数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

( |; Z" H$ p3 ?; V; G  c9 O3 W% id [(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?
) p, V( x6 o4 Z
. r& R4 n, [, u7 P4 K多谢了！

- b+ @# D8 Q% O: b! i) |[ 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！
* X" w3 J! {; a. i6 Y( i/ e
& ?# y8 w8 k4 t. Hd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
8 v" Y: m% {( {* e
5 s0 W& r! q$ x4 D( }' h多谢了！

2 d) M; n2 Q, w
! U7 k6 W% f1 yit is too easy:
: {& w. \4 N9 u- i. D7 {: v
d(a+bx)/dx = b

so

/ j. M1 Y( a) K& |4 L0 @* Tb=-kc+s
  v& H- d# i) |: E
1 L" K1 }7 p' ]" R* \4 Efinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
5 D/ q- z  P' `2 n! i4 ?( bit is too easy:

: ~% e* Y6 H& K# l5 Z0 ~; Gd(a+bx)/dx = b
" Z& p9 E; q) N; G; l
: @3 a# |$ g* k1 iso

b=-kc+s

$ S0 F4 `0 |0 B# v# J$ n1 H! F  kfinally:  c= (s-b)/k

7 |3 @. M" q3 n* |9 ?4 \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:
% {0 Z. S1 H# H$ M. r6 p请教大家一道微分题，多谢！
! [+ R5 j1 K& I, J) M
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?

( G/ k/ x: g! q: N" o3 j. b多谢了！

[ Last edited by 醉酒 ...

% K  }/ p( H9 {  \2 l' z" }* n' ssorry, did you post another question? your statement is still not clear.

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:
* g. F0 r1 q& U! P. B- Q" dsorry, did you post another question? your statement is still not clear.
/ E, N3 S% T( E: Q
7 F# A1 O2 \" q0 R/ o" ca, b, s, k are  constants? c means c(x) ?

please tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
- O  M8 o4 T$ y. {$ t8 y8 X
/ P# {, ?( i3 i/ V+ r: w! k' S3 i多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
4 X* ]5 |; @- D7 ^/ E- |+ I) M9 i对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
6 V, x8 }. I: B
多谢

7 m: A/ `0 T, h1 n' |! ~1 Y: ?2 `
do you mean it is:
. R4 o4 `9 u1 o* A3 `7 E7 T) _, I
d { (a+bx) C(x)}/dx = -k C(x) + s

) f0 J; A; w# b$ L+ Pwhere a, b, s are constants, you want to know what is C(x) ?
9 u+ S, a' e  P2 l, Y
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:
7 a0 w$ _& U$ [- z) L2 x1 [
0 g& J& P: U  H# C' P6 y1 td { (a+bx) C(x)}/dx = -k C(x) + s
9 H$ q1 M6 |7 o" W" d
where a, b, s are constants, you want to know what is C(x) ?
+ R' H0 \1 a* q
if so, this needs to solve the differential equation. please comfirm it fi ...

. w* j9 o2 c! Y' z5 u5 V
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:
/ `/ u9 l# N4 Z
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
5 B" G# V1 M% ^

procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
5 r/ v+ p! M8 K: ~
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
1 D( u0 `# k/ `% B7 S8 V& ~
2 t5 B; z5 ?- ~(a+bx) dC(x)/dx  = -(k+b)C(x) +s
, d, t% _* K1 S* ?. \

$ v2 ^6 C7 N# [9 D  @3 G! ]% W( E, }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
& c! X+ b) ~7 s* i! Ztherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
5 Q  K  f3 z1 D9 g; h( u5 Q
4 B$ r6 @1 [2 rfrom here, we can get:

. C) Z( P9 I" \5 B5 e, L( F8 d9 L% KdY(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)

1 k' A. a0 J. V& u& I8 L, K# Kthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

. P7 y, d" Q+ Z$ R1 C1 w( Z-(k+b)C(x)+s = (a+bx)^(k/b+1)
7 D( D7 M7 k  c- ?9 @, }
finally:
1 c6 z4 Y3 G6 Z# u+ a. D
9 @& U  n, P3 GC(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！

6 C  w* M1 T* Z* R& S% C  O
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。
/ Q& e7 A: {! a% o8 J- v您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
# A- B; c: i8 W- Rso:
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)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
( b* @' B( K& G' E5 ]. G4 [4 b4 g并不是   (a+bx)*C(x)   的导数除以   x   的导数。
+ [$ N6 Q& A/ Y比如说   dx/dy = (x'y-xy')/y2
- ^$ M) x- s5 i9 ?- L$ i% y: I& kx' 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。
0 I* f: m( N1 o
所以您解的有问题。
$ K0 ^- F! A' L1 D
& C7 G8 _* K7 U6 M1 j5 p4 l& {[ 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：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
: Y3 O" Q  F) {( g( {' R+ x  D6 RbC(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
/ T5 d( e  h& c7 l* K3 ~但这是不正确的
d{(a+bx)* ...

Please note here:

( m' ]# W, `! P" k( k3 u! nd{(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
+ [1 R7 ~# J% v  @/ G9 }! h并不是   (a+bx)*C(x)   的导数除以   x   的导数。

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
$ y. O" f9 C: o....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

- V1 C% p" P# }8 k
0 b% x4 L$ M1 l1 k) B# A这样算混淆了微分和导数的概念。
! q% e+ {- ]+ j" t. ?& o. G/ b  J- 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
0 X; E- u& B" M" t' h+ ]
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。

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 ...

1 s1 S: K/ c/ }: N. ^; {这样算混淆了微分和导数的概念。
9 s4 _; @! R3 {/ G
3 x5 `1 A" N/ i9 {- |7 h& Bd(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:
" z, k$ W: s% N) s, t* J
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
' {. d5 \) M& I6 B4 E, 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)

5 c5 J7 d& D) m  _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:
7 r# Q) K6 d8 \3 J这样算混淆了微分和导数的概念。
8 a( F8 f0 o; d+ A* P% O, f8 O: \" Y  g
0 Y  G& B( Z5 ]7 L/ k" R6 Dd(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
3 c* C" D7 U* S* T# T* d
; z; v3 U* |7 h3 ~Ah, you made a mistake: it should be:
% j8 l! [* b/ }! e7 ^( `
7 Q( m, j9 O7 s  I' bd(f(x)g(x)) ...

) d0 n# f) a+ m9 o
2 N( [, L4 {: J9 F/ E1 L5 |Yes, there should not be " ' " after "f" and "g", it is a typo. :D
, N2 R( O. L( }: X" l! p3 @
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:
, J8 H* q6 p6 p# D! `, q6 ]Yes, there should not be " ' " after "f" and "g", it is a typo. :D
: p* @' t) G+ W- B  B7 [
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

; o4 }, o& R6 E, ?0 }
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:

5 N* L7 @2 \$ F; W. 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 ...

) n! @2 z  u2 {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:
4 o0 U# j( H3 h; W$ n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？

2 @& x' V. `$ B5 C
. n% ~* V/ t0 `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。