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:
' K' y: @6 T) y2 z4 z- i请教大家一道微分题，多谢！
/ _. B. |/ @ t% H: r; ]) Y: k. P1 Q1 s- K m# ~
d [(a+bx) ] /dx = -kc +s
' d, k& u. T) h9 E, h, {" K9 iwhere: only x and c are unknown, others are all known, requiire c = ?
8 w: u) U7 u. Q4 g4 M8 o3 s$ q0 |9 |
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:) U4 N4 w5 k8 Z9 G
it is too easy:  C) O; y: {; A* @" M9 N9 a

2 W! O! \" s: p  |# u( |1 Md(a+bx)/dx = b  P$ ~6 u" D( P  y" x
" a2 ]2 x! J. j; ^; [( T- v
so
0 w' R  X5 H/ U  G2 T
% L1 I" h- }' e0 J, D- S/ eb=-kc+s
3 e4 H& }4 J4 C" o/ M0 Z5 }# f  `
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:) t$ Z! p7 w, m  ?8 x+ a+ ^) R" V$ {
请教大家一道微分题，多谢！, h% ~: v7 U, y5 [; s1 u

% _! P. u6 X: f. t. xd [(a+bx) ] /dx = -kc +s
- }7 _- Z9 J7 A- Fwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
3 Q. d& S, ]; B9 i% t2 A' b
6 L3 u7 Q4 g9 o0 ~0 ], U* a多谢了！- N: s8 ?  r0 w8 p0 g
2 e$ a3 L! u6 a% l6 O: O
[ 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:
$ H  }% z1 N* q& Y+ T8 l, l8 Asorry, did you post another question? your statement is still not clear.9 w: x2 ~0 y/ `' e% n% s* Y
4 Q) Z* q' v; e) N2 d  z" J4 w
a, b, s, k are  constants? c means c(x) ?& @/ Q/ ?( T0 p# E
: g& n* l; M7 z* G1 x3 @
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:- w& U" }# l0 T: g6 L8 g9 I" M3 ?% ^
对不起，写错了，忘了变量c,应该是
1 x& k Q9 z: i1 T: }7 U" b) cd [(a+bx) c] /dx = -kc +s 2 a/ U" G% D; E8 l/ P

. d9 ]; t G% W6 o9 i# \, y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:+ B/ }5 M- @& I t9 L# V/ G8 o
do you mean it is:
/ {+ J+ S& {$ U. K6 j6 @* J
: {" v& @6 W+ B; X1 l# }d { (a+bx) C(x)}/dx = -k C(x) + s/ T3 a; b& D3 Y- ^. U+ Z
% H) g' P( I- l R; B$ k9 O% j
where a, b, s are constants, you want to know what is C(x) ?1 O. ~6 M% ^( y9 a/ T0 z! |
8 l; s$ K" J5 R+ M' P
if 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 14:33:3 i0 X& }# x' r9 s) i3 Q: 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:/ t. b& u# Z, |; `  p
9楼的答案有点问题吧。
# f( y* t, R" V9 _' h2 x  o  Z, i您说：
4 k8 D9 ^7 g& y% d$ L
5 M; E5 q( j2 u2 a8 H  S, q# Gd{(a+bx)*C(x)}/dx =-k C(x) + s
% d1 P( Q* E2 l; G$ eso:
4 f+ D/ \  L% ~bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s7 l  b; S$ e: d- P& C$ v2 d7 o: r7 }

, P( A) i/ H  O1 I也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
1 I2 k7 N# E) f0 T' o/ [但这是不正确的7 [# w" i7 i( a5 w+ k8 k- D
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1 s+ y" X- o M) [
....( U' N/ e7 C( y6 W! a1 ?
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 20:30:5 q) I5 O6 b: Z2 f# I
这样算混淆了微分和导数的概念。. V( R6 U: w0 H

8 V: a1 F# c3 Wd(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; X, ?* J, p4 v9 Z

# E8 M" R. [$ `# ^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:, o! ~7 u# k4 b3 L
这样算混淆了微分和导数的概念。
+ |. \& s, L4 r) S! i! W2 @* L( M
. K" {' ^3 S; L5 I \2 {, x" Qd(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' O0 s# A) I3 z8 l9 `8 T) A/ J) M
5 U4 N- ~; s( I+ l
Ah, you made a mistake: it should be:
4 T; S- h0 |& m" W( Z4 H9 p ~4 q/ [
4 p* k; s% k2 @2 \' k* ^/ hd(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:% K- G% Q) {1 e2 e# M/ f/ |; c( A
Yes, there should not be " ' " after "f" and "g", it is a typo. :D$ r4 _5 s0 U& D( F
, l' b5 o3 g" o* g
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: K& Q% Y1 o$ R

% @8 j2 l) J5 x! ^- M( E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。
0 g; g+ e4 Q* d U$ f1 Q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:
0 F9 ~! @( n6 f- B) H$ a: X# C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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