数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
% e% R4 [: S+ o* j( F
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！
5 ?6 G2 D' w! z) q
7 u( f5 ^' S4 M) ^7 n[ 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！
6 R' m/ u# J6 b$ Y7 o
d [(a+bx) ] /dx = -kc +s
: K  ]6 m8 ~. m4 k, x. [where: only x and c are unknown, others are all known,  requiire c = ?
. B" E6 `5 C( U% R6 O. c6 a
多谢了！

it is too easy:
$ u2 A* }" W4 Z" a" i0 ~. j# n
/ `- c, G5 @: s  Z% Zd(a+bx)/dx = b
3 A" q% }8 H3 @2 D' e3 ?; R+ r# P
/ |4 M3 l* h( kso
! C4 z, A1 C2 ]! B- \  L
b=-kc+s

/ b0 q7 g; D. A6 Bfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
% s8 m6 Y3 v9 g( G/ ?4 m* S* vit is too easy:
9 V$ J4 Y5 [2 ~! M2 H
6 e$ u" O: V* k$ A3 i7 bd(a+bx)/dx = b

  E& j& g0 U" O5 Aso

0 ]$ o. q/ Z' |b=-kc+s
/ A1 |0 p2 N- G' z5 |: U
. A' `( \  q6 W# a/ L* \finally:  c= (s-b)/k

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:
* p# t: L# {  B8 ^6 f3 \& ]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！
  ]+ e' H9 X8 p" `6 j
[ Last edited by 醉酒 ...

7 m$ T$ j0 h! ^" r" J) W
: n% o! v& ?7 m! z1 H: F  @sorry, did you post another question? your statement is still not clear.
( S" f/ v! b  s: D& c
a, b, s, k are  constants? c means c(x) ?
3 N4 R0 @0 S0 k& f) N
7 n6 F0 z9 q4 W, W8 uplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
sorry, did you post another question? your statement is still not clear.
5 p% H) G# ?/ ?% v
2 ~" B  G1 f9 b1 c9 Fa, b, s, k are  constants? c means c(x) ?

3 _( G8 m* O% {; R+ Nplease tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
2 r( [: r; e  w5 P( D6 V5 z% P对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
( U: J' O% ]' b" M0 m9 Z
多谢

do you mean it is:

$ L# {# L, x) {; Q; zd { (a+bx) C(x)}/dx = -k C(x) + s
4 G7 Q, [3 I" i; v# z6 E
where a, b, s are constants, you want to know what is C(x) ?
6 ^& B3 @" E- x8 K
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:
  q7 s4 W" t& B2 e% Ndo you mean it is:
6 X9 s' ?- Y) ?5 U# C. v6 G: K
6 ]# i$ J- U4 D* [d { (a+bx) C(x)}/dx = -k C(x) + s

" [& H% t, I5 w" l: f+ K0 l4 Uwhere a, b, s are constants, you want to know what is C(x) ?

1 `  F9 W% h- E9 N& B1 aif so, this needs to solve the differential equation. please comfirm it fi ...

2 k* R( j9 V' e* Z
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 b1 j1 [. ]" O: G1 t/ b" b& e
2 U! T7 R+ f6 q, s9 uprocedure:

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

, l* v3 r& l( I. E* ]3 ObC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
& @% C: }' J: Y" M+ O7 b' b! W% T: E
7 s5 i- G! [2 E% m9 G$ Y(a+bx) dC(x)/dx  = -(k+b)C(x) +s

( }5 ~) `/ k2 ?1 I* A/ {# C
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
2 M6 E2 l2 i6 u) s4 Zwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
, H7 }# @3 @/ S) a  {7 K$ T; L0 S
{(a+bx)/K} dY(x)/dx=Y(x)
+ ]2 h4 t# b; s9 |5 W
from here, we can get:
4 ~  S3 u( R9 ]! L* u9 O. C2 s
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)
3 i. T# v( [1 l5 [8 i* e( e
this means:  Y(x) = (a+bx)^(K/b)
* e4 I. ?8 X; O' nby using early transform, we can have:
/ Z, K& r0 t2 R5 Y1 X; O9 Y
-(k+b)C(x)+s = (a+bx)^(k/b+1)

& N4 H/ F! [: N( b+ t8 h1 [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！

  ]6 b+ W. {. a; h1 l: Q& d" 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。
6 S, Z% K( H. R' z9 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。
: f/ {! L4 h5 c  |您说：
; a, H" m; O  d6 V
d{(a+bx)*C(x)}/dx =-k C(x) + s
1 N6 W% M) e. R  i& h0 G& @so:
# Y, I! ~1 u( |7 r* i2 N3 ]1 x1 }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
+ S6 ~% F1 ~2 e- m8 v7 Id{(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：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
( p  Z  o  v5 ~4 i/ Q最右边是x的平方。
5 o; ?4 z2 q+ n
, f8 U) v, l* W+ @- F% j1 X0 |所以您解的有问题。

# B; l; h6 s! G6 M. c8 f" 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：

2 \) _- f+ f3 |2 wd{(a+bx)*C(x)}/dx =-k C(x) + s
8 l3 P8 S( e! ~1 b$ Jso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

) B+ {& L' j+ v8 ?# k) Y/ o也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. v  q/ z7 V+ Y2 Z6 W6 l( R* c8 Y9 U但这是不正确的
d{(a+bx)* ...

6 M! v; |8 c6 {. A/ ]+ E/ n6 P
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
  n! i8 ^8 b) ?# B8 Ad{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
$ k* I1 ]4 D/ L0 R5 B! ^- x并不是   (a+bx)*C(x)   的导数除以   x   的导数。
$ X* S7 c9 j( b. c
: Q% P7 G' G8 ~% \8 S( nno "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
....
5 C) Q1 z, _: D7 P; Ad{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

9 H3 g' E4 y+ g, p9 \1 G6 x3 `
" l3 w" [7 {7 n3 x, d+ i3 g6 m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

( L- z" a7 S7 Q% Q; k7 X* sIf 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)

9 U# H5 [$ }. b/ X2 o/ F[ 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:
  w% @5 M8 D; V5 L  l这样算混淆了微分和导数的概念。
, Y- z; i: f( P* 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

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

# Y' |3 t* _- F" F; s- k$ d6 m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
( t8 T! q7 K( f- }, B. ~1 g
Ah, you made a mistake: it should be:

; F5 d  B/ {( P0 zd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
' I# A; D- b  p& y" n- N+ ?
' J- z" C: k3 h' B7 j# g, v6 j
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。
( F7 \% _% ~/ {7 S" `7 o* ~- I
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
: p' {+ j% c  L& q, C7 c- E, \
# ?4 P: ^- R1 t  m. _) z, qAh, you made a mistake: it should be:

) O0 w& R- q( }. H; Z# S3 T4 r5 _d(f(x)g(x)) ...

6 J6 C' l* R! j$ H9 S) W
Yes, 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.

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:51:
( n! z+ f6 }# w2 e4 `Yes, there should not be " ' " after "f" and "g", it is a typo. :D
& k9 P- I( b) d/ t7 C3 @# G4 i7 j
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

. K; H3 X( y- mThanks, 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:

, f4 t8 C* n" u, L" X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 ...

: o" n0 }7 B. \- w) L6 f- T+ w4 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:
: s% B' V- X) q8 x) x+ I1 {- G/ b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？

- V6 Z0 U: @  ]" C' N$ a2 Z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。