数学高手请进，多谢！

醉酒当歌

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！
0 s5 G. Z) r7 w" ?7 C0 M2 g4 N
) K* m, d& T' P$ J  U[ 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！

4 g+ O  a9 m, p2 k! n# X* v9 Ld [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

it is too easy:
  H8 L* W' [) n: R, D: }) {
d(a+bx)/dx = b
! B  |) I! p0 \# s' x; V8 C+ |
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:
it is too easy:

d(a+bx)/dx = b
- S; {3 s' v9 Q# B
so
' d2 L, _, j6 N6 m6 p3 y( |( k+ B
( o- j0 b" g' z0 R  fb=-kc+s
/ p0 B4 ~. T+ o# E' u1 W: N
7 Z. A( G0 A9 V0 ^4 H: \+ mfinally:  c= (s-b)/k

' q5 P/ W, D: s3 L1 a/ a/ N  F) r
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:
请教大家一道微分题，多谢！
% |, W$ t! Z  q. q  h' G% p4 D
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！

[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.

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:
* G% [/ @5 u8 c5 @& c' Z, L& ysorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
) ]" a3 |/ V! N
please tell me.

- B! X& A: s- W. ^7 D对不起，写错了，忘了变量c,应该是
9 d4 ^! J5 C2 ], yd [(a+bx) c] /dx = -kc +s
+ `5 ?, i/ E* X, l! c- F
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
4 {6 j9 z8 u! s  [: @7 ~; `# W
+ s$ G& s- t2 |  _/ \+ h8 W) O; H多谢

. h  g. X/ }9 N- S8 E
9 j7 m4 k3 P- H: I$ udo you mean it is:
# }/ B5 I% p2 ]
% _* r- h1 k+ u, b+ vd { (a+bx) C(x)}/dx = -k C(x) + s
/ N1 m4 @" j% V* m4 H3 G
where a, b, s are constants, you want to know what is C(x) ?
+ `6 m5 p: Z9 E2 ?/ O6 P
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:
% Z7 F9 d( K8 [) b$ W7 ]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 fi ...

& @9 R& l8 w7 j& R
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.
8 y" Q# `+ ?- B

4 V' A( g2 M; t) S" ^procedure:
2 {5 {0 P# T( @0 b, g1 E4 i2 S
) l7 Z4 P1 r: x% K2 ]. YFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
% v, O2 I+ W& u$ V9 `7 ]" f( i& cso:
) Z* P1 m8 }! p9 C7 Z( r3 t
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
' P7 Q, Q* w" ni.e.

6 B* \; u: X& G, X. S' @5 ]- l- ?(a+bx) dC(x)/dx  = -(k+b)C(x) +s
) e3 g. M$ u4 v$ @( Q' b' `( Y

- r2 R2 f" F! F9 Ointroduce 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
0 T$ O- E% |+ R4 Ytherefore:
' y: H! Q1 S4 N7 y, [
' e( `# z4 S6 @9 D{(a+bx)/K} dY(x)/dx=Y(x)
4 r) m4 n% J$ q4 L1 _+ T
1 A, u7 }; @$ ?. }5 Zfrom here, we can get:

/ w) D/ x" l' a; k9 P2 ydY(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)
' j' _- e( G3 e2 t* U, R; \& R/ W
, F2 f( @: X: f  R5 Mthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

5 g# ~, T0 Y6 {2 y' x  R-(k+b)C(x)+s = (a+bx)^(k/b+1)
. f7 D/ t& V8 z( I* ]) q" s
5 I8 T! j; D+ S9 d. u: O. N! afinally:

5 n) u: Z6 N  u7 w) I! U8 gC(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！

! W/ Z0 ~; A9 N+ y: 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。
' s& D7 E  v! ^9 _/ m, x4 u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。
5 @6 ?6 u3 X. I7 T9 H8 g您说：
% _7 f. v3 W5 x! U+ V# W
d{(a+bx)*C(x)}/dx =-k C(x) + s
5 K# A1 ^  F# S! ^+ w: V' {* E: Bso:
1 P! y5 Z  C, v- N! I9 ubC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

. a2 b6 F3 u5 l1 v* ^- W: n: 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
6 }" X; D& E/ }/ od{(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。
6 H! e4 D4 @* T0 [6 u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：
+ D2 g; M( o; j. ?9 B$ _
4 C/ A5 x3 z$ ~# K/ l" ld{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
5 M# J2 p# }" l6 e! Z最右边是x的平方。
6 }. Z7 _: `& o( ^1 P; M9 `' |
所以您解的有问题。
2 L" c  U5 C% G' Q" B, N- Z
[ 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。
您说：
( i' z& l% a* K  k! d, {
9 a$ k' P" Q9 T* _d{(a+bx)*C(x)}/dx =-k C(x) + s
( ~' q- g4 g6 I6 Zso:
4 t0 x$ j: U' ?* |8 u& P1 x, KbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
' x: D0 E: _6 K) E- O  a8 J6 X
/ h/ [4 ^8 D' |也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. g- |2 M. H9 r7 f& o( b0 i7 f但这是不正确的
- m: \) S/ `9 c1 Bd{(a+bx)* ...

3 B/ Q% ?8 o* s  j" N
Please note here:
* E+ p* W5 P0 j2 i2 H
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
7 I; v# c0 k. r2 Zit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
6 C5 y& K  e5 Gd{(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。
  Z6 N) }& C$ t7 n7 |) H
9 z2 K! C3 E& |2 L, X2 Mno "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
9 y7 j2 ^, @. C' F! w....
* ~0 `* \! _7 e2 t6 g- g# Rd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

+ t6 C% b0 N! B, Y3 @
4 N" V0 a, t9 b3 e  m; o这样算混淆了微分和导数的概念。

9 F( ]& x/ z9 Y- f4 q' md(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)

& N' O* h- `1 t  Z, C[ 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。

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

# f" O; ]' {9 u. tIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

+ g) Q% x' H, R' _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
- S7 k+ g' @5 b
Ah, you made a mistake: it should be:

+ q8 Q6 s& [" id(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
5 X6 p" g& \& z! c6 D  [& g% u6 ]& a

; |; B3 @" s3 [! s8 MIf 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)

for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

+ P' `, ?! Z) e3 Y6 Oh(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。
7 H8 p( }$ j( M" s$ ?4 j# @' h$ T
& c1 f1 c: L1 k& @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
% G2 `- O, C" ?4 _0 i$ {
Ah, you made a mistake: it should be:
0 o" {- {' \1 y+ I; s
( |3 @! T  w) }0 x6 W) ud(f(x)g(x)) ...

# `9 m! X$ n8 TYes, 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:
9 `2 @3 D! M- \3 \0 zYes, 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.

! B+ p2 ~; Z! \- y9 M' c  _
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:
3 w5 [9 P2 y1 k4 p/ z3 s+ p2 H
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。
, ]# d2 Q6 F$ F, k' {* ^, g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 ...

# A& T/ Y9 c0 G; V: F1 _- a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？

( @2 R3 J; E6 f0 P* o
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。