数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

. ^+ p, v/ I4 H1 Q( p# jd [(a+bx) ] /dx = -kc +s
" m% [1 V7 O0 q/ Cwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

4 B4 W. }7 W+ ]; H7 B- S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:
" c% V6 o# Q4 ]9 F5 l请教大家一道微分题，多谢！

+ R6 g! j: R6 d0 O; Wd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

) ?$ s5 P% ]. g- q4 j# n- u多谢了！

it is too easy:
( D! N7 O, X% i/ |
d(a+bx)/dx = b

' |2 Z* W: j( U+ N* _% Fso

b=-kc+s

' c. ^) `: e' I" Ifinally:  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:

! X$ P7 m7 u/ R9 c; V  K. ?3 y( td(a+bx)/dx = b

, J1 e( h7 ^' i  N0 u. W' Bso

b=-kc+s

finally:  c= (s-b)/k

3 B5 m  ]- l/ j6 ^% [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！
0 f; H" {- k2 ~7 l
d [(a+bx) ] /dx = -kc +s
) h$ {) R# B2 [; |/ wwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
( o, B) l$ W# U' z' ?
0 [- u1 r  W* x& n6 O[ Last edited by 醉酒 ...

  q3 [( p0 m/ ~) B; f
sorry, did you post another question? your statement is still not clear.

9 B& H& p' T) I3 F7 z8 t) qa, b, s, k are  constants? c means c(x) ?
. H7 G9 q, U4 X6 k, L8 \5 |7 |3 K
; X  s7 E; W3 R# ?7 cplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
8 ~) W( Q! a/ z) asorry, did you post another question? your statement is still not clear.
: t' v3 l8 B1 ]5 x: f5 H7 _
a, b, s, k are  constants? c means c(x) ?

! P$ Z& ?, f, O" m/ |* A) p& V: rplease tell me.

对不起，写错了，忘了变量c,应该是
' T! o& M2 x: e! e6 T' d/ b  E7 Ad [(a+bx) c] /dx = -kc +s

多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
. g$ ]( a& [  _1 p3 c5 D0 V对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
) q! F# z' Y* e
多谢

6 l0 }, W, F3 `5 p+ ]. b/ Z1 w
do you mean it is:

. G: d5 @0 M/ m, F* l" r' R# \d { (a+bx) C(x)}/dx = -k C(x) + s
" _: X) c- G9 a- a0 F8 F: ]
5 N( a! w! s- ?6 cwhere a, b, s are constants, you want to know what is C(x) ?
- [0 N& a3 ^6 E6 u3 R4 w. c
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:

2 w" J- \% h$ q3 Td { (a+bx) C(x)}/dx = -k C(x) + s

0 N8 p5 S  X5 T2 fwhere 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 ...

! V1 I# c0 a. z" K. X/ }9 @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:
" y& L9 P' f0 P8 |4 F& j: Z) }6 m
$ O3 w1 y6 n4 b. z" U% A4 QC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
- d1 \# J' j" {
+ T% d$ J( n( r9 ]- _" {# A3 L4 u
5 U: s0 F' D# l: H( w: ^: Oprocedure:

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

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
* e& F1 \2 c6 c4 K# Xi.e.
) M5 p! K3 f! u
0 s& Z! v/ A% ]: I- C# O(a+bx) dC(x)/dx  = -(k+b)C(x) +s
0 w/ n0 {' q- R" S9 R9 k( o
. b9 L3 K+ r2 `+ E9 J
: z1 D+ C) _( F5 {introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
+ J% x! Q' O* @- T, cwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
3 E  I5 }% O% ctherefore:
; U2 ?7 b( Z" t8 |, b; q3 N0 V
{(a+bx)/K} dY(x)/dx=Y(x)

7 r# G6 p$ \7 H$ U6 Ffrom here, we can get:

" I' s% Z5 f  h$ b3 K1 g6 z  {' _dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

- i+ l# t% ^+ D' y( Tso that:   ln Y(x) =( K/b) ln(a+bx)
( r0 J& c1 g0 M9 F! h+ M
. K9 S$ ^7 O! c9 gthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
! _# z" g, ]1 p7 |0 F$ i, m
: x. d& I8 F3 E) c  k3 R-(k+b)C(x)+s = (a+bx)^(k/b+1)
, l2 p4 M  L- }0 p7 ^. y2 y
" f  _: Y8 g4 d& V- [finally:
5 c( g1 Z) [* @' Z9 Z
0 w" U) @& B0 r* i( T5 DC(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！

& c. R+ y: l& a
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：
8 D8 ?/ h/ t* o7 }
d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
0 H: Z; o( B! p- AbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
9 ]: _6 T- L% j" \
  |% Q- N' [1 _  L: ~也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
  g5 J. n) q( X1 J% E但这是不正确的
+ ~' b$ u+ I6 Jd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
* z$ w5 N2 `+ x/ s" E! D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：

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。

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:
' r. a+ P4 A* C9楼的答案有点问题吧。
6 X: j  O3 E; _2 t) j  G您说：
2 p3 {2 R8 V+ Z* Q8 q" I
d{(a+bx)*C(x)}/dx =-k C(x) + s
* V4 Z" W  T' g# ~! kso:
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
d{(a+bx)* ...

8 \' o& ?4 }9 r" w& Q
Please note here:

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
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。
8 V: D4 x8 K& [( [6 T/ q+ \
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
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

# h# n4 P7 L" E; u; Z9 e
这样算混淆了微分和导数的概念。
; \, Q# {, N+ ~2 {7 m* u
: M8 B- w1 j7 N: 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

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)

[ 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:
这样算混淆了微分和导数的概念。
5 u; ~% y; N6 a0 P& ?) A2 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

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

% |/ D" E0 j7 I8 {7 V2 m( J1 c: k+ ^) F
这样算混淆了微分和导数的概念。
; `* A- N6 a' A2 ?5 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

: \& v! H- V4 f5 ^* h1 GAh, 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


; k9 H9 T, f! Z* q' U! O2 }* fIf 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)
$ b. v0 n+ n- {  X& A( h9 u
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
; m2 Z: i2 {, A8 e" i' @  j* o1 V" \; E
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。
; ?6 ~* M4 j9 ^% {1 n/ ~
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
2 e6 `, {  c1 p
6 ?" ^4 a8 I* P5 C6 m. rAh, you made a mistake: it should be:

- O$ E' ^+ b: l. T" D: y3 Kd(f(x)g(x)) ...

  _6 A6 j7 {& t* h( q" ?" Z
; F  c) h4 K8 i4 zYes, 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:
+ O: R; g$ B) fYes, 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.

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:

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 ...

! [' M5 F6 g2 E. m
5 G0 W( T" [6 i% l$ x* R( `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？

; o9 H9 U" t; h9 E3 f2 G% Y; |4 O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。