数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
& s0 l  v/ b# ?' Z8 s/ Y
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?
- i/ w6 O7 i  f7 \; k* ^4 O# Q2 l
4 T, U% P- x, N' S2 u. l多谢了！
0 C# d! ~, z: e& V, {% b
[ 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:
- `2 n( B' O& H2 k9 T( H请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
2 F; f! P# F' {+ Iwhere: only x and c are unknown, others are all known,  requiire c = ?
$ x2 m9 X8 g! Q$ }6 D. x+ n; v
- d, Q9 {' f) Q$ Y多谢了！

" U+ L7 Q' r3 c' q* \: t8 s2 t0 Uit is too easy:

d(a+bx)/dx = b

2 B! j9 l- X7 J$ I, v! pso

! h. U. |! V. k: J, b8 Db=-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:
) E, J; ~/ q& c: J& sit is too easy:

d(a+bx)/dx = b
! w- G! ^2 j5 `4 x0 c  H  X% J
1 {) _8 i, h% @/ Pso
, `: ]0 w+ U$ W6 U  g4 X& ?
) Z* O4 h3 K* I% ^b=-kc+s
1 R+ B' y$ u1 y9 V2 f
finally:  c= (s-b)/k

) C  Z6 L0 e  u2 b  y$ K, J5 Zthe 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！

d [(a+bx) ] /dx = -kc +s
# N7 w  b. L$ w+ Jwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

4 u. d  `, w$ w( W多谢了！
# Z% B# f/ s$ U! E+ g
[ Last edited by 醉酒 ...

; E4 l6 i" k2 W; F! }  O; N- j. Zsorry, 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:
9 v! H/ c( r& s  V. bsorry, did you post another question? your statement is still not clear.
# a  E8 w4 m7 w, {
1 q, S& L" _# K9 R, T: Y% qa, 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
% h5 l& \- z' d# m' C& ?2 [d [(a+bx) c] /dx = -kc +s
! z1 I6 u+ |$ R! F! {6 k* e
& E0 J3 f$ m$ j2 j* C( G; e% C9 U" U多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
7 G) i* u6 N. l5 X1 ^8 L; q对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

; q; {* K7 w- @2 o: I4 X/ C多谢

do you mean it is:
8 b! p& t" H% ]5 Z# M$ v/ R; \* {
9 G8 o. H3 _, a, J! c* fd { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?
. ~6 T. C% o. @/ b) N+ E6 v
% {: {2 g$ _* d, J: ~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:
* n% p* w1 q) ?0 Y" odo 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) ?
/ E  ]3 Q, |& ~0 _! E
if so, this needs to solve the differential equation. please comfirm it fi ...

; B; `3 t. }  \3 X
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:
7 u- P4 S9 {: Z( `, U
* [1 ?3 N/ w- e& N: Z* D4 a! k, |C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
" c. |; ]! n0 F6 c4 B
3 g! h7 T6 Y* `) E0 i, b9 ~+ ?  sbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
1 A* u7 ~4 d$ ~1 G7 P1 V" Y5 ki.e.
) B$ g5 l4 b) n8 E0 e
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 ~. N( U3 t5 b% v/ }+ ]/ \& ]# w$ _

5 `" v) B2 ]0 N1 L% {' wintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
- I5 Q# o* i: a) K+ uwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
' i4 j2 J' P! V5 b& Otherefore:

3 }+ V  T$ i- d8 a4 r3 S9 z8 Z* b0 |* D{(a+bx)/K} dY(x)/dx=Y(x)
8 N/ h- D" `- }! x5 l
from here, we can get:

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)
! v/ I: k( E0 V& \& e
+ {" F2 x% T( ]0 }this means:  Y(x) = (a+bx)^(K/b)
7 z: g6 }# _1 s0 |) z$ O  sby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
7 N1 @8 ]" l! k; M" d
finally:
  d+ B1 @) Q, L
# x" N6 r, m$ r5 {1 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！

/ Q+ j7 z% K( w0 F3 b9 I! k
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。
8 `) i/ u- E4 }: [. F* T+ Z3 b您说：

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

8 _! I: E2 h- y$ x! b! V# I# }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
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
% s" ?% |: S% d% A  u$ [8 T: Q  Px' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
( y$ ]2 S( o6 P% |0 l# d: G1 {最右边是x的平方。
- d- C5 b6 v6 P) Z
/ u6 ]6 D- ~4 L6 `4 N! p  \所以您解的有问题。

. c# D7 S5 ~- ]4 n$ ^[ 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。
# e- t7 u* \/ V7 M% x. L您说：
9 t/ o* Q6 }. x
6 \% N6 t5 L. Z# O% ?7 k8 sd{(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
  b5 V8 Z; ~" d* k( fd{(a+bx)* ...

Please note here:

d{(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
并不是   (a+bx)*C(x)   的导数除以   x   的导数。
  O6 f$ X9 f0 e* K8 h/ b( }5 i2 u/ t
2 r! {5 q: S" ^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
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

  ^! ~- n4 c3 k+ x' f8 w$ Y
9 _1 l: Z8 g6 T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

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)
" p0 d  [. j& c1 P- v
1 p  o6 J9 p3 K3 }[ 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:
% q$ p3 F) N2 j2 [; G* w这样算混淆了微分和导数的概念。

* c3 w3 @) Z, {! V* [+ W2 _- ?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
: B' g1 E) i7 l' P: `  M
( Q0 D+ f! a# M7 FIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

3 E/ X8 [; h. n# z: D: k5 A$ ]$ u- C
这样算混淆了微分和导数的概念。

% p' V8 [" Y# X0 {8 s& r( q- Od(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

0 n/ i5 L% J/ kAh, 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
4 Y' H$ F0 s6 n& K

2 H( k8 E! j" i( H% c; 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)

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

; V. p( Y) I+ Vh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
; T  q6 N- f$ D7 C/ 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
% H3 Z$ t/ v$ W& D
Ah, you made a mistake: it should be:
! |# x+ r; M* S9 R
d(f(x)g(x)) ...

- ?9 w" Q" S5 Z& B7 o9 A; O
1 q  B0 Y0 `* l* z. h9 n- ?Yes, there should not be " ' " after "f" and "g", it is a typo. :D

5 o2 K* u8 r3 GBut 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:
+ T* I4 l1 s7 I" 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.

& Z0 L8 A+ `6 [! t* l- b1 D
/ I5 O, w4 W% d1 u4 V0 h; IThanks, 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 F6 [" [0 Y1 ?8 R" T. u
$ b* a2 @8 u+ S- @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:
, Q* Y1 Y$ ~& b) v2 g1 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？

. _$ {- ~1 d6 m+ x2 C1 q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。