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:
/ }% S. a* Z* @请教大家一道微分题，多谢！8 X2 X, y7 E7 d+ S6 x5 n

: h, u' Q: n+ |8 W7 @0 ]d [(a+bx) ] /dx = -kc +s
8 B/ U$ A7 V3 b1 zwhere: only x and c are unknown, others are all known, requiire c = ?
. I) M: \ q3 i7 y
: J! [2 B( ^3 R5 P( e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:' l1 `3 n3 E% V! e( B
it is too easy:
4 v2 `8 A, M  I/ _1 Z1 p
+ l  ?9 K; g2 {% f  Z  cd(a+bx)/dx = b6 n, N5 ], u) x3 o) x3 L2 S! i/ D3 D" [
! [) e! z" x  Z, j2 z5 E9 }7 N
so
& k* p; o+ a: _: ?8 f; u3 C7 v: T' Z! q" Y; X5 I
b=-kc+s
) Z. V, [  p, l, |
" E& r( D' `8 ?' hfinally:  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:* Z. e: d4 b3 X( J% L
请教大家一道微分题，多谢！
3 s+ H5 j0 z! J; o: K6 S% L% C& h/ L* F( J( c
d [(a+bx) ] /dx = -kc +s : @" r9 ~  s9 y, u) z+ c! I
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?1 B! Z' L  N7 i2 a5 n
7 Q% |. E& w. S
多谢了！- \' X  I8 J: j( U* P$ u
2 y7 O' |4 v; {% i& 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:
% [7 Y' i9 ?1 u3 C" A) bsorry, did you post another question? your statement is still not clear.8 ]. F% |) P. e4 d: R) m. T8 s
' T: X4 X9 b9 h. M1 @+ M, `1 z' l
a, b, s, k are constants? c means c(x) ?
3 V% ?$ A5 d1 q4 \) b9 h8 f/ x, |- n+ q, P& f1 {9 U2 F
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:
7 C9 @ p' s p: Z7 g ?- E对不起，写错了，忘了变量c,应该是
5 J/ W, O% u; }' `9 A4 i5 K8 Yd [(a+bx) c] /dx = -kc +s : x0 p( E* z8 V
6 p, e. U K, V- X; O
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:/ s9 t( e1 `  R- i; X1 y
do you mean it is:
' A6 |. U. Y# a4 C: C
! C! u6 q4 S0 N+ Y# {1 o, hd { (a+bx) C(x)}/dx = -k C(x) + s1 ?( L8 B( ^3 D+ C* ?4 ]

$ n4 l$ M$ g  z. p$ bwhere a, b, s are constants, you want to know what is C(x) ?
0 g5 o. z9 M/ J7 n; g) s3 ^; Y7 o  B- F
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:
]$ q# u& i& A3 l# b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:" ~- k1 O% x* S2 {
9楼的答案有点问题吧。# M% r. o1 }. Y# T
您说：; W* |0 X: r' t- G9 ], ~

- l* }2 S1 m6 ~1 a2 vd{(a+bx)*C(x)}/dx =-k C(x) + s8 B3 J, Q) D5 M9 n* [, P
so:
; A* p8 P+ p5 W4 BbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
& O* ^1 P5 ^. W- q ]! h# \2 [8 \- n
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx' X. w1 h8 L! A- ^
但这是不正确的6 I- ?2 k/ Y+ x% Y" n( _1 M
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0 i( p4 O- w) b5 \5 A
....
0 N6 u4 P. |5 L" _& o: Jd{(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:# U' `% E3 v9 w; [* z6 ?
这样算混淆了微分和导数的概念。
! F, ?1 V0 y( ^$ ]( G( n0 M2 h) u/ k& H% h+ \9 W$ }
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
" F# {# I( g3 A: x8 w0 z4 K' s. c6 s/ y* v5 N& z- `0 {
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:/ ^/ T4 `+ J# P1 p/ I& Y2 @
这样算混淆了微分和导数的概念。+ A/ r+ Z& P7 U- J! [0 K- z
, P% Y. O- e( O
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
6 P3 ^9 [( z) e! o" ? a7 J- o1 Z; ]( y+ G1 S( f
Ah, you made a mistake: it should be:
' Q K) R& p! }: r+ t4 o% s, c% R* P$ y+ I! J, q% A
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:
$ f6 i7 U: o; M( V4 X6 ~! ?/ bYes, there should not be " ' " after "f" and "g", it is a typo. :D
7 Y& J+ f: n& `& d* _9 H" z& O j. N/ h
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:# G6 _; ^+ S8 L# {/ E7 ~2 ?
1 Z1 I8 L3 o4 b1 @" E: 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。
6 Y+ k# e" K: T4 G4 o! c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:# q0 n2 Y, F% ^9 d% ~ h9 d3 r
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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