数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
# ~% S2 D2 Z8 D: C+ T- e. |
- E# n$ q0 f4 O# W- z! E' ad [(a+bx) ] /dx = -kc +s
8 \6 ^, B* B# p4 k% c& ?where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

# h4 ~' g" L5 t9 e, o- |多谢了！
$ C; C; B" C1 N4 w
- m# d; ?" J. I+ {  ]4 j3 e[ 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！

d [(a+bx) ] /dx = -kc +s
/ D1 @* `: A4 Lwhere: only x and c are unknown, others are all known,  requiire c = ?
, T  V* v1 u4 B* t
多谢了！

/ U# a1 W* q/ A5 D7 b2 uit is too easy:
1 Q/ A: g3 H6 d+ Z- u
d(a+bx)/dx = b
, Y6 }4 [) s; N- L
so
: w9 ?- `7 B! ^
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:
6 s" a' G3 U+ tit is too easy:

2 E: e% p; ^5 Bd(a+bx)/dx = b

) F) G/ E" Z4 Cso

/ b: a$ w3 X% ^+ t2 p7 j6 mb=-kc+s

- U* M5 G3 [% J4 Ufinally:  c= (s-b)/k

' L' k6 Z$ F4 U4 j: q. 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:
/ B1 K" }3 q9 N/ O( T4 B/ X请教大家一道微分题，多谢！
# v' f; A4 z) n" v0 R$ a. w; o
d [(a+bx) ] /dx = -kc +s
- y( B0 W* d2 s1 J" Mwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

0 ]4 |) v8 S* l多谢了！
# E" a0 b* u' i9 p
[ Last edited by 醉酒 ...

4 e2 e1 W0 q" y3 {  Rsorry, did you post another question? your statement is still not clear.
& K2 G2 T- U" w
a, b, s, k are  constants? c means c(x) ?
5 m; `/ M! B  N; Z+ }, }" z
; q/ M% ^* D8 S4 z1 t2 s9 _6 Bplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
8 ]- N7 z/ S5 e3 tsorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?

please tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
2 l( s$ _0 b7 l; P6 N7 B. t6 f
/ b& h0 q2 `$ W% U' V2 k: K多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
% M/ D3 g5 F7 p$ p6 g$ ^对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

, _, h# a" J; A
do you mean it is:

  l5 b' a! l2 G0 q! \9 Id { (a+bx) C(x)}/dx = -k C(x) + s
. |8 x. z& F# m! I  U5 q0 w3 f- k
' K! J+ ^6 |; r! H" ?, iwhere 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:
" l4 a0 C* S, h8 N' Ndo you mean it is:
/ c! l' _0 N6 l
! H* h  b  s/ M: L; t) Dd { (a+bx) C(x)}/dx = -k C(x) + s

+ j4 j8 C0 R8 V* K! {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 ...

* e+ K  j% ~. Y7 t  ~2 _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:

! u" J8 w' ~! d  A& r- p$ tC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

- k3 F9 U. x5 K$ _
% D9 o0 ?2 ~5 d5 b3 J2 h+ }  k) Eprocedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
4 s8 m1 n' d, O& |
5 N/ X+ x% R! o! w$ _$ B4 ubC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

; H( w1 h; U9 ]7 i(a+bx) dC(x)/dx  = -(k+b)C(x) +s
7 J1 A  I3 H6 M4 x- k& i% v; e
) I; _( ]/ f( I  |
0 M6 o- h% e9 z7 `6 K) f' tintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
7 W" q) U  c7 L& s# Ewhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
  n$ ^' U( r, T
  M! D* L4 O/ \, x) ?1 g{(a+bx)/K} dY(x)/dx=Y(x)
9 M' f* Q. F# ?5 C3 X& B
% ]) d* g$ v/ y9 ]7 Z% y3 }; @  zfrom here, we can get:
+ E) i$ ~, y+ Q+ f, c. q
9 o: t+ W+ c/ g& d4 }' E8 X' ]dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
% m) Z) y+ W1 g9 \- }/ f  Y
so that:   ln Y(x) =( K/b) ln(a+bx)

$ r. D. K1 @6 k# U. {# K$ tthis means:  Y(x) = (a+bx)^(K/b)
3 h* V) `; I$ Z' T& F5 yby using early transform, we can have:
* I, @1 ]0 w* H
1 c( k. y5 ?) Z$ w-(k+b)C(x)+s = (a+bx)^(k/b+1)
- f2 e" K5 V% _
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:
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！

4 v0 O' Z% L& s7 o. T
- }; U/ J6 o& F; I5 C7 ^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。
$ ~- N4 ^( Q) s  W1 W0 @7 H8 c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。
. Q" Q' L8 N* y3 W. a0 o9 x& L& \您说：
  d6 e  M" G' d: i( `% c) N, x: u
d{(a+bx)*C(x)}/dx =-k C(x) + s
+ h4 C6 [( ^% R$ U4 X  W6 |so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
* }! ^5 f' m0 O3 W( L
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
1 ]+ S# Z6 ^! y但这是不正确的
8 S' j$ t" J; S; X/ U: i6 B2 Pd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
- r7 d& `9 z1 k! x# S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。
( x0 \5 ?! |% F' l2 @7 W! N0 v因此：
- \" s8 W& W/ C4 d& Y3 D9 Z8 g
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
& c9 X: Q* A0 ]! P- y) y最右边是x的平方。
: D4 ~! w( }+ C' s- {9 }3 W
5 a8 ]) u1 a5 w所以您解的有问题。

$ ?; T+ R' w$ e) y* r, y[ 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:
% {& K# E8 N. m+ ^0 C  z+ X9楼的答案有点问题吧。
您说：

- `4 C5 [' E. ?" u& Bd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
' f7 w9 U6 h7 L5 c! j) O& kbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
; s# P/ s( q5 l  R! T
  S! L! j4 n& b$ U/ V8 i9 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
& ?9 L4 [' G4 h8 ^* Hd{(a+bx)* ...

" g* a' k0 X% E5 Y- m
9 E" g" P) G; |' u% ePlease note here:

- L3 b9 [! |0 a0 W) ld{(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
3 q2 a  ~3 @/ P, \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
& V5 v- E% e/ o; `( E& @....
4 \7 F5 W* t& }& w4 d5 Od{(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。

6 w3 b/ i4 W' I' ^0 fd(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

+ O  K4 P6 [8 H) o2 _" cIf 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)
7 I& c) M$ l  n0 z) [
! I5 k; @; K' d) |% @[ 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:
! D& F- u2 S& ^& x这样算混淆了微分和导数的概念。
) f/ }( V, U* e' P6 y6 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

. K4 E; j# F0 i0 I0 L% e: @. gIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

4 i7 d4 _( v7 {
这样算混淆了微分和导数的概念。
" E$ q; |1 _0 j! E/ C
# w- t5 u' i& [6 f# z3 ld(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 ]: t8 i$ T4 ^
Ah, you made a mistake: it should be:

d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
; h2 S0 m' N9 {

  P  Y+ O; k* N: p& W; h4 aIf 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)

  ~4 X1 A! C3 }4 `for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

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。
, v: o& U  m& ?! i* v
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:

) O1 I( P' C% m& O, kd(f(x)g(x)) ...

/ ]9 d3 y8 H+ ~0 t
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.

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
9 r1 F3 c% a/ p0 N& V
1 J( ^$ R" }/ G/ h$ A/ k2 O) b8 PBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

  t. o& d$ |7 }$ K- R) N
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:

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 ...

  D0 C3 h1 p0 f) @" c* _/ Q
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:
4 @  i0 r! ?7 f" f" 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？

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。