数学高手请进，多谢！

醉酒当歌 · 发表于 2005-4-3 19:41

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_ · 发表于 2005-4-3 19:52

Originally posted by 醉酒当歌 at 2005-4-3 20:41:  ]: y2 n- {4 T" b8 K" C
请教大家一道微分题，多谢！4 A) e3 f& }/ H7 r% P. K# d: A

) t+ Q; l1 c  r; S7 Fd [(a+bx) ] /dx = -kc +s : I0 @  `" I0 l( A0 ^+ K/ a
where: only x and c are unknown, others are all known,  requiire c = ?: u- M& b) {" |$ J; R. O

" j& `' l7 e1 X& d1 P/ o多谢了！

it is too easy:

d(a+bx)/dx = b

so

b=-kc+s

finally: c= (s-b)/k

醉酒当歌 · 发表于 2005-4-3 19:57

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
' q7 C7 a  M+ U0 A# N5 Sit is too easy:
; E# j" L9 b* b8 O/ G6 @
* Q, E+ O" U' x2 J4 c. ?d(a+bx)/dx = b+ W  t( B  h0 O+ i, N
) _7 P$ o, p9 X  ^- V- D' h
so; ^" ]6 @- H6 X' `& v/ p

  i( a1 u1 c9 N/ _! @! R1 k' Tb=-kc+s. t" M& d5 ]( ?) G! W

. o3 n0 P1 d% _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_ · 发表于 2005-4-3 20:51

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
7 E/ d3 j7 R' B9 E: \4 v请教大家一道微分题，多谢！
# V- q2 B3 _* I% \/ V7 ?8 y/ c+ L: c" X  T; l. E8 l
d [(a+bx) ] /dx = -kc +s
, Y# b/ W9 K) b' |4 {. i1 K4 Awhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
. a% W! h- k+ J1 G, Z3 L8 ]
" }+ R3 Y* f0 A' L多谢了！
; W; V& G* q1 o& Q3 [
" _) j7 E5 y) ?% P0 `: w2 t5 |  B[ Last edited by 醉酒 ...

sorry, 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 · 发表于 2005-4-4 11:30

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:( ?3 e! r' ~* |- y2 F9 B6 f
sorry, did you post another question? your statement is still not clear.
. x. z. I" L1 T* v" }
* p1 l: p& K: G& a% G, D- ]/ ya, b, s, k are constants? c means c(x) ?
4 q A* \8 s1 ?: T0 z: Y- c4 q. W; `# d& j1 w1 C- U2 v* e
please 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_ · 发表于 2005-4-4 18:36

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
: X+ s/ L/ A* H. [. ?& D9 i, ]对不起，写错了，忘了变量c,应该是. A5 X4 \- C' }3 ^- [
d [(a+bx) c] /dx = -kc +s
3 Z& U7 e& K( m$ l
6 T( K; P% n. m0 G多谢

do 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) ?

if so, this needs to solve the differential equation. please comfirm it first.

醉酒当歌 · 发表于 2005-4-4 18:43

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:- G p$ @ ?4 @# R/ f( A/ }
do you mean it is:
5 ]: {# G9 s; W( h/ Z) ]4 G4 D0 N- d8 D% S9 H
d { (a+bx) C(x)}/dx = -k C(x) + s
$ s, e [5 a: v! U7 N8 u" P% K" y$ w' ?& d( t
where a, b, s are constants, you want to know what is C(x) ?
" ^( _0 c& y# A0 E
5 J" A3 c0 L: B' J1 c/ Qif 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_ · 发表于 2005-4-4 19:17

answer:

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

十年移民路_ · 发表于 2005-4-4 22:26

this answer is the good one.

procedure:

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

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:

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

from here, we can get:

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)

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

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

finally:

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

醉酒当歌 · 发表于 2005-4-5 13:33

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！

十年移民路_ · 发表于 2005-4-5 20:41

Originally posted by 醉酒当歌 at 2005-4-5 14:33:
0 W/ R! j& E6 Y# m0 J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！

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 · 发表于 2005-4-6 18:34

9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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
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
最右边是x的平方。

所以您解的有问题。

[ Last edited by 龙舞九天 on 2005-4-6 at 07:35 PM ]

十年移民路_ · 发表于 2005-4-6 18:53

Originally posted by 龙舞九天 at 2005-4-6 19:34:) {/ m' d0 ]2 ~/ A( `# `
9楼的答案有点问题吧。
, }9 {' @0 r, X) r$ _' N您说：% C7 g- h' I7 L6 L
% S+ Z8 {7 r$ k
d{(a+bx)*C(x)}/dx =-k C(x) + s
* u8 h) s, W0 s- W$ E7 {( Q2 o0 Kso:
3 _  ^2 H+ t; VbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
2 L& `! X6 ^  B( Y0 H2 x0 M
7 g% s4 r5 E, q$ A- v5 x也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
2 C# ^- N' p( X1 O* C( S但这是不正确的" r. c1 s) W$ W2 ?" u  g- n
d{(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_ · 发表于 2005-4-6 18:56

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。

no "x" in that position.

gnome-kde · 发表于 2005-4-6 19:30

d{(a+bx)*C(x)}/dx 意思是说对 (a+bx)*C(x)/x 整个代数式求导数
. O }# ?4 j$ [6 N4 V....
4 t, G; q; V! r) v! }$ 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

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_ · 发表于 2005-4-6 19:45

Originally posted by gnome-kde at 2005-4-6 20:30:" H- g' P( {+ w) P- t
这样算混淆了微分和导数的概念。8 V1 S2 j1 ~: c, V9 I2 N9 \
* Y- b1 d4 \* G' f, e+ P
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8 |- {$ O5 k: u c( Q
! I% H% Z9 V$ j+ I9 e' T/ v G
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

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

Ah, you made a mistake: it should be:

d(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)

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 · 发表于 2005-4-6 19:51

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:7 b" X& u/ Q) o) a) |9 q' Q( X3 w2 u
这样算混淆了微分和导数的概念。
9 i$ y. t6 r O8 a( T. _# Q u- B" o! q/ T
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2 [! n! I& @! e% ^
& |* i+ K$ ~' m" r
Ah, you made a mistake: it should be:; n$ I+ |. |" a- N y# ?2 m6 n( n- ~$ i6 l

7 V3 j: E, [: O- \3 u/ i9 |" k! h* [d(f(x)g(x)) ...

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_ · 发表于 2005-4-6 20:05

Originally posted by gnome-kde at 2005-4-6 20:51:
( U1 b9 G4 o$ o6 Z. tYes, there should not be " ' " after "f" and "g", it is a typo. :D2 j) ^; R" a9 i( N
6 ^/ y9 K& v& ?" |8 ^) N
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

Thanks, I am happy here to discuss math.

醉酒当歌 · 发表于 2005-4-6 22:24

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:% r/ p) c9 J7 a
0 D m8 \9 r. ^& C9 z- l ~' c9 c' M
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。 u! W% u) ~4 g7 x
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_ · 发表于 2005-4-7 20:44

Originally posted by 醉酒当歌 at 2005-4-6 23:24:/ X! I" S9 I9 ?9 u, Y4 }: N9 V3 f
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。

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