数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
) F7 i# T6 ]7 g; _
- n3 G0 k3 }( m7 Y4 m$ {d [(a+bx) ] /dx = -kc +s
6 e5 P5 [% v( o! Swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

; ~8 {: J$ {% K  q: p多谢了！

[ 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:
3 Q: f3 J$ q: t! _请教大家一道微分题，多谢！
' ]7 ^- a) X1 j5 I5 ~
d [(a+bx) ] /dx = -kc +s
. c/ F! h0 V; |2 `7 x! `% ?where: only x and c are unknown, others are all known,  requiire c = ?

9 x; P. x; @' A% H( ~9 ]2 |; k9 l多谢了！

% W8 _' M" T  e/ Z) R6 n: |$ _% p/ m# u, e
: P( B- D+ i: _: Z; V: J7 Tit is too easy:

1 F' \" \/ Q* Hd(a+bx)/dx = b
1 G( J/ X+ l) [6 o3 ]1 o5 k
3 ?$ D: i% y) D1 Y2 A9 Hso

b=-kc+s

7 `2 @4 o3 b* A3 ?finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
& x. P3 C- c/ T2 }7 pit is too easy:

d(a+bx)/dx = b
$ R: w: X2 l7 [, o8 @* S
3 U& O4 L, ~4 [' ?+ c! ?1 }1 R$ W; \so
3 C) Q1 Z! e5 c4 Y6 q
9 u* w$ Q: U: e5 C' a( p+ u& Xb=-kc+s
2 H2 n- z4 j6 g
) l2 C6 [, \* I2 @) g0 G2 y7 Ffinally:  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！
! h& x- Y+ z% t1 G: |' i
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！

0 K" g$ s5 u: ?8 T[ Last edited by 醉酒 ...

3 O+ y% \$ F0 n3 U" r7 ~! @4 F4 d
sorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?

3 v2 n* I8 m& K, |% }% _please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
/ |2 I: {) f5 S8 [5 s) \' e+ dsorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?

% f8 J& R" x  c. o8 Cplease tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

, Q) q! F! ]2 g3 o多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
0 ~" v# r% c& U0 i对不起，写错了，忘了变量c,应该是
8 n4 |: x& _: r+ {; i  gd [(a+bx) c] /dx = -kc +s

多谢

5 I/ \5 h8 K0 c: a
9 C! G' e, x/ t) ?do you mean it is:
; {" R  c& ~0 R' ~- m
0 {2 |$ Y, q0 W) e3 bd { (a+bx) C(x)}/dx = -k C(x) + s

# z; v! i+ w2 p3 {3 w' W- S1 xwhere a, b, s are constants, you want to know what is C(x) ?
4 h4 _* G% I( O$ Z1 {2 v
0 l6 E  v' t$ p$ w5 ?: Sif 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:

6 a: m& K' s. [1 |0 Z6 E# vd { (a+bx) C(x)}/dx = -k C(x) + s
* {6 Z! N% v+ h9 i1 q1 {
where a, b, s are constants, you want to know what is C(x) ?
6 H3 ~8 k+ g' _' R/ D$ G0 ]
if so, this needs to solve the differential equation. please comfirm it fi ...

- _5 ]! y; Y2 _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:

$ q) Z; Y) C( I0 tC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

: F0 }' ^7 R! O1 G- @
6 s. U" a6 c, r. e  |1 h8 F3 U* m. j6 \( _procedure:

, g' r) K4 |7 F: sFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

+ c2 u6 J1 L1 m+ n" u$ Q' C3 SbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
* E+ Y: I/ Z9 T' R
6 M% G' g* e7 w/ f5 r
" ?6 @5 ?, e' Tintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
) G! D& V+ R# l" x, N; {* F: nwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
5 J6 }' D9 n, v3 H! X4 t$ ^therefore:
! _* G1 W$ X# d0 s' Y- @
( d0 N& @. L# B! P{(a+bx)/K} dY(x)/dx=Y(x)
. t; r( l8 I/ @7 g6 g; ^
from here, we can get:
3 V$ r2 O  h9 @% A! c* J% N
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
& o- ^4 e$ d8 ~/ i6 x
so that:   ln Y(x) =( K/b) ln(a+bx)
; U# p, x# h& B* V
/ f3 Y3 \) l% F# }9 \: W$ F" R) G% ]3 mthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
  I  Y5 x4 R- u9 u* E/ H1 V1 P6 }
-(k+b)C(x)+s = (a+bx)^(k/b+1)

