数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

0 N& N% o% k% gd [(a+bx) ] /dx = -kc +s
, w; U  z/ E$ X( n3 iwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

- B" p- W3 e: A- D2 l% w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:
请教大家一道微分题，多谢！
, a- q- ?2 b1 n
d [(a+bx) ] /dx = -kc +s
( Q* n& Z: I3 r1 ^where: only x and c are unknown, others are all known,  requiire c = ?
8 A6 a0 [0 }* E2 r
6 V2 f0 f# o3 c; t# q$ [# t7 z多谢了！

  @) L" k6 r  K  i' z
* Q% e* g2 k% ]1 xit is too easy:
$ X( W% W. d- ~( ?9 y
5 D9 M6 O  ~0 [9 M+ ~, n& yd(a+bx)/dx = b

( G* E- E; N( d. t: H3 g4 Mso

b=-kc+s
( P0 J1 p4 o7 p5 E
- C- V7 h7 N* K; _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:
) o$ O! S6 K' g- n) T7 U7 Q& b  A
d(a+bx)/dx = b

7 H: p/ ?- \2 f5 Y  D" Lso

b=-kc+s

finally:  c= (s-b)/k

# c2 R* Z& i2 `2 }1 {
. G, ^3 }5 Y. S* H& \1 H; a( ~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！
1 |! _: P/ l( [0 Q' `- N( C
7 P6 {. s  Z& A: {9 J; n( cd [(a+bx) ] /dx = -kc +s
& ^, G0 [! v3 Qwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

# i! k- ~0 O- B$ j$ _" c[ Last edited by 醉酒 ...

$ _0 K5 Y9 n1 y& o) ^9 fsorry, did you post another question? your statement is still not clear.
8 n  ?$ a1 `3 n# B
a, 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:
$ ?# B5 d8 v6 v  B% l7 Isorry, did you post another question? your statement is still not clear.

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

please tell me.

- L" U( r5 _; ~; ~' h对不起，写错了，忘了变量c,应该是
, f- I: c) U1 x* x6 p; t8 q; W+ j. L5 W' Od [(a+bx) c] /dx = -kc +s
) _- M  e' O6 {
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
! w( |- l2 \* v" Rd [(a+bx) c] /dx = -kc +s

多谢

7 c: `* Q* F0 |4 Y3 Q: v
do you mean it is:
+ o) J2 L) g1 E$ E% A  ^
4 z: }( P5 u' @4 Gd { (a+bx) C(x)}/dx = -k C(x) + s

' b; L( W. y4 T2 g& \where a, b, s are constants, you want to know what is C(x) ?

  e: W7 Q5 V5 k' W7 hif 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:
' F$ U* Z$ f+ Rdo you mean it is:

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

; m. {3 g. V5 wwhere a, b, s are constants, you want to know what is C(x) ?

1 }: b5 o" @; I: h$ c1 }, U# Zif so, this needs to solve the differential equation. please comfirm it fi ...

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:
/ v( J) g' e' C/ i
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

( t; v0 S$ ^9 T- L+ M% _
( q+ S/ M! D  F" k1 r9 Y/ wprocedure:
) h: ]* M+ W! W. ~2 R+ P
+ t3 Y8 d2 J) i  B4 H) xFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
8 a1 K1 H+ y- _; H, `9 ~i.e.
8 B; t, j5 H# E, h6 M0 X: V/ v
) o6 V3 T* m* {: O+ h2 B(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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
therefore:
, ^5 t( L2 a2 v& h7 Y
: Y) t$ Q$ B% T; F1 ~{(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)

so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
' R+ z5 R" E$ Y6 ^1 ^
-(k+b)C(x)+s = (a+bx)^(k/b+1)
# @+ r6 R; \6 H1 G( Y7 I
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:
4 E/ n4 Q0 j0 r0 G: J( E% y$ Q) ~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。
. E$ E  o1 x+ G7 H* 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）。这种题真的不是难题。

龙舞九天

9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
  k4 ~+ y1 [* u, Y8 yso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
& u, q1 ^+ w( o! y; 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
& ?8 i( E& D" C, w/ Ud{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
  @9 ]# c1 l' ]) B, c4 n! Q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
7 B4 F: {' s/ L5 H" ^; Wx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：
6 H5 j* V% p* G
( m# B5 G; a# u8 ld{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
. ?8 x$ O  s/ [  Y# P8 j$ w% j$ G( n最右边是x的平方。
( c; {- q% z4 ?; T6 a/ b6 C: U
, k. [; g4 S" k所以您解的有问题。

( H2 N1 G& A( H3 i' G: b! q9 O+ h[ 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。
您说：
. S: b' p( m- @' z/ n- S0 L
/ N4 b8 S% E" B) G6 Kd{(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
d{(a+bx)* ...

( X2 u' `& I) L: s& A1 D. z/ z6 p
3 F) R4 M/ }# U& y& m  f* DPlease note here:
( k5 |8 b8 e1 }' E
- L) i" g6 j6 Z+ d/ ad{(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
: b2 s& ^6 H; R并不是   (a+bx)*C(x)   的导数除以   x   的导数。
) {- x4 ^' }& M5 S
; A; ^( c( i, s, I, _: tno "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

+ ~" W5 p* d+ W% P7 C7 h6 ~6 H2 G
( b& Y0 B  Q+ e. O: |6 q. g这样算混淆了微分和导数的概念。

, \# \$ w/ x$ h1 w5 r1 k3 O8 xd(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
0 M% C7 w* f7 p$ H- ?
6 b7 J3 H8 F: 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)
3 a7 d/ w; z( N0 Z
[ 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。
9 d0 E! g$ `+ {7 K; M
6 d3 A- b* a/ A- Fd(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 ...

5 ~9 M4 @2 Z* m1 ?6 B( i
" n6 w. K$ {9 c  j( p这样算混淆了微分和导数的概念。
( J9 i) W( P6 }$ ~5 U) g: j
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
6 j& U* J5 M2 B/ D( `  T
3 y+ w  T6 F0 ^Ah, you made a mistake: it should be:
8 I+ B; E$ E) [* n
. k. R0 {& p- `: b3 gd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx


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)

3 V0 i6 H/ U# U% X6 Pfor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

. |$ n; a6 C+ s- ~& q" th(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。

+ w9 l0 T2 U; I8 i5 z3 a  Gd(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
+ i9 J8 V0 t- f4 [
0 s2 h7 T  m* q7 l3 n2 {% |3 BAh, you made a mistake: it should be:
+ M+ d( y. P2 g" f' b
+ [$ J: }5 n! ?; `% w( td(f(x)g(x)) ...

* Q# K. Z7 W" Z% s/ MYes, there should not be " ' " after "f" and "g", it is a typo. :D
! r2 T0 T; ~, ^+ c
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:
+ M, t4 l7 o+ {5 XYes, 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。
, _; w/ H3 m5 N4 [7 _# `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:
" W5 c$ `" G- H; P% R. A$ 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？

% E4 Y# z: j+ V' Z* b, h5 l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。