数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
. Q) k1 W5 u3 u
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?
, X' A& }8 S# ^: c
: |" K. @- j6 v8 g  h) w; Q多谢了！
$ |0 g5 g6 r, w. z- W4 m
9 r+ ~9 V: p& ~& k  Z  w[ 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！
2 n' M+ S! e& z
d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

, y  Y. c' V% F: C. i3 \9 y7 w* w多谢了！

9 n6 g* `& ]( R" C" q! F
it is too easy:

d(a+bx)/dx = b

so

b=-kc+s
) p1 _+ ]3 J2 h+ I4 k0 v
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
* p$ ?0 ^6 v- l; F4 Nit is too easy:

d(a+bx)/dx = b

  V5 ]" H/ |! z( Mso

$ l8 G7 ?) t4 ]! e4 db=-kc+s

7 ~+ I+ K/ F5 x- a$ \2 vfinally:  c= (s-b)/k

/ ]; @. T, b( c+ |1 kthe 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！

/ Y& _1 ^% B. |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 s9 H0 f' {$ V6 b- m9 G, c6 d
+ v* f/ t' Y0 q多谢了！

[ Last edited by 醉酒 ...

2 z" [% A; e1 \
sorry, did you post another question? your statement is still not clear.

9 j* [$ E) O) A6 Xa, 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:
4 h; E& k! i0 d4 g$ B5 b+ p1 Osorry, did you post another question? your statement is still not clear.
/ G) g+ x3 z1 A# U+ T" A! B
3 D: A3 b1 A# ?5 W$ j& Y& ^a, b, s, k are  constants? c means c(x) ?
5 y2 C& k; c3 }$ h6 z
- `+ U. B0 ?! d- W* Y4 Eplease tell me.

- K  T" v; l" I0 U- h9 ^& I对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
. L, h% \* c( c
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
$ ]! U, J6 j9 wd [(a+bx) c] /dx = -kc +s

多谢

' o$ F6 k# c* g+ @' I. Z# P6 X' N! |
2 k6 I) h1 o3 udo you mean it is:
: B) B, u. ]8 S. x6 H) L0 e
d { (a+bx) C(x)}/dx = -k C(x) + s
1 x. d. d, O- h, k
" O  |) i5 p* N% T, Ewhere a, b, s are constants, you want to know what is C(x) ?
1 \" a& M/ `8 w8 K' t
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:
0 I$ [2 f6 R2 }  |do you mean it is:

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

" }/ F0 x, [7 q/ t9 u7 ^where 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 ...

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:

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

十年移民路_

this answer is the good one.
5 @7 P" x+ C2 P5 w1 o% {8 k

; U2 J2 X$ B  Aprocedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
9 b2 i/ M# q7 z- B; qso:
! a  l8 x  F3 V, W$ d5 t0 ?
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(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:
* R8 u% T- Y4 L: g& q2 ?
{(a+bx)/K} dY(x)/dx=Y(x)

# C, Q- x/ t9 F5 u3 U8 b% qfrom here, we can get:
2 Z7 |4 U/ u% Y6 w9 r" F% {% ]
6 Y# s' c: Q4 g+ ~! C6 TdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
3 Q$ b) f2 P# z4 L! y
so that:   ln Y(x) =( K/b) ln(a+bx)

4 j) T8 _. C" }: qthis means:  Y(x) = (a+bx)^(K/b)
' i: M6 C3 K, {: Qby using early transform, we can have:

! v& j: J: i5 m6 W-(k+b)C(x)+s = (a+bx)^(k/b+1)
# Z# u2 i& m2 \; [7 s6 e% L; \
  s3 Y8 b8 }/ H% Efinally:

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！

) z: u/ q+ W( O) 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。
5 u" K, S8 H. N: t& y8 b5 B" e您说：
3 [3 h& w5 }% K) F, Z% R( X4 k
  s+ h, G/ i+ Ad{(a+bx)*C(x)}/dx =-k C(x) + s
' y8 [9 x: d! |4 F0 t4 C0 i, lso:
% u5 E! s: I* GbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

4 p% w% Q% G+ N3 Q% o$ }3 [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)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
8 w2 s  A" u) t0 t9 J; W3 B并不是   (a+bx)*C(x)   的导数除以   x   的导数。
& ^6 M  X; n* l% Y3 d' a比如说   dx/dy = (x'y-xy')/y2
) ?& s9 p2 B9 h2 o9 W9 |x' 意思是x的导数，y'意思是y的导数。分母是y的平方。
0 b$ U% [. B' o4 X0 y. a因此：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
" n$ o" a  c$ l: B$ w. P4 n最右边是x的平方。
  j$ ~3 V' U9 z7 U
所以您解的有问题。
# ?3 U! m, F! O- n; j9 a$ d, ?
[ 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:
- y- u- N( U4 [5 U1 G" `9楼的答案有点问题吧。
+ W0 s! B+ j2 \; e/ R) Q7 t" R/ w2 m您说：

( |/ h9 y6 H; L. ?- \d{(a+bx)*C(x)}/dx =-k C(x) + s
' h$ I; y- N- w6 J# @2 O" c! }so:
9 {- T7 e% e4 y3 ]% u' pbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
& Y7 e9 c6 F- k5 P
9 Y2 O* b8 F$ ~也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
( t5 \# c/ K: `; j' C但这是不正确的
# p2 _! g0 A3 ?. O2 U7 e+ @! Jd{(a+bx)* ...

/ J* M5 S& n5 J5 ^  W9 }( zPlease note here:
5 z; Z& ^: H$ I& y+ i6 K% i
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
6 c; D% l; L1 p6 V4 \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。

! m9 k! V  E1 U/ E! g1 i3 P$ 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
" p4 T6 Q, m( y7 e6 K6 b....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

这样算混淆了微分和导数的概念。

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 ?8 c0 f8 I- S8 a
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)

" a/ G+ F; w/ w[ 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。
( X; ~7 C, I, N& W$ L' e3 R
& `/ V6 c& e1 F9 ?# B+ |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
- ^0 G+ B' {0 `) Z3 T
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

/ |3 M. m* Q. {3 x这样算混淆了微分和导数的概念。
+ q$ ?, C4 U0 [# E4 D* q9 g! 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
# {+ L+ m6 M( u2 P& D( K- n: T
Ah, you made a mistake: it should be:

/ [" t; _/ P0 R0 Ud(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
7 {/ c6 O4 {! p" q$ ]6 g, x

  n; X2 u: b5 U1 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)
/ a% Y6 w4 N6 i! R. }% E
* H# {1 l0 t: ufor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

+ T  w3 ^+ `, I# t" Yh(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。

, I0 f6 b  Y* Y0 T/ J. B; u# X1 C' Ld(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

+ F* Z2 d8 [; }/ Y/ u7 OAh, you made a mistake: it should be:
3 N5 [2 ~6 A, R9 `
d(f(x)g(x)) ...

% h; @6 _: b* l( y; f. B6 k/ j8 |
' `; l& C8 `8 b. q1 qYes, there should not be " ' " after "f" and "g", it is a typo. :D

# X1 |# C4 A8 j+ ~. YBut 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:
3 d3 M0 M4 f' {/ G' b& ?, T. D1 Z, UYes, there should not be " ' " after "f" and "g", it is a typo. :D
( ~# `" Z9 N) @- _' z
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

% n; C1 X/ b% ~1 c3 U6 T. ^
2 Q: g+ _8 g( j9 c9 iThanks, 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:

# n& C- b( Z; D" ]4 ^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？

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。