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:
2 v8 {7 u" p" U1 a请教大家一道微分题，多谢！2 D# K. P& ^% }' |5 C

* x/ ^, t p4 E, F* [d [(a+bx) ] /dx = -kc +s
* `' V6 s( D6 q% P6 kwhere: only x and c are unknown, others are all known, requiire c = ?
3 V6 y- F1 o6 o2 u; `" |5 ]: b. s! D9 v2 |" A/ X9 r% \! Y
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:
" x. l& @+ \# T1 k6 vit is too easy:, M$ T4 V* N7 |; j" z5 d

" m+ l: e2 }; }* y0 Vd(a+bx)/dx = b
" C& a, [: x( Z# ~
* R) z4 S2 q9 E4 O: @: @. L# U, Q' Cso
. c6 Z- T: B- R7 U% W* g8 j, f* N' _* a- E! I/ ^6 \
b=-kc+s/ \( |8 m6 u4 W4 N. R
: p7 K2 F8 n7 ^! t! S9 G4 \2 \
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 20:41:4 k4 n% _* B+ |! V) n
请教大家一道微分题，多谢！
; q5 m* d, y7 m% ?" k  e' T% M' q: B
d [(a+bx) ] /dx = -kc +s
. i( ]" p; H6 ]! m! P) swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?6 D( ^6 H# D, {* a, \( P$ }# _

' G0 Q  D5 i9 f' l/ s1 O多谢了！0 z& v" C9 n$ E' _8 Y5 a$ A
2 [$ S5 N  v8 I% r6 d4 J2 X
[ 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:
& [' s: S- X* q3 c- l* ssorry, did you post another question? your statement is still not clear.
% B2 B; o, u4 C0 `
8 L3 R+ z+ ]5 `* \a, b, s, k are constants? c means c(x) ? E8 B" n% M) U
# c8 n7 N; J( 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:
2 I" Z+ G' q5 w对不起，写错了，忘了变量c,应该是
. T6 L1 S& n& i# M0 g" g/ W( o7 V9 m' |d [(a+bx) c] /dx = -kc +s
8 F7 j9 r- D" ]8 E0 }
4 p& E5 ^& ^$ \2 E& Q4 I& r* b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:
) O- @" w, c7 l5 s3 N5 L2 gdo you mean it is:! }1 ]" V2 ~! d" Y

9 K5 M2 y# [, Ud { (a+bx) C(x)}/dx = -k C(x) + s8 V: K' z, ~- k3 D

5 n, J) k( L2 f, twhere a, b, s are constants, you want to know what is C(x) ?
: P0 c2 y8 O# ?) d: I% l* g m+ m W3 w3 C
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:$ u- ^4 h& q1 A
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 19:34:2 j$ l9 t4 c0 N) x* v
9楼的答案有点问题吧。* E3 j7 I# m! a' f4 j
您说：0 u+ K  Q4 e! P3 G

+ Q0 T1 h7 |5 Cd{(a+bx)*C(x)}/dx =-k C(x) + s
; @5 v8 {2 ~$ H6 i* `7 u& W7 @so:
, ]! D1 Q( x" g, ~- CbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s5 c  W, e! p% u+ o5 k3 s% Q; K9 H

. T6 v- x! t1 @也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
" s) N  Z2 `! f' X% q但这是不正确的
4 M4 [8 }8 C: Jd{(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
* w4 [- ^! T) Y5 A7 h....! ^" k' W" H* C) y1 P3 E6 x' e
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:& }, f5 `3 c3 A/ a( x
这样算混淆了微分和导数的概念。
& A( T9 q0 n! j6 }0 b/ S1 m+ w: |" d
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 K) _" G- l
% l. V! K1 d ?& |8 o1 o$ h
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:
! J0 t( p2 g8 M9 K& {$ Z9 ^( j! |; t这样算混淆了微分和导数的概念。
1 [; N& x: t4 K6 Q
+ l" K: d2 l, \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
9 v7 U; h5 c$ U" _* o/ G: I( Q$ `1 M P* h: g% Q% _& g
Ah, you made a mistake: it should be:
3 w4 G- g. I& l) y/ n2 m# D2 P" D3 [# q( w# ]+ G5 t
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 20:51:1 ^" w. T* O- M1 E* i
Yes, there should not be " ' " after "f" and "g", it is a typo. :D0 N, w9 I1 b1 D- y4 I+ V
2 v/ x( p+ r F. d
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:
% j' M# |5 E9 A; R. ^5 |) y, F5 \0 e+ \0 l' a
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。
; v5 I7 Z% o+ S+ R; y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 23:24:5 R8 a# l/ l P
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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