数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

1 G! o. u+ ?' R, \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?
& U0 \5 H4 ]2 I& M& R7 w: F
/ C( r* _% P9 W) t4 P多谢了！
7 K- ?, m- |7 L/ @+ C/ d! C; V. d
& _# d/ @. j7 K[ 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！
$ x9 g' _& z- Q0 U- N
3 b" C$ I  f2 a+ `; l/ `d [(a+bx) ] /dx = -kc +s
$ y6 e0 X$ j3 x# hwhere: only x and c are unknown, others are all known,  requiire c = ?

2 ?6 q5 v/ n: c* G# k多谢了！

it is too easy:

. g$ a, z, V( r% {: G  U) Td(a+bx)/dx = b
8 R0 Y. d$ W" V3 E/ E" v  Q
3 ?( R: O5 T/ c( X5 M& u0 e' _so
$ j; `4 T5 ]: \0 p; F8 Q
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:
- Y# j3 i0 k5 Q$ Nit is too easy:
; G# t1 T  ?0 K  V' q8 t
# E5 w9 P( G  L3 K( e) v; [8 `4 ed(a+bx)/dx = b

so

( M$ j2 @, i+ k: j! j) Jb=-kc+s

2 Z# K2 J, ^0 T' n3 vfinally:  c= (s-b)/k

* [$ O# K# M5 P0 w' A+ z
4 h: K# \8 Y+ M, {8 f; q6 V' w4 jthe 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:
2 J1 o$ h) `# \* y+ v请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
8 v. @- d: ~: I) O  ~# twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 p1 L! a! H! I, X5 N" N3 y
1 Y3 ~) _& L$ j$ m0 C多谢了！
1 v$ X- ?0 |7 o' k2 v2 d- m9 T
% K; @' p1 n) B  O- y- m[ Last edited by 醉酒 ...

5 F& r. c( [& ~; A. K8 @( F
sorry, did you post another question? your statement is still not clear.

1 Y: |+ P3 T8 R, {7 a, Sa, 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.

a, b, s, k are  constants? c means c(x) ?
! P* b1 \+ M5 |) Y+ R7 B5 ?
please tell me.

' G! l1 Y$ Z. o7 k# S对不起，写错了，忘了变量c,应该是
6 u" e: x2 A$ Wd [(a+bx) c] /dx = -kc +s
  q+ O  U( T, f) r: p) ~. l6 C  _
1 F! N/ S1 p5 @$ [) g. L* A' A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
+ Y% }9 K9 ]# T1 {% Z9 p1 d# k
0 b& A3 t0 [8 h3 P4 d8 P+ P多谢

do you mean it is:

, c' B. {3 }0 g7 l9 P( gd { (a+bx) C(x)}/dx = -k C(x) + s
0 e0 G+ z- _4 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:
. ]4 e/ U6 E7 |& }$ n% @7 Y+ Q
d { (a+bx) C(x)}/dx = -k C(x) + s
. x& A6 t/ u% O/ V1 B4 ?3 |
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 ...

% V' o1 b1 s2 R7 }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.
' J( O0 Z. [' A2 `* y
+ [+ y/ v& ?" r" O
# H9 h$ @$ s2 o3 F1 Aprocedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
2 V+ B+ E# Y& s! Aso:
8 R4 h7 R+ D, [1 E" @
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 b6 p" A8 B* F4 I3 v, r6 e$ ni.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


- z) N8 e! y6 `( L6 z* C: bintroduce 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
+ n) A2 q: {5 c9 u, Qtherefore:

: Y- R) ~# a3 x* y# O{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
) \6 J+ i( `: F
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

- d  n* o( ~! x8 Qso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
* t( J3 ]  g, _4 I7 h0 k( G& z& Uby using early transform, we can have:
, F) M9 }# t0 d
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
  U6 ~3 X9 E% i
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:
! j7 \' n; r) x" R% ?( }% f4 \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！

7 e" a5 V% l' B, r- B% a: e) f
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。
$ I8 @$ o1 k9 F" n7 `7 y" G) r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：
, h5 S7 Q8 V: J# d' j; D" p6 I0 {
d{(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
9 K- w3 N' }& f" X. c1 b: G5 P/ ?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。
1 N" C/ d# ~. ]# T% j6 v9 i- G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：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
2 H1 m% W7 ]- Z( E" w: x! Z最右边是x的平方。
0 e; Q* ?: Q; u
所以您解的有问题。

/ p# F9 R7 K& {- r9 b[ 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:
/ R/ K  w* C7 u9楼的答案有点问题吧。
您说：

& ]& V" m- l, G4 p$ N( Rd{(a+bx)*C(x)}/dx =-k C(x) + s
" A% Q2 {$ _! s( ~1 \  X  I6 pso:
; c+ u$ V& \2 C5 j3 F! mbC(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
* u1 y% \1 L- S! p& j9 e但这是不正确的
d{(a+bx)* ...

5 A/ c2 @9 x) p4 `. n5 yPlease note here:
0 L3 {* U% r- P/ P! k3 C7 K* i
! h; z' l6 W$ Jd{(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
  _& S2 d5 y1 s1 ^2 ]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/ S: Z* G+ I- T
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
- a0 _, o8 b0 h0 O# z....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

  P0 l4 c' ]5 T  z* W: Q( S这样算混淆了微分和导数的概念。
1 H* n6 e2 |- |, ]) B3 N
0 {% R7 ~. [, C% Z1 r: p% {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
  b; {/ r/ }& _. v: q+ O
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)

$ N- i3 g3 L3 m0 l[ 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:
( V( W# C* u/ _, u这样算混淆了微分和导数的概念。
  r1 B2 ^) b+ u$ }
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
# ^: a: g# y) u: g. V0 }( R: A
$ J+ ~. q* c2 C" Q0 lIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

+ b8 X& f" s- D: h% J! y" G* \, s这样算混淆了微分和导数的概念。
/ Q$ d: P" w) f7 S, Q4 V
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
9 V! h! B( J. R: V' ^4 J
, n' Q0 t0 a9 h+ m& M4 @Ah, you made a mistake: it should be:

d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
6 D! c, }7 V$ ^+ u9 b* Z
; {, }- x: r" y3 n) Q4 e& l5 W& m
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)

+ L1 G( x) t5 p1 R' Hfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

/ ?$ ?. `0 U+ Q7 u3 i9 {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。

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
8 A; c% ]  U+ n
Ah, you made a mistake: it should be:

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

2 L$ q7 X' o8 F: {# s, m+ L& `8 E- I
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
! O( Q1 ?. |" w+ E# x: e. p
. r) i$ x2 a3 Y8 hBut 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:
  B) k1 W+ y9 {1 f# g! O/ DYes, 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.

, ?+ F5 O1 L) f- ^
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:

0 A2 n2 [" F% f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。
) r. K7 f  B; B3 W  m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 ...

, A% H3 t/ {2 G; H/ u9 L
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？

2 _5 x, C+ Q2 f
; d$ q- o0 m: m! l: t6 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。