数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

# m* p0 b9 y% o8 N( j  Q  ]$ |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?

多谢了！

; P( W  ?  Y6 N  ^& D5 y[ 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！
0 t! {$ u6 l! s" A6 `) z. x
7 y- i7 R8 P0 {/ u: w! dd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

it is too easy:

d(a+bx)/dx = b

so

b=-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:
it is too easy:

d(a+bx)/dx = b
, l+ H! m% n3 ?% y( r. ?# j
so
$ q- Z" e4 u2 v3 E, g$ h' ~2 ?. @
b=-kc+s

finally:  c= (s-b)/k

& N" W/ ]9 d  b6 D6 d4 |+ [7 ^
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 [(a+bx) ] /dx = -kc +s
3 d: U3 Y$ \1 uwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 B5 D4 M/ Q. t, R8 j7 ]( a6 Q. X
' ~) ^9 N2 I! N; _多谢了！

[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.
) y4 @, o2 U; @$ F1 O6 K- M
  |5 c5 S1 E0 [5 @a, b, s, k are  constants? c means c(x) ?
% J0 L# ~  y, w6 s
& r$ I( a9 o  g4 m/ ^" s7 a% [: G, Yplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
* g& f, D/ H: W# `sorry, did you post another question? your statement is still not clear.
' A: N' j0 f! L
0 y: ^7 [  C  Y7 l2 d7 ya, b, s, k are  constants? c means c(x) ?
! n5 y  }( b" V6 C7 l1 }
please tell me.

. h/ g" Z# a, N6 S8 T+ K& `' u1 m" z对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
& C1 g8 e% O" s8 \( \  F* w9 j9 n! W
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
d [(a+bx) c] /dx = -kc +s

6 \8 z: d7 z% m* p1 O* T& J6 {6 I多谢

9 q5 n" z" b; @" J2 }  ~
do you mean it is:
7 Q0 O$ ^# K, z
1 k$ C* i4 R3 v9 w1 o' i0 od { (a+bx) C(x)}/dx = -k C(x) + s
) R0 h+ L  u; T+ N! _. P0 h9 t* F9 L3 \
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:
# N' t5 X+ L0 h" Y$ ~8 mdo you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s

, s6 K" H4 @: r' G' Q% xwhere a, b, s are constants, you want to know what is C(x) ?
% P4 }  l7 |' a- r$ p4 K  O
if so, this needs to solve the differential equation. please comfirm it fi ...

9 L. a1 z, p# x$ u
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:

C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
& r7 x2 g( G' C; \

procedure:
/ _6 E# _/ P, o* Q$ x3 ?
+ x2 P; S2 ?$ i/ w0 q8 lFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

: _/ a5 o% {9 ?  \/ E% n4 G0 nbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
$ g+ T* V5 \/ w: q7 y" Si.e.
! M9 ]: p* U2 u" Y' Y0 W
4 U" l0 y+ M* d& z) a(a+bx) dC(x)/dx  = -(k+b)C(x) +s

5 _; e9 d" A  Z3 f
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:
8 ^$ T  X4 A. e3 y; z5 h
, V# o* C& h5 _8 ^, v6 o5 c5 b/ N{(a+bx)/K} dY(x)/dx=Y(x)

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)

" d3 D8 w  t; Qso that:   ln Y(x) =( K/b) ln(a+bx)
5 \0 L0 [9 |6 d: d- L
2 F4 a7 ~* F# G. jthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
' p) r( e4 n2 Z+ o/ ?4 Q& \
finally:
: X3 R4 r8 i& o+ ^
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！

" X) g: q# c) k9 O# `
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 c9 y0 H  |8 P. v* m" P/ T' i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：
7 l( ]9 ?* X8 u+ \8 c& P4 Z
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
* f4 X' S6 q3 o( mbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

. B) M4 Z$ E) n& X6 P也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
( O/ H5 p& P& g/ W/ V但这是不正确的
- m* D  P9 [5 L$ f8 J4 Od{(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
$ ^' a2 k3 Y7 q: U0 wx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
- T8 }6 E( v! [" S
" |  k. i7 H% F( L- }d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
( Z1 g) Y6 I" ?最右边是x的平方。
$ y  W0 d8 F( B) W
! r) `  I& C/ U) C4 I; Z所以您解的有问题。

[ 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。
您说：
* Q, X2 ]) P! H
d{(a+bx)*C(x)}/dx =-k C(x) + s
/ @1 {/ f. t* y9 pso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

6 \3 ?" n* P" N, F! ?; e也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
$ u: {' ]7 v& M1 [' d但这是不正确的
) h% D- y! ^. p( V6 Bd{(a+bx)* ...

5 p+ z; `  E1 F" D& f& ^% m
# N0 x& S* @3 `+ G$ U3 W9 F0 bPlease note here:

/ G5 m9 s2 m7 s- U+ D! qd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
  [. ?+ R$ P7 ~/ q0 b2 O/ X# w( iit 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。

) r/ X( ~/ A& j# f. N9 U! nno "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
6 R9 T! L( _: k4 L) }* e) A8 t....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

& a* a( [) ^$ y5 q
这样算混淆了微分和导数的概念。
' |5 R* k0 `5 M3 j! W3 B
6 D: V4 h$ ^% O6 r# O# P# D- f: U/ P5 jd(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
/ b1 G9 ]% `3 c
+ C, ~8 r: F+ B; ?* s+ yIf 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)
0 @+ z/ u" K% V1 p" U- f
) m( f3 \- C  {9 t& F4 G[ 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:
) H2 m- Q! _/ I5 c# H) `这样算混淆了微分和导数的概念。

) ?1 G% Y6 A2 i) R, 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
0 b( d' l7 ^, |0 O
( h! O2 W1 ?+ C. l: I; `7 ~3 WIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

- @+ H5 m% `7 L
2 v+ s5 _) \* g( j$ ^' z这样算混淆了微分和导数的概念。
' K' }7 Y, \) j- T
6 `3 W) j8 F' K% C3 b3 a$ L& P" Wd(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
6 s" X' ]' l- S! x' ?
* [7 X! t) c) mAh, you made a mistake: it should be:
+ \4 M1 }6 S; Q; H
% H: S% N+ g" e% X# u+ Z0 vd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
3 h8 l7 C3 E0 }0 n3 [" _$ o4 y

9 ^1 d$ M3 Z0 K' `6 W8 eIf 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)
8 E% f  r' i4 C. a& Y
+ R  C- e# Z. S* zfor 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。
' Z, @4 i+ [3 L! O) P
& `1 \8 C2 e1 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
; T: i$ h3 ~& w2 P; y
$ A5 H* ~  `9 V, H+ p) eAh, 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! o+ Q5 z4 ~" {  K5 a0 D2 `
4 q8 i: ]0 w% XBut 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

But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

! y& Q# G& \: C1 k8 B6 d
$ ]6 X" }2 E8 J: w5 ^1 U0 CThanks, 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:
/ @: K, ^) @7 N7 l2 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 ...

, B: K6 Y* d7 N6 |3 e5 o8 v5 P
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:
1 O5 m0 p: U1 e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？

9 w# \& E6 o0 H+ J  U, ^
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。