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
6 k: b4 N  c. hwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
: u( [! M1 n1 ]
' i$ H7 I+ r7 [- h多谢了！

[ 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！

( w0 l: t+ S' K" S. ]d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

, K0 o/ d" o+ P1 L9 E5 C3 i多谢了！

4 O& g# k( |. f- c6 |
it is too easy:

! Y) J; @+ ~" s) A3 qd(a+bx)/dx = b

so

b=-kc+s

) |0 o5 O: N, n" g- Bfinally:  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% v! W; J. L0 |9 x
+ }# X1 Q7 O( n3 P4 [7 g: ~0 K/ _d(a+bx)/dx = b

so
% K4 L& h% O; b
b=-kc+s

  w% C' h; \3 j$ |9 Xfinally:  c= (s-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:
请教大家一道微分题，多谢！
% ]# d( c0 b- e; p1 T
, s3 @8 B3 [) p2 }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?
- N: o( j1 V: o  I& ?) w
多谢了！

1 k1 z2 `9 j7 p+ ?( p7 i[ Last edited by 醉酒 ...

! o1 r! s" f  U% L: \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:
sorry, did you post another question? your statement is still not clear.
; q! q' r0 n7 ^" n/ h7 x9 w* }
# h2 [1 D4 @9 I) p/ ?) X5 n; ~2 wa, b, s, k are  constants? c means c(x) ?

please tell me.

  C$ \6 |: E7 f' V4 a8 X$ l6 o对不起，写错了，忘了变量c,应该是
* Q  a/ p- M: m* ~) ]" ^% ad [(a+bx) c] /dx = -kc +s

; i3 ^2 H* ?, N8 B3 m& K! b9 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
3 t3 F% T- n' L- Zd [(a+bx) c] /dx = -kc +s
7 y- d; D% q/ L8 j$ G  j+ c1 w( S$ K
% H! x; o' B& q1 U多谢

) q( Z5 w- u; m  @% f; D
do you mean it is:
6 v4 G3 r! H2 R2 q6 v* K( L
4 P$ M* U/ M& h7 pd { (a+bx) C(x)}/dx = -k C(x) + s
* y9 O0 `0 \. s4 |- \
where a, b, s are constants, you want to know what is C(x) ?
7 s0 y4 o% f& M( I, }: {; Y" }
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:
do you mean it is:

  u$ q4 h3 n2 w* hd { (a+bx) C(x)}/dx = -k C(x) + s

- k  y# j4 v% y5 N& mwhere a, b, s are constants, you want to know what is C(x) ?

0 G% c( K1 `3 d+ j; Dif so, this needs to solve the differential equation. please comfirm it fi ...

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 n" \' v/ ~. [# K$ ?7 W8 `5 O3 L
9 C( _% o9 {, [# n  wC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


procedure:

# D# F0 ^8 S2 I" `) e; uFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 t: p* Y  U9 O2 s6 @so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce 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
& k1 y, R+ c  S6 |9 ^1 n: ptherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
3 b/ u6 u$ \/ J% b
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)
- ?9 x' Q8 ~% A8 L: E7 o
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

8 }0 `$ p2 ?( h, w, W-(k+b)C(x)+s = (a+bx)^(k/b+1)

% ^+ ?. G, c9 X1 C3 Q' b, ]$ Tfinally:

( N) t+ r; i# p/ r: g' wC(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:
- Q4 d: g# g* ]+ j3 [2 i' F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！

+ m3 e* ?5 g+ g( B5 g5 E2 C! V5 ~+ e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。
5 r7 s# ]4 F( m% l0 R8 h5 U  Z$ e您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
& q+ n$ g6 L( W, q+ X  U, qbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
" W- k+ i8 a: S7 s" c2 y
8 W6 Q/ A0 I0 U9 b  c+ h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
, V- W% M. ~4 w1 i7 cd{(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。
' r. G( H) r4 C9 x比如说   dx/dy = (x'y-xy')/y2
" t3 w% d# P, {* f' C& R" Y8 f4 W6 kx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
7 G9 F" ]* v: I8 V, y6 b因此：
- }9 a7 {% W9 L1 c- ~  S
# D/ @0 S* z* [  Q1 Gd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
* l0 z. @; s$ L4 W7 U5 z2 p最右边是x的平方。

. q1 L) V- c0 z. m3 |所以您解的有问题。
5 s% J5 ~$ x& O  }
[ 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:
  ~% ]. ]+ h+ k1 L& Q" e9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
4 p# J' i7 Q' S! a9 yso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
* N1 G0 F: ]) w' C4 {
* K0 E% x7 R& K也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
; e8 n5 R- d' e+ y! x. D6 ^3 U; S但这是不正确的
7 [0 A$ x1 P) }& R. Y* Q* f$ jd{(a+bx)* ...

" M  _( d) e- n4 i" s/ ]0 }4 n+ Q
Please note here:
  B9 U8 Z& e9 M7 C( C
7 s0 C7 B( w6 _' m  X7 [- ~d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
$ L3 J& W9 w* @$ O! x! k1 dit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
6 x  C# D* I$ w  Ld{(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。
% b! Y$ a; m0 [0 v5 \4 A, j
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
/ W9 x9 T& i9 H3 t' }0 d....
. ?+ }+ m6 p' b% f1 Z/ Bd{(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。
8 U* l; G: I2 l. w
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 U$ C0 J4 O* B9 \6 [) QIf 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)

/ d/ r7 n9 `  A8 P[ 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:
! h- A! l2 q1 N% x# r& o这样算混淆了微分和导数的概念。

! ?& C9 m+ a4 X. }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
, U3 ^- O0 O  C7 o' j
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

4 Z0 y$ H# V4 \7 S
# `9 [- _% {% D+ ?! N这样算混淆了微分和导数的概念。
8 h4 |* Q7 z( }* k; M0 t7 d3 Y1 C
$ }- H2 j6 L4 c# C* Gd(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:
2 y. y: a; `! B% ]6 ~
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

* ]" b$ z% x! W8 A2 ^' w! E2 d% o7 V
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)

/ D4 j( ~4 Q/ X9 Sfor 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。
$ h# s! D/ K7 W. N* X5 O1 {
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:
# m0 p+ z# n$ N
/ d$ Z4 K! t8 n0 ]' F' Vd(f(x)g(x)) ...

  w0 q4 i0 A6 g; U- w3 W
6 W" w4 h% O9 s# i0 J; o/ vYes, 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 x4 l1 l( Q& S* u; \$ v! F$ aBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

& h5 T: k' |( u- ~+ G+ ]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:

9 p( Q4 |- m6 h" d+ R0 a, r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 ...

" ?+ P- [$ V% ~- L9 a/ S* P
4 \1 m7 H- y" g8 z! Z; ^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？

) ~+ h$ m# [# c
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。