数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
( ~2 o% I& p- I
# |3 V; M# [( N* S4 S' `5 z( ed [(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?
3 K0 u) b) I  `
多谢了！

3 {8 C# `2 Z4 P/ X6 S4 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！
" j/ _) s  B( C' w7 \. W3 U( b
2 M  ?$ j: Z5 S6 F5 k, b) e0 qd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
5 U) t! V8 {) r3 [2 E
多谢了！

) K6 l' T$ F" W. d) t! k
it is too easy:
3 M- N0 \# p4 M* z) s+ s5 I
6 Y' v7 z. b+ nd(a+bx)/dx = b

so
) s: X/ U5 o& z3 c6 ^
b=-kc+s
+ m3 o7 ~) J: y: N
1 S5 ^" E: y7 q# {) z+ l' gfinally:  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

so

b=-kc+s

4 Z. x/ ]1 _0 w5 I3 ?, q+ @finally:  c= (s-b)/k

( j* c* z% _6 s4 C- Bthe 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！
, ^& O/ S6 m$ G0 @* S, t9 c
d [(a+bx) ] /dx = -kc +s
. g2 g& @9 t, K- P) P  Qwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

9 t( s2 R! R* D7 |0 b$ a多谢了！

( {& E! ?3 K% s: ?- k- }! m6 ^/ s[ Last edited by 醉酒 ...

2 T- R$ Q$ x$ m% @. y, m
) H$ ~5 B: \3 G* rsorry, did you post another question? your statement is still not clear.
$ J; o8 C$ s( h' {
& T2 B# |8 V: c# f; Za, 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:
. f* w; a. {1 c: X- H7 wsorry, did you post another question? your statement is still not clear.

0 M& y, l0 B' ^3 i/ d5 u. ^3 qa, b, s, k are  constants? c means c(x) ?

" e1 k2 M8 C/ K3 G$ P; d) |please tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
) G/ l: B) V. ]  I. _
多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
  Q4 y8 _5 [7 `/ x% v对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

5 t" t; A! T$ F
/ z# x$ H+ {2 |' edo you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
$ l+ F) v. y3 L* g4 x* }
3 e5 `- n  [* E2 Owhere 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
) Y  |; l0 t. j2 c0 d1 A
where a, b, s are constants, you want to know what is C(x) ?
, i- t$ r" U! D, F$ D
2 L* v+ _9 c0 L! Vif so, this needs to solve the differential equation. please comfirm it fi ...

  [& S2 u6 m; t" q7 a9 s* d
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:
' ^6 T0 c9 g, m6 f0 p4 R) m
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

. X& ]+ c! L9 M! h! K- v$ H
+ i' g1 D$ y4 x& z- R3 T! S6 fprocedure:

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

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

2 S& `4 _% ~4 d4 b; s(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
, q8 X& X& c9 N: s% t/ @3 `which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
- O* g4 s/ c+ L3 P; |. T: ]% Qtherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
, m+ x- [) n' ]9 {
' V- p1 s% a/ tfrom 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)

- m, h& ~2 H' e; ~$ Sso that:   ln Y(x) =( K/b) ln(a+bx)
& v% e' F0 P7 x7 `5 r* I
this means:  Y(x) = (a+bx)^(K/b)
! l6 g, V& B* p$ k4 rby using early transform, we can have:

  F6 Z8 p5 F- M( r-(k+b)C(x)+s = (a+bx)^(k/b+1)
4 ?0 n2 q6 b" s
finally:
$ Z4 d. M1 I" J. J+ f+ K  p) R
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！

' D: A( e) n  @2 N: D' U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：
9 J$ k' V7 y- {6 ]( ]2 g* v. Y2 k- W
& Z4 Q; m8 p) q3 rd{(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
( J5 j, v( [! I( k, ?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。
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。
5 s) \6 ]! a; W* P. j: D因此：
- ^  e: w# l5 L1 Z
d{(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。

; h) ?/ U0 \( b" T5 u4 k% @* D7 r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。
( m4 K" L% w2 e7 |您说：
" ~% e/ ]. W: P; A
( e. R$ S2 U* }8 u' Cd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
0 ^: a9 t7 z8 P. A4 R1 ^bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
* Z! t. Z: H' ]+ e  a
- O8 [9 Z- |3 y. R% v* @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)* ...

Please note here:
; H+ F9 Y; q' W* P  p
d{(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
5 r7 D& N1 C9 A0 jd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
7 B% T7 ]2 O9 J6 i! E6 A- s# m并不是   (a+bx)*C(x)   的导数除以   x   的导数。
5 b+ g+ y1 y' g- {' |7 U" P7 F$ v
9 a6 u( Y( j8 q# ?; J5 Rno "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
5 }" f0 ~5 U  V2 k....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

) }. @) _( Q. `1 k9 z6 O5 W& y/ f
这样算混淆了微分和导数的概念。
# f3 K1 ]3 ]1 P( u/ A
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

' S8 X" D  y; Z/ z0 cIf 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)
) S# [, T: X6 s5 J+ Y
[ 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 T: J* i# M) s& z; d( X这样算混淆了微分和导数的概念。
7 U" v. y3 P8 W5 n: |: B, d" w
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
5 ^! x$ {% p; }. @" a
+ \1 a5 T1 R/ I- y8 YIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

0 q/ p% Y2 R2 U' h
这样算混淆了微分和导数的概念。

, }& r- w- G0 z/ Td(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# G; u  p# v: T$ ^5 f6 ]3 u& D
Ah, you made a mistake: it should be:

) n9 R' |. h" X2 R5 f8 w2 q2 ld(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
0 i2 i6 P/ t; V+ B( X. ^* k7 r, r) X

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)

# G/ B) D  @+ ?' k; ]for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

0 X1 ~7 H3 m2 {: M6 Sh(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。
( e+ m* q, m! Q6 N7 j
. r' M) l: a; A- B. 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
2 @: _1 V* F- u+ D
Ah, you made a mistake: it should be:

/ q$ P5 t+ h# Wd(f(x)g(x)) ...

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.

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
- I' [: f# |4 v) M
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:
+ V8 T9 c1 S' v4 y' S$ b0 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。
- Z) }3 h0 ]6 Y1 _5 d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:
; j4 n& E+ e3 O, ~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？

( f# y7 J# ?! Y/ d) H/ z: G+ x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。