数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
% Q* X( W) P5 \7 ^3 V
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?

0 Z6 r$ ~" m" ~) T$ k) f多谢了！
& H+ l$ \  r+ U+ G- U6 k, p- z, 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:
请教大家一道微分题，多谢！
, _3 m$ R2 U' O* ~% B. h
* H4 H! B. a- B+ Rd [(a+bx) ] /dx = -kc +s
: @5 Z6 A# s. l$ ~( Z* F) ^+ ^where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

it is too easy:
; A% D: X( ?. X4 o8 D5 q' D: B  [
d(a+bx)/dx = b

. V3 W$ W' w0 r% K2 nso
- g, N0 K) _% E6 ~
b=-kc+s
5 I+ @4 a+ @1 N6 F% m
3 ^) k% ~' G: G, Y9 K: |1 Wfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
: l5 d2 {( V3 h. Eit is too easy:

% O4 `- o! U" Y! N( S. t, A+ Ad(a+bx)/dx = b

5 z5 _/ @( y7 H4 u+ {so

b=-kc+s

, Y- d: c) @1 o1 ifinally:  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！
! q' @, H4 c: k! p& c  m0 P
3 r4 [( G% T' w" r- ~/ G9 gd [(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?
, l6 F: z! ~1 G* H' L% S: i
多谢了！
$ D; e: c3 D, l) K6 H- d8 Y! n
, Q6 l; R" T! r2 d" b# |8 ~[ Last edited by 醉酒 ...

, @2 j4 n" Q" K9 i* P6 Q/ [
sorry, did you post another question? your statement is still not clear.
) Z7 |2 p' `4 Q
4 H1 @, k) `: o$ ua, b, s, k are  constants? c means c(x) ?
0 I5 Z$ {& G$ B( h' _+ C! C( h2 z
please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
1 K6 q6 Z& r  l' v& E. }sorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
& L) j' F8 n' {2 Q
( b) Q$ f( g: \6 bplease tell me.

+ V% y# H- f+ v9 ]' C* n对不起，写错了，忘了变量c,应该是
9 b* W" M! q8 g: H4 H; ?# id [(a+bx) c] /dx = -kc +s

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
+ P* H  P! G3 {# w. @, Xd [(a+bx) c] /dx = -kc +s

: Y7 n* T+ [( ~4 V. ?$ ^; c+ d* Z多谢

6 t+ _' r; H, ?& f/ S/ Y4 ?4 h
do you mean it is:
. }- d$ b  G% t$ U/ o- l) X+ g
d { (a+bx) C(x)}/dx = -k C(x) + s
( U' x$ z% k/ q- t0 e
  Y8 R* m0 ^' [$ E+ d! @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. Y% C0 ]1 V/ {# ~7 }5 A
d { (a+bx) C(x)}/dx = -k C(x) + s
5 ?7 N7 r7 E6 X, [3 `
where a, b, s are constants, you want to know what is C(x) ?
. W1 M; [0 U. ?( Q4 ^9 m
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:
, H) l' ]" A. S5 V) D
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

) ]+ q* p) c7 X# M
procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
$ u/ g7 \: X; a- p2 o' N* o* T7 Zso:
' T6 n5 B0 [) n  w- w1 d1 I( U# ?
& A9 X2 u8 I+ g4 B- W4 GbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 ^, p2 k: X3 P: ]* A: oi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
! i" i0 k. @/ `! a

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
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
# r8 O" u4 M! ~/ X% i
from here, we can get:
5 \4 g$ \$ b# c  i
: u3 }' E4 j0 _9 _& `7 b% adY(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)
' c( X; P% S9 a% D% h0 K$ U
3 ^; j( C5 L' K: e5 Y8 ]4 [( Z4 i! dthis means:  Y(x) = (a+bx)^(K/b)
- r: K. o  Y+ k" {8 K0 hby using early transform, we can have:
1 m& h# x0 s0 i$ U0 r# h
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
: d. v0 Q( o7 A1 t  X! T+ o
- M$ [) c" D# `4 u6 ^3 lC(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:
$ t: W. h# ^4 B) @# S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！

0 s$ c6 ]  Q* q. e' ]
# ~6 s: \. v$ m/ G% r" f/ H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。
1 t( \) f/ M8 O  B$ 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）。这种题真的不是难题。

龙舞九天

9楼的答案有点问题吧。
7 Z9 y6 M$ [4 s- F& ~1 A您说：

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

7 c' \8 r( L! `/ @( Z8 n- }也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. g5 I5 H1 k+ l/ [8 r  W. D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
: H" {' m/ _3 t, l( O- c9 ]5 e并不是   (a+bx)*C(x)   的导数除以   x   的导数。
' j2 i$ h$ }, H. p' X9 m8 \! d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。
因此：
1 s$ z1 I6 E9 N' B3 |, [
; I& [- ]; G: m- r' o7 Kd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
最右边是x的平方。

! O+ D5 }( V8 F& m$ X所以您解的有问题。
! u  a/ y8 n6 }, Z
5 {+ p9 q& X) I9 b. Z[ 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:
3 q4 {5 c- a! t9 i/ D7 T9楼的答案有点问题吧。
您说：

" G2 F& N7 Y& Z. ]# Ud{(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
( g( i4 t" e2 S; R! U! `; dd{(a+bx)* ...

+ o6 _* y7 F, f3 b* VPlease note here:

- h$ u$ R4 d8 r) C8 [( sd{(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
7 q5 Z: h$ ?- a6 G: A+ b, j并不是   (a+bx)*C(x)   的导数除以   x   的导数。
8 G( ~3 o7 y% g9 v9 o7 `+ K
$ u9 ?2 [% m* w8 |0 A* Ano "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
8 p3 d& M, L9 ^; \  }....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

- r+ U: m4 X; `) l) C4 R. _/ @+ ?9 g这样算混淆了微分和导数的概念。
! K  z6 y2 r- J' N0 n( a8 ^
/ T& t( P7 j3 b) f+ Cd(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) ^" o- C. y) G  k# C
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:
' @9 ~4 Q. d& {5 }7 E) V5 n7 u这样算混淆了微分和导数的概念。

8 k. s  n8 B0 N( B* g# x( C5 e' sd(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
% V$ u9 i  Q0 Z2 j
: f# o" z) Z: R* mIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

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

; M' R2 `  t; V: Z" D: x& wAh, you made a mistake: it should be:

( Z. S+ d: A/ r" L0 Id(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


9 j% K! p, n# r" _! f' 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:

h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
5 ~0 p/ N1 Q% w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
, N# c9 R/ f7 j' F' d8 O
9 F6 k0 M, q$ ~! j5 g' g, `Ah, you made a mistake: it should be:

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

Yes, there should not be " ' " after "f" and "g", it is a typo. :D
$ v; g1 z( z/ z) ~5 f+ z, {
5 m; I6 c5 \2 X# ]$ f$ P7 [. eBut 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:
0 m4 M# p1 h  Z2 @  g4 lYes, there should not be " ' " after "f" and "g", it is a typo. :D
* y0 @" R, H- Z4 I* N! ], q
- m: k2 R  g- B6 u# qBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

" r1 v& `5 K6 O# r- R& J2 a
) b% q2 ]! r: w1 x! p: ?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:
8 i& P9 n$ `5 J" K' G3 n: y
9 C" t3 x) ?/ J, p* n6 O2 H7 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。
' g+ V& ~' x6 `4 N, e& 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。