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 20:41:1 U& j% O6 P5 @; p; ^; ~2 Q
请教大家一道微分题，多谢！
; e3 O+ o1 B; ?) ~2 l
0 r5 z# [7 H. g$ J; Xd [(a+bx) ] /dx = -kc +s # y0 w- d7 X$ O, K3 r
where: only x and c are unknown, others are all known, requiire c = ?7 ~' p* R7 |& w* u7 _0 `7 M

/ p& u; R7 y+ X0 {5 |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:
# F# |) w, ^& A6 s9 f. J. q6 Eit is too easy:
" n8 O4 T/ Q0 S' b3 C9 j, ?5 d- n9 m' r
d(a+bx)/dx = b$ j2 x* e0 `# F) i% f

& ?* T. @4 b0 u1 z( U) Sso
3 n1 O) s6 K" Y2 A! m) B* M' `; A2 |
3 F$ O8 ~) E+ M, Cb=-kc+s1 c2 l+ }  }* _
% B  E' P4 T% T
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:
" x. q' s3 y8 Q2 d请教大家一道微分题，多谢！
# W/ q. I6 p4 \1 E' z+ w) q- ]! N% S, Q# m; r1 W. E
d [(a+bx) ] /dx = -kc +s , J- c1 U: e1 }' D0 N9 X# B
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?4 z2 J0 U- G/ N3 [2 ?
& ?' U( H- M# z Z( o
多谢了！8 B3 t+ k4 a6 D
( K+ @7 l$ V( K" F2 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:
) f0 z. `5 W" A4 h! msorry, did you post another question? your statement is still not clear.
. g7 |0 H- m: |6 a5 b, c* z5 d
- D) H! e" a ?# T2 K; H, C- E, ^a, b, s, k are constants? c means c(x) ?; N1 F" {1 U3 e, w: J) V$ n2 ]

) r# C9 d3 t+ `' X {/ `# C& yplease 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:
- r* ]% T: h$ i9 Z: b对不起，写错了，忘了变量c,应该是
) z9 k8 Q) I7 F7 r/ n, Cd [(a+bx) c] /dx = -kc +s
X$ `4 P! J7 b8 u3 b( c
: S1 o# a" |- t8 E; I# X# h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:. R! p3 a0 p/ G2 F2 C
do you mean it is:
- i4 m8 w% D' h+ N
- w7 \) V9 c; wd { (a+bx) C(x)}/dx = -k C(x) + s
' X& |3 O' J+ Z7 H- g8 }5 X9 X+ B
* P* s0 p9 V/ @: F6 C4 @9 w; xwhere a, b, s are constants, you want to know what is C(x) ?4 O" {2 s- K" J# @; Z
8 R, g; C1 k, E4 G6 x n0 U
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:, M. B/ K! K, v: ^* K
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:+ q, o/ [9 x* l
9楼的答案有点问题吧。
! M( D% J5 a3 x7 i# t1 ?您说：
7 V9 }2 V7 d& c6 b f+ d J* N- f' u6 }) P7 y# I$ u
d{(a+bx)*C(x)}/dx =-k C(x) + s
4 ], ]) l/ p; j+ wso:2 I! O9 Z. r! S* v
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
/ v, [" @% s" G& ]
6 F1 o4 s" @7 b) G7 q也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx$ `# n& }: j2 [) n9 f1 n. g
但这是不正确的
1 `- U/ X: p7 E4 w5 dd{(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, ]( _: i1 n0 z! `
....
1 m3 i$ v# Q, g* `' p; Td{(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:
; M$ h j) m& q& B这样算混淆了微分和导数的概念。5 H; J C( ]; F: \- x, |4 P

& H2 E" p4 ^, K, B9 O: a* ^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
+ ]& t! @4 l$ [9 e- v/ O: d1 e/ ]& _* {7 p2 X$ u( ~; g3 o
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:
6 ], j8 d. `7 i9 u& T) }这样算混淆了微分和导数的概念。
9 g3 `' w# T9 b! \2 \, `$ i$ H+ L+ |+ J- ?9 Z2 N7 `
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 G5 U9 D$ [6 X  I2 g' M* |# n* V0 f! ?! W0 q4 Q9 K* K" i8 f" v
Ah, you made a mistake: it should be:
, j6 R& R$ z( O. X  R
  r" _# K' [! n$ ud(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:
% Z+ j. r. U) Q2 Y N( \Yes, there should not be " ' " after "f" and "g", it is a typo. :D
" ^" `' G" o7 u2 E& {- v# j" }% a5 f- F4 j: P# [
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:
/ g. P; p( d9 w% `. p" C
! Q3 m* L, Y) u5 b3 a$ s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。2 V0 H5 k5 \( j+ b% F6 s! {
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:' m1 k+ \3 t& 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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