" S! c$ s" W9 k) |finally:

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！

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。
" l5 ?  x# [( o# 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）。这种题真的不是难题。

龙舞九天

9楼的答案有点问题吧。
您说：
" }8 `3 k6 y2 Q! g: h
0 L: X8 @* Y- \. K: K( ~1 Gd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
0 Q  s+ e  H7 O. X  t5 B' _
/ k8 x$ J/ \  R也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. i: ^" n, j7 M+ I! T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
- i; V3 Q0 U3 s2 _9 H3 X并不是   (a+bx)*C(x)   的导数除以   x   的导数。
; D& y  z1 k7 p( h) P3 u; r* q! j比如说   dx/dy = (x'y-xy')/y2
8 J9 i+ I( A  ~- x+ ?: z' hx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
. z- F+ o  k& z) q% A5 ~$ z( f: s
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
0 W& M/ y6 d  F- s2 R最右边是x的平方。
: K  J2 a  Y8 ?" {8 |
0 C# `2 M3 f2 C1 _* E所以您解的有问题。
/ N7 m3 n+ q+ n3 l# y
/ b0 f6 J! q% v2 F% q[ 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:
) ?' E8 a9 \6 J9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
; p& o- F5 R) [$ lbC(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
4 W+ v5 w. A# e- _  `0 |d{(a+bx)* ...

Please note here:
) C: J+ ]. ^# L2 }
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
6 g% S% e3 E2 Fit 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
5 ?( Y  S) T* p并不是   (a+bx)*C(x)   的导数除以   x   的导数。
( Q) r4 \# H9 z* e3 v9 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
- U, o0 \  K8 s2 @+ v% L# A5 m. W....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

' B% _' X" H+ ^) ^# A这样算混淆了微分和导数的概念。

% B% {; g8 n) ~, ~4 N' h" E8 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

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)

: J& b# B/ L9 H. k3 e[ 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:
: |! z- `# c. l$ [' ]" @- M这样算混淆了微分和导数的概念。
1 G; t& o3 `' B! U% B3 ^" c/ |& c5 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
; v5 L5 N" z7 r5 o+ o; d  r
7 I' e- T* P2 V5 D$ P! PIf 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。
9 L* u; B' X$ ]2 K9 X. n- x+ g) Y3 f
  W  I( |  v" Ud(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
+ P$ ^+ k6 G1 m4 u
Ah, you made a mistake: it should be:
7 f7 E4 y- H$ F: J% j
7 H8 D0 o  `' i9 t) Qd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


; p% p8 F# Y- _8 O6 Z, k6 XIf 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)
( c) p) s6 {% _, Y( `) y% R" g
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:
/ V2 e+ E' o4 [; D1 W* T这样算混淆了微分和导数的概念。
, ~: _8 M6 Z, D; ~& M, ]; {) H
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
$ W# r# Q3 p! s1 x# t
Ah, you made a mistake: it should be:
% j/ O3 k( |& S" T
  u2 X* @0 ~3 B& g$ j0 V' W. Md(f(x)g(x)) ...

/ Z9 h5 T0 [3 s0 m/ bYes, there should not be " ' " after "f" and "g", it is a typo. :D
* V0 t. f8 S3 h/ ?. I1 }
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

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

8 [$ B6 t; T2 Z3 V- H' ]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:

6 p) [  q( I/ I& g0 J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。
) W+ ]' E( o6 ^; y% F! 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 ...

! b) v" Z5 ?) r1 ]- M# k% a2 S
" A+ Q6 m6 c8 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？

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
5 ~% c. m5 `& K! z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？

/ K" |# e, A2 k3 m
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。