# e+ `6 l

  Z9 Z1 N& U# K5 P1 @; [% q, ]introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
7 q2 b! M7 w- P1 f6 twhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

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

from here, we can get:
5 H# D3 u9 d& F; d' U
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)
; A* u5 X. k' y2 Y' t- I4 S
5 I: P( w& C' ~3 vthis means:  Y(x) = (a+bx)^(K/b)
, f/ L6 s  I1 _% bby using early transform, we can have:

0 d; p% T; b& ~8 n% d3 s4 _* E-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

  Z' K9 w: e- D! uC(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！

5 O- H8 ?/ d7 q. N: ?2 b' W
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：

* n1 G+ N5 t$ c! b5 od{(a+bx)*C(x)}/dx =-k C(x) + s
" a( o8 q/ x* [) o* \: dso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

- X4 R: c  }- P$ F3 t: H: W/ I. R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
: X0 z/ K" r9 T7 y2 M, F并不是   (a+bx)*C(x)   的导数除以   x   的导数。
; Z, `& o1 S/ D8 l7 p1 Z+ }7 m  B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：

* P8 V0 I7 J- |2 \4 V2 R! C+ I8 Q, D7 n  Bd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
6 y7 m" Y- |8 a7 D$ W9 Z- w6 @最右边是x的平方。
" H. {7 K! t* G: i- k0 }1 c
所以您解的有问题。
1 P$ X) Q* E' O& N; m
/ c% ^4 K+ L2 z/ K9 {( e* r' T* P[ 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:
# c7 j9 G6 [/ O0 @9楼的答案有点问题吧。
+ s& ?0 W- t- H; X( T( c6 c8 G您说：
! Y! @7 A6 E: F" `1 b
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! V0 _0 B" l% n. y# H0 ]+ ?
2 h! T0 N) A+ L3 D4 C: l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
! X" D' v9 \( y+ Ad{(a+bx)* ...

7 [" p1 V6 E8 D* ?3 x5 N, rPlease note here:
9 D0 v0 B) K$ O6 N9 d: J6 Z$ H% B
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
6 q6 R6 Y+ V4 t3 B$ tit 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
并不是   (a+bx)*C(x)   的导数除以   x   的导数。

, `/ G, B" z7 x% c7 Ano "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
....
  B/ Y2 L" _7 L0 a7 q0 L2 U8 T2 qd{(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。
/ ?' X/ b0 X9 ~8 u
5 O- d* z9 [7 E3 R# B6 Q3 Z: 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

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)
  n3 q; Y+ {+ {- T+ K" c! q* t
/ M4 Q7 h4 b" }9 W3 Y8 @9 t9 ~0 b[ 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:
' b/ n9 \$ w0 |3 b; z. ?这样算混淆了微分和导数的概念。

1 M- }, I+ A9 p5 rd(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
2 B" a3 B$ U/ b2 ]5 j) c" Q7 S' p0 t
- {) z4 t" e! L. m$ JIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

, r' ^0 B/ y7 i( u4 B! e4 b
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:
& F; k' k8 Y' W- @
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


9 \6 F, K  z" Q% Y/ M6 zIf 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)
" A; M# S5 }! g2 B
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

) N4 `) {. L* J! }' D" xh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
) s6 I. a7 J) ^" ^这样算混淆了微分和导数的概念。
" c: W' G" n0 k8 M/ O
2 V0 H* p; w2 c4 ?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

2 Z  G& K# `8 r0 d- M: tAh, you made a mistake: it should be:
' S8 U; }" T$ S, F. S7 b/ L
d(f(x)g(x)) ...

1 d1 b/ r' u2 W2 S7 O* ~" BYes, there should not be " ' " after "f" and "g", it is a typo. :D
4 _* r/ M, T+ h5 N0 f' ^. b3 L
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:
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.

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:
2 Q$ U5 p& k; u0 _
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 ...

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:
9 c3 j, F7 E* Q5 n1 w- ~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。