数学高手请进，多谢！

醉酒当歌

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?

4 i) Z5 {6 c& m& H5 Z9 ~9 v( C: Q多谢了！

- I/ C/ e9 g0 E( G- l[ 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 F( I4 Q5 p1 F
6 ~) b# V6 H6 ~0 e4 ]( U0 _d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

# {0 z. c! q) J
it is too easy:

d(a+bx)/dx = b

& A5 e  W0 }/ Z, ?8 f1 J% ^so
; y4 `( u: x; ?+ O% r4 p3 G% L, _
5 m- v% T7 X% E) Fb=-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:
- E7 p$ I- Y  U, M# k
. m( e( g. P- P' @6 I4 Fd(a+bx)/dx = b

so

b=-kc+s

finally:  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！

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！
; q* q; A. R. E4 q7 ?" n- c
[ Last edited by 醉酒 ...

" t+ J+ W2 G3 W, O% B, Qsorry, did you post another question? your statement is still not clear.
# w" V% A9 g2 E! y, @0 r" T
a, b, s, k are  constants? c means c(x) ?

& x, _- m8 H/ ^& r" oplease 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) ?
; S8 _0 Y  [! c$ x4 W
- W  D# k1 N& n' ~please tell me.

对不起，写错了，忘了变量c,应该是
8 L2 ~8 ~% y( y  o# A3 h8 l* N! R% Ed [(a+bx) c] /dx = -kc +s
4 M0 Z8 l" B8 F% _# T8 A: D& h
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
3 r3 ?1 C* E! l) n, x8 k
多谢

do you mean it is:
; d- |% }2 O) G* z
d { (a+bx) C(x)}/dx = -k C(x) + s
7 g! t' z/ P) b
/ r# C- I$ V; |where a, b, s are constants, you want to know what is C(x) ?
+ j/ d- ]7 C* D7 V+ \$ L
  d5 |( ~8 U' H6 kif 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:
- B; t& Y- ]; X7 y8 t: L3 ado you mean it is:
, V1 d' z" ]# b2 Z: u% @. B1 Z; i, C
d { (a+bx) C(x)}/dx = -k C(x) + s
8 L2 f/ |+ j% i+ N+ E7 T. T+ d# L
where a, b, s are constants, you want to know what is C(x) ?

: x- A% a9 S: y! ?+ Fif so, this needs to solve the differential equation. please comfirm it fi ...

# L% h( `& T) C6 I4 b& H" x% X& j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:
) U0 h. c7 c; v" A$ D
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
, w, c6 D! E: W- H% y+ Y
, x. `( `1 `: l- y8 c5 ^bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
, D" ^- d* K; a& A7 o, |! [- Ei.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
, j8 k' q# A+ L
8 N9 [+ {( z9 B7 t
8 k. m) B+ Y: i9 y% e7 Y- I' J' @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
+ k4 D4 E& H4 r" D" i7 Qtherefore:

* B% ~6 K1 Y  L3 R! Z6 d{(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)
' q* ]( y8 Z3 N9 N
so that:   ln Y(x) =( K/b) ln(a+bx)
# B1 g* S; g6 _
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

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

  I4 s" A5 b4 @6 ~+ Ufinally:

( F) f0 o+ L) X, z0 ]4 pC(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:
/ C6 [* \# e" l" k& R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！

6 Z( ~3 c- R$ l8 s1 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。
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。
; k" X# m) n7 E1 S您说：
- v, k* D  X& X
  p. m7 l9 g4 G1 jd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
7 k. W* s: q2 H4 c: q8 @/ lbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

# i7 X9 a  f& F; N) e' P4 j也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
; L) p% D$ @9 J6 A" U1 z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
- {. L2 M+ m/ e) ]* {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。
因此：
) ]; s  }, S, n! P8 t
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
) U7 t. T. E% W3 |/ y最右边是x的平方。
0 C# q* o! U0 s/ j% R1 f  W
; y2 e7 M/ {# N* ]所以您解的有问题。
7 H5 x0 R$ X% j' ?
3 B# j) Z- d7 D. E[ 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。
- x& w8 [- [! T0 I5 v+ s0 u您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
+ t, j; x, y; F# b
3 H0 t% ^* `; e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)* ...

, J  A6 k8 L+ x6 x
- P+ A# o1 {& ^Please note here:

6 }" z0 [$ M5 C& Kd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
5 m4 v  a$ Z6 Z" y; E% p: I* _* {1 [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
' @* r' F# U+ ?0 h% O' u并不是   (a+bx)*C(x)   的导数除以   x   的导数。
5 l5 @& ~) j/ X4 U) r- u  ^
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
* r: z3 m/ `- o" t....
5 Y' |- }$ A7 u. ^6 wd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

0 `! O) n+ l4 |- S: c这样算混淆了微分和导数的概念。
* J8 x" ^. C: g% A2 g& `" Q
. ~' E0 R% @3 Hd(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$ ?1 D3 a+ j! c, 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)
1 I' q9 b8 l  d1 Q5 W
( F& H% M; T1 |9 T* C[ 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:
5 ~/ F! }3 O. t# d/ \这样算混淆了微分和导数的概念。
) Y) X  e7 {/ t5 L# X* `6 Y. h2 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
& f3 q1 Y; i% ~! ~# E4 A5 U  Y5 V
9 i. h* M5 H! y* g9 ZIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

" O1 e  I3 F0 C4 P* n) D( {
这样算混淆了微分和导数的概念。
: Z3 h8 ~! q5 g; e. E0 [; g
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

/ l  z7 r! M- N. u0 K% zAh, 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
# i8 ^$ G' `& o
9 L3 s) z9 C+ |  M* R
8 }2 J5 }3 d9 N- a/ w0 lIf 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:

8 x; |, r6 T9 U2 P/ i5 F3 `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。

# [1 ]6 x% A2 j% 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

Ah, you made a mistake: it should be:
" M0 R) G$ u( X# I6 y
  Q: g( t  m' q# y& U1 Bd(f(x)g(x)) ...

Yes, there should not be " ' " after "f" and "g", it is a typo. :D
) X; n6 f' G$ c: e6 e
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
! \: e7 j: Z( B+ H8 Z! ~5 C( \/ Y9 A
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:
/ ~% L  A3 \- U1 f: \( _
5 H& ~0 D4 _, I- t! X1 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 ...

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？

7 m( m# [1 ]# n$ B; ]& V# 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。