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?

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！

d [(a+bx) ] /dx = -kc +s
9 l/ K) H2 `) t' D! ]/ m3 Jwhere: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

0 y0 B& D# w' mit is too easy:

d(a+bx)/dx = b

so

b=-kc+s
  l8 y6 j, ]. D2 `1 ~; I! k
3 ~/ {% f" H( ^: T- [+ O5 mfinally:  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:
2 \  o7 }" x  d! A  H
% ~  d# }) I. h+ p. z- y4 W* n2 md(a+bx)/dx = b
% e) E' }9 [0 `
so

/ K" H  R0 D' i) C, {4 o: ]b=-kc+s

) c, _+ |  ~8 W% V& Tfinally:  c= (s-b)/k

1 w' ]0 k3 U0 S3 H5 g, bthe 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:
( c& w! @* P7 [请教大家一道微分题，多谢！
: T4 D4 b0 ^  C. W* M$ I* x- F2 W3 n
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?
& K: r7 k; F, C5 y
多谢了！

2 X' h8 I# V, k! v0 o0 f* M[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.

% I  w8 d. K$ Ca, b, s, k are  constants? c means c(x) ?

8 N4 ~/ k; e* q# m0 `! |/ aplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
- H0 Q7 U* V- A0 ~sorry, did you post another question? your statement is still not clear.

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

! k- s0 I) _* [! `& vplease tell me.

' h+ J2 e( `' N. I对不起，写错了，忘了变量c,应该是
  L! ^) C' J# Yd [(a+bx) c] /dx = -kc +s
' K3 A9 C: a$ L! |
: i& u0 H) X4 t# e/ g1 `多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
% i; x1 b) Q7 g- |) t) \3 R3 T对不起，写错了，忘了变量c,应该是
# ^5 q3 h4 F, r. e% {3 ad [(a+bx) c] /dx = -kc +s

多谢

' i1 q. T3 d! i- L8 r
do you mean it is:
7 O" w0 Y" H' E+ ^. x
5 l7 H& k& v' Od { (a+bx) C(x)}/dx = -k C(x) + s
8 V5 N8 u6 |3 t: }/ i0 r% v8 v! ^
where a, b, s are constants, you want to know what is C(x) ?

+ A  j8 ~% I* c8 W0 S7 u5 c4 bif 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:
. c0 K, p- F1 A6 ~5 Q
7 {5 z6 z8 v. b, }( V; o- F1 md { (a+bx) C(x)}/dx = -k C(x) + s
' ]8 T5 v- P8 |  A4 y
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 ...

) \8 ]2 e+ a; N) n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. i: h) t( T3 vC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
$ U8 r4 A& n- H7 _7 a+ m. j1 `# X- ?
! D) y7 ^3 Y+ L: N9 n) h  Z
procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
( G$ h- A# ]  O# j6 Z$ |
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
/ j/ N6 \2 i" s3 ui.e.
! d( ^3 P, H8 e$ H4 U' x$ k
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

' ?) F  T$ C5 M. b: h, _+ i
( [. b+ c9 ^* I: _* K3 Q  dintroduce 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:
, k' b3 O9 T2 _0 U5 R2 u( P
4 `4 Q+ v' i8 a9 l+ K1 F) C2 J{(a+bx)/K} dY(x)/dx=Y(x)

8 i0 @6 m9 N7 wfrom here, we can get:
0 M( u9 U& G- ?- L$ ]6 O9 S/ l: i! y" `
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

/ ?$ u1 g) T5 {- w, U' T, tso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

. d5 \- c4 v8 D( x8 P* \) Q4 M) `-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

) P6 b# [' V9 T) \# c3 MC(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:
$ ?, ^/ V2 ^; M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！

1 b) y: s' j1 v; m' B
3 c; F5 K& M0 |' @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。
  \9 r% t& L. _! m2 }8 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：
( r4 e/ u" l1 W4 X6 z
# X- ?; Y6 X; Td{(a+bx)*C(x)}/dx =-k C(x) + s
! x: W& z* g6 Y% M! 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
( S( i: K$ p: @% y% l6 f* C5 t但这是不正确的
1 d6 L! B/ v9 v- X6 g' w* L4 _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。
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：
7 D# [: E2 b5 L6 b0 W7 C4 x$ @
8 N6 u" u- G" g! w; B4 G) td{(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。

1 O6 I' R9 Y* p6 W所以您解的有问题。
" [" r! M& [% m2 Z, K# `0 {& z
/ ]) N6 W$ N; g9 Z/ q% y7 m[ 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:
+ G; k5 b2 F9 X* k5 c0 W9楼的答案有点问题吧。
您说：
& U. c3 E7 `7 h* _1 g9 m
d{(a+bx)*C(x)}/dx =-k C(x) + s
+ K6 Y( q7 K: W& X% Jso:
6 p- j# m) w- R9 K' ]. qbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
7 S7 U2 {8 u$ |0 Z, S  M% O. \
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
: C8 k. F/ `, a9 [. Z但这是不正确的
d{(a+bx)* ...

8 U! V* M3 b2 E5 p$ R" L  V: j0 W
3 K0 I* n* \3 R+ A7 w4 X9 |Please note here:
2 C3 w4 n% U& c$ j5 j
7 v6 D4 H1 X; E7 o( t$ M! dd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
+ b' W/ b* ]- R# J3 d3 ]6 mit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
8 d: A' [+ ~6 S. Xd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
  q$ J. A/ v& S$ y- ^) }: @并不是   (a+bx)*C(x)   的导数除以   x   的导数。
& k6 C! x% H  C0 ?" e: [
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
....
8 T+ f% P  @- h) ld{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

; N1 T. M+ e# B" S7 j2 [  l这样算混淆了微分和导数的概念。
5 U7 z6 P9 V% t6 n
% l+ P; \6 A/ U4 K1 r0 X, Yd(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
! K( C  T$ C; c1 j; I  U9 D. d
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)
) I- s- d" d* Y2 {* \. \* H
[ 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。
1 @; _, g$ F% r- u
- A  @1 H; l5 D& ~; r  ad(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 |: P( [( ]5 ~. i& wIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

4 y) S6 V# O4 ?: N! g9 C这样算混淆了微分和导数的概念。
' V7 I' ~# d9 F" X0 t- Y& E
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

Ah, you made a mistake: it should be:

% Y1 Q7 b* C1 b& m: T  W( ud(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
5 M2 V1 ^3 ]3 p- i! d5 N

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)

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:
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
+ C7 r# O$ ]3 v& \9 a! N8 }4 p
Ah, you made a mistake: it should be:
( Y2 r! }% l1 S- D7 s% {
  A. c, k) x  ]& B; M- K7 z+ sd(f(x)g(x)) ...

Yes, there should not be " ' " after "f" and "g", it is a typo. :D
+ X5 B7 f( f5 v- d
- Y  b! L6 h+ g% q% ~, gBut 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:
# M8 j" l; p7 H$ tYes, 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.

+ f* [5 b% Q8 |4 x3 @+ h: W/ gThanks, 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:
# U  d; g4 c# T
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 ...

# c, _6 J) z* e/ A$ M1 s0 L
+ L" v6 L& |1 t5 o% B: S5 |) }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？

( t* T5 A2 \2 K: l" [" D2 t
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。