数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
- N9 p8 K; f. ~# }0 l0 w3 @3 z
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！
4 |3 J( `1 j" z; U% _( B2 y
[ 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:
6 |! m7 ?9 z6 V请教大家一道微分题，多谢！
: i, _8 z" F1 r0 p
7 B: t6 B9 J2 r6 K1 b$ nd [(a+bx) ] /dx = -kc +s
6 v" ^8 c6 A2 F( E6 P- Q$ _+ dwhere: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

9 [3 z# r( T' k6 e  V' Jit is too easy:

d(a+bx)/dx = b

5 n9 p) {9 D# M1 cso

b=-kc+s

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:

" o0 @0 q5 E. ^% Yd(a+bx)/dx = b
2 D. E/ Y7 E8 z1 g9 K
so

% _$ G( O# Z- B. [+ b; I/ pb=-kc+s
6 L$ }: q6 j( O3 ^" ~. ?
finally:  c= (s-b)/k

! E5 M# u2 y# H1 T
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:
9 w5 O" _; [" K/ F' @1 {请教大家一道微分题，多谢！
% J2 N& R' ?2 v- E; ^* Y. ~9 k8 I
d [(a+bx) ] /dx = -kc +s
3 T+ E9 c  W' I, b1 |where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

( Z) a2 c# G* j% [( h5 Z" Y多谢了！
# y& q( E6 l" L! A0 p3 l7 ~
- r% U# K- V5 e- v' n+ I[ Last edited by 醉酒 ...

% a1 U* {* j9 a* ]: \7 R. N
sorry, did you post another question? your statement is still not clear.
" E. b: j- A; {
" c' u( ?1 D9 X2 L/ F" o0 Qa, b, s, k are  constants? c means c(x) ?
& u6 n9 }, K. Y' Z" j% U1 ~% R' j
9 Z/ P5 Y+ _; [. Q5 ^please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
) O0 b" j3 E9 ?& T+ x7 Esorry, did you post another question? your statement is still not clear.
5 P5 X4 J1 G- _+ }9 a' m& z+ ~0 ]
7 a! k) W$ k( la, b, s, k are  constants? c means c(x) ?

+ f, Q4 R. P5 n7 r, M5 ?; f5 lplease tell me.

对不起，写错了，忘了变量c,应该是
8 o6 r: t. w8 [2 u7 l- zd [(a+bx) c] /dx = -kc +s
9 n% |3 {- n4 ]" i* s
& i/ ~3 W) Q' \: j多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
$ f, W# T9 C' Q$ S- T1 I9 N对不起，写错了，忘了变量c,应该是
. O# g8 ?+ T, h! n. Q9 A5 Y+ K4 I* O  Yd [(a+bx) c] /dx = -kc +s
5 s0 W6 _* @- k0 {' e  }8 k
多谢

4 k3 }8 V$ Z3 kdo you mean it is:

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

where a, b, s are constants, you want to know what is C(x) ?
5 A3 Q; m/ G, i, x% O' h8 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:
do you mean it is:

' N4 Z4 _  }' g+ {" h% q: md { (a+bx) C(x)}/dx = -k C(x) + s
; `1 {" _. a0 o( Q# A( u
where a, b, s are constants, you want to know what is C(x) ?
7 \9 |; y% \3 |- \8 B# Q7 x
if so, this needs to solve the differential equation. please comfirm it fi ...

4 `- _) i7 Q. L. C1 n# z* M
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:
/ A8 Z: O0 {' n$ b
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


procedure:

4 r" H2 Y" a( A# H6 ^- L3 b4 VFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
" e. d% x0 Z6 k* Kso:

1 B+ o6 {  J$ b9 e# x# e' dbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
; o* d& B' c% k. ai.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


2 l1 n) u' W! h/ y3 `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
6 r5 ~3 A# n0 Z# Ntherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

5 l2 g$ Z; v  sfrom here, we can get:
% w, ^, H) O& a3 n+ J4 S$ G+ G
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
7 `) w, O. Y7 A1 z$ B: \$ [. Q
so that:   ln Y(x) =( K/b) ln(a+bx)
4 }  C1 i. h1 Z; j0 r  c
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
* \9 G6 X, l: K8 V% r( A$ B
. o! w' w& r/ B-(k+b)C(x)+s = (a+bx)^(k/b+1)
. k/ Z. ^/ _; o8 O- n% V
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！

+ a& G5 m; Z) J6 [+ S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。
% ~, ~& I- P: o/ p. @6 _1 v您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
5 w  a3 y( Q/ F+ [- U( E. QbC(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)*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。
" P  v$ M1 R% s8 b: D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：
* Z' t. l& n& b, ?2 m0 P
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
0 ?. b1 d) u) s5 R最右边是x的平方。

  x# M! q3 ]  S& ?所以您解的有问题。

* F, o1 p" G9 a[ 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。
7 u! K+ ?8 Z# h3 R您说：
+ |7 u( W8 ?) m3 {9 J
5 c. Y% W. C3 K5 L! E4 h& zd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
% f* g/ j; W! D4 u7 Q
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
9 o) P' d+ n- K; d6 Q' ~' ]3 Vd{(a+bx)* ...

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
+ I: g  V$ |/ \  Cd{(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。
" k& J& N4 f4 k$ G3 U
7 t3 \9 E3 |" t* \. W1 vno "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
....
* M- A; ~5 \+ L: w+ t/ Y+ X$ kd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

  {" w& A7 V5 Z) J这样算混淆了微分和导数的概念。
( o: C, L6 ~3 |2 H/ t( t  |
" o. g) p$ f) r$ V) O9 F2 Q" g* td(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 T2 Z2 a% f: F$ I' j- W
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)
2 T3 r7 W% a% [1 Y0 ]! w( X
3 Z! J: J1 W5 s6 F7 l[ 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。
4 c0 C0 W% M4 Z! ]- v8 G# X8 {
! n5 _3 Y& A0 j3 d2 u% P$ l; Nd(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

- I8 m! I: C' \5 Q; k$ M. MIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

9 R2 D/ ~6 e! z' i5 Z% S% w
. _/ Y7 T2 X, M! i& g( y2 x- S7 _这样算混淆了微分和导数的概念。
% U9 r3 s$ W  E9 J5 t- x" x& 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

Ah, you made a mistake: it should be:
7 f- W0 W0 g% m) m3 x
* {) w7 |3 e. J/ @3 V" dd(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)
% m% A4 j( v- J. A& R1 B2 {2 S
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
. P7 z# p( \% |* m( H
h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
$ a4 A& r1 N' V( I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 E) Q4 z+ Q9 l6 Q* r# h$ TAh, you made a mistake: it should be:
7 O+ Z2 R/ a8 x/ P! Y( m0 ]
d(f(x)g(x)) ...

+ l! Y; v8 f2 Q6 uYes, there should not be " ' " after "f" and "g", it is a typo. :D
2 m4 Y! L+ T# B( I$ }% w
5 m0 R# K( O, Q6 U2 u' \1 D$ mBut 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:
; W5 L- R, Q/ b, j' u, ^* {Yes, there should not be " ' " after "f" and "g", it is a typo. :D

7 p8 @9 h7 R, z( {4 b- aBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

. c- m' `3 D; G1 NThanks, 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:

0 @6 A0 o1 H- c0 |' d! Q# g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 ...

1 S& G1 j8 D) \
! Y* U, K0 @% H$ d1 i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。