数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

& Y4 K4 ]9 T" Cd [(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?

9 }+ y5 O, N! X1 a" f" u多谢了！
% d- j7 z- p# J7 ]: A9 F
5 \6 T: P- z4 o' h[ 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:
- U% j3 I) z: f( D' b6 P) `请教大家一道微分题，多谢！

0 d  [5 f; ?0 I" v/ B8 Yd [(a+bx) ] /dx = -kc +s
7 c) u9 E# T! y% ]% [, ?/ iwhere: only x and c are unknown, others are all known,  requiire c = ?
) U1 a/ T  }1 ^- n7 E5 v
多谢了！

4 H" p9 ?& I/ X/ ?. o2 z
* a- [! F0 ?0 E8 J+ v! o& o3 D4 S( zit is too easy:

' l5 o0 G7 o# I* `  M: p- id(a+bx)/dx = b

) z8 A5 A' m$ E2 {so

b=-kc+s

; y& S9 b. T( pfinally:  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:
! q7 c; P6 H5 O
  `/ W2 u$ F9 X9 Zd(a+bx)/dx = b

) g( x) w# s+ ^& @) zso
) a. ^0 k" h/ a
' `0 ]0 P; s& }% Fb=-kc+s
" Y% |0 V9 U1 S( B! P: z# q* w8 I
$ |3 h% ^. A! K2 \: J2 [finally:  c= (s-b)/k

0 N* G7 g3 O% K3 Pthe 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:
6 w# k% F% Q+ j3 F4 a' R5 A0 A请教大家一道微分题，多谢！
# m, Y" {/ i1 v8 Q3 u9 ?$ e
d [(a+bx) ] /dx = -kc +s
" ?2 W% D: f( b. ?, W. ?where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

- D6 l, w9 \' Y4 L: Z, L& f多谢了！

" R# L  L+ [# _. c2 C[ Last edited by 醉酒 ...

6 D5 Y0 B9 x$ H5 s+ G
9 I) P/ I( u; B5 F+ b5 U! u3 r8 W$ Rsorry, did you post another question? your statement is still not clear.
8 ~5 S) O  Y6 i, A3 V
% `. `8 Q& ?) Na, b, s, k are  constants? c means c(x) ?

9 L  h! g% x5 Tplease 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.
, ^8 }* X- D* Q" U7 o3 G/ J3 B
1 k3 Q6 K, ^2 T9 g! S7 La, b, s, k are  constants? c means c(x) ?
% x% a; ~6 [. Z7 O( `
/ T4 V- x9 Q0 L6 yplease tell me.

对不起，写错了，忘了变量c,应该是
+ I5 }; @5 J- z2 _( b* `; s% td [(a+bx) c] /dx = -kc +s
! b# m* ~  ?8 t! n$ D9 x/ Z
3 o; q8 Q: \' L. f1 ^' c7 ?! Z多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
& b6 k  G4 v/ ?  ?" u7 P! B对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

- r7 I  Z* i, Z1 E6 D  `% U8 hdo you mean it is:
0 |7 q0 ?" Z8 q! r. v
, {" C: W2 x( m9 e' b3 O' nd { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?
8 W% V( |" N7 [2 v
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:

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 ...

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:

) }0 J+ {: W. O$ j& pC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


procedure:

9 O+ e6 t8 E; Q/ S; T/ H) ]From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
7 N4 H5 J6 J' @. i7 N0 Pso:
' ^5 n& ^# t. w3 G  {/ S: p2 v
# |. M7 q5 ^( l6 TbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
& @  R. O, M4 m: R4 u
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


6 w) n' h  T" i, ^$ jintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
. S4 g2 N+ g5 `0 Q& |. Jwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
" l7 L) ]; l8 M
2 U+ \; \7 v0 G  h3 U7 V; o: x" Mfrom here, we can get:
1 Q$ A/ m* p6 L( y
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
3 P$ I$ o% Y- T. K7 `
so that:   ln Y(x) =( K/b) ln(a+bx)

  \4 t: `0 x& f+ i  u& jthis means:  Y(x) = (a+bx)^(K/b)
: R# L" B* L, \" L' \by using early transform, we can have:
& L: V2 l8 s$ {8 n+ a# Y
+ s8 u2 e) @( F( d7 I- L6 r  p-(k+b)C(x)+s = (a+bx)^(k/b+1)
  m# {+ [$ I, A8 g4 G- O4 r
/ y" Y! ?1 K; @' T, Nfinally:
  l1 o" K- i: k) `5 `6 q, {
+ v) @2 q% i8 U* p/ s6 I7 tC(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！

, Z1 n$ D& Z4 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：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
: C) F; R0 a& v2 X. xbC(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
0 T1 a+ q- y0 T6 X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
% s$ t* ?* s3 E: I% b- V/ }% u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- q1 k- r! M8 l/ u" Hx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
6 ]5 M0 d! |1 |/ o7 ^4 F+ I7 n) `, n
8 x. [- y' z  od{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
, q* `& f* }4 m" f6 z* w最右边是x的平方。

$ s. _+ e. U$ \* [+ c1 B所以您解的有问题。

' Q; f' m! `3 E7 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:
( e2 R0 _' J' B1 x( O9楼的答案有点问题吧。
/ w% ]* L# B/ Z" j- N( D您说：
* N& {0 B6 b4 Q" \' R; m9 S
d{(a+bx)*C(x)}/dx =-k C(x) + s
+ }. h# A. t! }7 W1 S& Sso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
9 {6 f' T2 I0 C1 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
% i/ K2 Q+ }1 j$ Z; ud{(a+bx)* ...

Please note here:

d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
8 y6 t6 o% m% k8 eit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
# @+ Z' Q4 w3 a0 T7 Zd{(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。

  T: @, ]% V5 _  _0 c) d0 E6 D& f  Uno "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

5 @# f, [) g5 x2 x3 ^
/ y- V1 A' m8 D) L这样算混淆了微分和导数的概念。
4 B+ u9 Z0 G3 I) a+ H- y. G# f1 d, O
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

- A0 C% h" ?) i5 OIf 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)

% B$ b$ M/ b2 J/ l4 \; i[ 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:
3 R4 ]# ~& B' |& X  L这样算混淆了微分和导数的概念。

1 @/ Q& g8 X0 C' T4 B1 S1 kd(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
1 N# @) S. E% J/ c4 ^% M# b/ q
5 i$ u. P- ~7 f' q* vIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

# q) B! o( t" d* 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

Ah, you made a mistake: it should be:
4 r: w& F* @8 \5 c, E  C
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

  j) g* f* e* Y2 [6 K8 S
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)
( z4 J$ C# q* V  n8 C
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
) B0 b2 T0 P2 v
7 `1 Q9 ]! |2 ]+ _3 |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。
, C% ~  ~# L& B% [% l3 J4 k' J2 P
9 Z- L1 i2 a% f4 @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:

; G6 a9 n3 u/ h5 i# c) Jd(f(x)g(x)) ...

- }" H  P! m* p% wYes, there should not be " ' " after "f" and "g", it is a typo. :D

6 F) C) Q" P  d1 D4 x5 IBut 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:
" w3 r1 i: [& a7 C( JYes, there should not be " ' " after "f" and "g", it is a typo. :D
. B7 [* h. @. P0 `; M: g; p7 l
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

; `8 t* ^( ?; e( e4 }
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:
$ N2 r' h. O# x# _# p/ \
8 P' r( o8 y: t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。
9 @: F! t" D/ S/ ~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 ...

; m& O/ j# p  I7 [. r$ 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？

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？

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。