数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
( {6 i5 g6 H  h# J! i4 Y0 `' F* T
- }/ i3 \5 Z2 [* [, J& u2 y" zd [(a+bx) ] /dx = -kc +s
5 Z% H) ?6 i+ r; Xwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

4 G0 L6 p  P# u% l/ m多谢了！
) X0 @8 b! n0 j  N/ K: D( P
1 [2 s' \3 I/ K4 _. n6 _$ A# R[ 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！
2 d, v; t/ x) h0 G; I
d [(a+bx) ] /dx = -kc +s
1 g+ z- w3 w* S6 d' L7 Kwhere: only x and c are unknown, others are all known,  requiire c = ?
7 w$ N- L  W6 b+ }3 g. B* R
多谢了！

# l6 F: i' |0 c0 X  pit is too easy:

d(a+bx)/dx = b

" h5 D4 _0 E6 j+ Vso
! ]& v) v+ i2 l+ J& `: k- X
2 O1 a5 p9 ^' N  D$ W/ cb=-kc+s

3 [9 v4 I9 \/ Q$ [' vfinally:  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:

: U. }* k9 o9 I# n" C0 Sd(a+bx)/dx = b

5 b" a/ d" V7 z$ n3 S* o; H. d7 nso

% \' E" P/ }$ i) w% j; sb=-kc+s

6 o2 p/ Q' \: a- \- Zfinally:  c= (s-b)/k

  G* ~  X. j; f3 c1 y. Tthe 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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
: S" @& S$ f! X0 X3 L1 T
多谢了！

) D- i1 w0 y( }& q' N0 t+ r[ Last edited by 醉酒 ...

9 A: q$ D$ e; w
sorry, did you post another question? your statement is still not clear.

* S, V; t& Q  S6 ea, b, s, k are  constants? c means c(x) ?
% G. r4 K& _& o5 a& L$ A& S" y
7 d! N( ^9 v- Gplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
+ F  t, Q( ?3 {, 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.

6 o/ z' X8 [& v8 \" p对不起，写错了，忘了变量c,应该是
" _8 K. q  H) a# ed [(a+bx) c] /dx = -kc +s
" @: Q+ t- E! Y8 n( ^
多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
. ]5 D2 v) {  b' R7 f2 q对不起，写错了，忘了变量c,应该是
1 J; p: W5 D7 L3 H/ {: yd [(a+bx) c] /dx = -kc +s

+ e1 U" G- Z9 J  S( w+ z2 b多谢

; X$ X: w) b* U: \2 J8 m. a9 {
- Z% r, Q" Q& u' t/ w" p7 Z" \do you mean it is:
5 K# W; B/ X6 V% X
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 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) ?
* J& ^8 ^( b, X5 t: ~# _# l. T& y
3 I, k. H# Y+ `7 w4 T0 x5 E5 \( \if so, this needs to solve the differential equation. please comfirm it fi ...

. {  p& h& f3 `9 M6 T+ w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:
: J' H- c. x) s$ y# i; N7 W. \
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

6 {2 a1 ]$ F; ^) T
procedure:
( W/ \! g% i4 p+ `( M
( }% p/ ]/ D' x9 gFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
7 M8 W/ H1 @) i- g4 F  }. u" _: s7 G
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
: A0 p9 P4 Z# r
6 Q3 S' s6 P& x3 N1 t9 W$ t
1 ]( Z. F" v" o4 ointroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
& d5 k! T& m" T: ?# r0 |which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)

* j& {" v$ `' H8 v: Ffrom here, we can get:
. ~+ C# a1 z* X5 t5 J, G* x
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

8 W  o+ v* [6 H$ y$ z0 N3 Cso that:   ln Y(x) =( K/b) ln(a+bx)
: ]4 C3 f, K$ c, O; |
. Y& ^) N! ~) U7 ?& othis means:  Y(x) = (a+bx)^(K/b)
- Y* E9 y" m: I+ g! Tby using early transform, we can have:
9 ~2 \7 E3 `; T0 {- r8 O3 s
-(k+b)C(x)+s = (a+bx)^(k/b+1)

1 [0 o$ R0 d& {finally:
; H* r/ P9 r5 H- K9 I5 A
3 j7 Y5 f# A3 y) I) v$ V( ]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！

+ Q, @0 D" _# V% D* B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。
$ H$ q  j8 \3 H, }0 @6 s, N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。
) g" P* G+ k% ^! T6 g3 }  p3 s您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
% M+ C$ f) G6 Y2 WbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
3 ^2 ^+ y8 e  \9 \
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
6 @4 \( S' p" S' g$ D# a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: {& B: a. r  o# nx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
4 O1 X# U: [# Q; f
8 U. m% R7 E3 w/ O1 {0 p" I$ bd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
$ m" P! @- A( [% e2 H# n) p! `6 j- k最右边是x的平方。

  J" e7 |9 b% V所以您解的有问题。

1 R6 T$ B7 X& P7 q[ 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 X) t  D) }/ f+ m9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
6 v& e0 Z5 k  PbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
- B0 o8 ]1 R6 P1 y! L" `  h6 x" M
9 U& T6 J8 [3 _5 v7 R也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. ^& t; Y3 u- a4 V( o6 e3 b1 D但这是不正确的
d{(a+bx)* ...

' M7 a- g# b& \$ S
7 k' \+ }6 j! M& X4 A3 BPlease note here:
! w  ^) D/ e0 f4 D* k7 \7 G
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
) s* |2 p& f) u+ C3 m' E" ?0 m2 v" nit 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
9 |. y* B3 `& s3 Y7 P, j) E并不是   (a+bx)*C(x)   的导数除以   x   的导数。
/ k0 O, l8 s/ C3 r
; d% P6 \8 y$ O' L' z4 pno "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
3 K4 A- y+ b( {: i3 h....
9 `4 y+ z( H0 k+ gd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

. U( k2 }; h$ J. \
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
/ Z" F/ {& C! B+ R) R( e2 ~4 Q
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)

[ 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。
- z$ v0 @8 |( K' n* f$ s
; d- l5 H( |7 vd(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

7 g$ d; f3 M8 V$ y2 DIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

' ?+ }- p9 o" G4 @8 y6 F1 K4 r- E这样算混淆了微分和导数的概念。
! Z6 E1 d6 Z2 l8 x# h6 @2 \& X% q
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

% G5 ~+ M. H0 |: ?5 {  E( xAh, you made a mistake: it should be:
) w& W  Y2 Y1 H  x, N5 U
& R5 A( T4 n! I4 e  \' ]& F" Ld(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


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)
$ C6 }+ n- N( ^  K0 T5 @$ a
6 b$ P: O: l# j4 tfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
! s  J) u& n* M
$ y) c4 ]8 |& A1 Mh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
$ G5 B' ]4 P1 N( F4 m8 x3 P" o: N" O这样算混淆了微分和导数的概念。
  ~4 T7 O# M3 E/ W" P& C
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

/ y: K7 K5 y' |* t7 Q* \Ah, you made a mistake: it should be:

d(f(x)g(x)) ...

+ a6 u+ d& }7 Z1 _; S$ ^  lYes, there should not be " ' " after "f" and "g", it is a typo. :D
4 _( V  M* V5 O, I; E
0 r) t( s1 p5 c9 \1 Y6 \( D% DBut 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 U# T- X6 m" `Yes, there should not be " ' " after "f" and "g", it is a typo. :D
& Y. P6 w: E, y( w: W! t$ f
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

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:

* F+ S- ?. d  c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。
5 _* u( X4 J" w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:
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。