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:' S0 E) L% S0 q$ C) y( _* q7 ]8 j: [9 x
请教大家一道微分题，多谢！! j; ]) ?$ b- f
1 K! h: Y% U( a6 f
d [(a+bx) ] /dx = -kc +s
/ Z) n# t, H; M: Y+ R9 Lwhere: only x and c are unknown, others are all known, requiire c = ?) j3 o6 z5 O# x* V0 {; |; L( ]
7 @- a p, x' S
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:  K& y3 Y4 U  p0 Y# }
it is too easy:$ M( r  b5 U' h5 u6 a$ @/ h4 b+ [
3 ]. S* T5 e! t1 X) T2 t
d(a+bx)/dx = b0 W/ M  j# P2 C) I+ C1 @
0 d) w$ D% P) y
so
' P; w# z# c0 x+ s( {8 L' n
4 e# l2 ?& w5 x7 s3 |" l7 Vb=-kc+s
. C. y2 \: ]( `7 \5 B% H9 t0 L" M, w2 y( q) ?
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. u9 U1 t8 g# m  i9 J6 d0 f请教大家一道微分题，多谢！
! f, r- D4 s& ]* s+ S
! y6 E8 {7 U( [% rd [(a+bx) ] /dx = -kc +s
" }6 E4 h# Q  m# Z- {1 A. Zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?; g( _: J4 e7 j1 `

; o. O: |$ Y9 |) U. k: k$ p多谢了！' x  z+ ?& C  x- A4 K9 Q9 f) m5 t
3 x, x9 X) y6 t4 d5 E! Y
[ 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 L0 V9 m. P+ o7 f# Rsorry, did you post another question? your statement is still not clear.% [+ u* m0 k- S3 m5 D( b3 O
6 x" f7 Y/ p' H5 g; T
a, b, s, k are constants? c means c(x) ?
. k; y7 N# |( c. P# j* ?, a' y! M! Y; A9 n/ s3 C3 G3 ]2 b
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 12:30:4 j, m) M# W$ e0 {5 f
对不起，写错了，忘了变量c,应该是
7 Q6 Q0 x+ c6 b8 E3 t5 Wd [(a+bx) c] /dx = -kc +s & p1 V( j D( W' \+ l; {

# J# I2 d, U( @+ m4 {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:
# U7 s' F, K r; B& r( Wdo you mean it is:. Z5 w' G8 j$ T3 X# s8 ~

8 s9 G" v2 x: [4 jd { (a+bx) C(x)}/dx = -k C(x) + s
7 i4 X5 z0 |; L. ~2 D) [1 ^0 R$ j& l
where a, b, s are constants, you want to know what is C(x) ?
" \: x0 k0 H! ]/ `( l; J: c9 H/ c$ 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:" ^8 V' t. R2 X8 |& f+ E9 x
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:# H( a: ?+ e0 d; F0 E2 L3 N8 L
9楼的答案有点问题吧。
# Y/ h# e$ L5 @( R- P/ s您说：
2 c$ _+ ?9 f! s" r  z9 E
' e5 N2 O" q5 ?, u3 E$ Ed{(a+bx)*C(x)}/dx =-k C(x) + s
# ]2 x9 ]) t2 j2 G* u# S  n- Jso:$ `* ~( s# O" s$ W5 `
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
/ P" I1 m3 I8 |, z  `5 v5 C2 q1 b& T) D7 L, e0 z
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
0 n" g. K# A  {, N$ X但这是不正确的. i& L2 v& }6 H: z. @% h# S
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" Z {; O& v& U8 V" ^) q# C
....
, E/ X9 x5 a8 q; g% k3 wd{(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:
+ r2 O1 x6 W9 a% \0 ^这样算混淆了微分和导数的概念。0 b5 F/ D4 o( D% U: O3 B

! V9 k ]& j3 q: B: E' s' V1 |8 Fd(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/ W. j* {" m7 N2 Y; q3 ]) G5 C3 ~
6 D: e9 Z: W' t) y- q& 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:6 i3 d; q5 j$ C8 Y3 @
这样算混淆了微分和导数的概念。
y, R, B7 C# C0 Y" t
& X3 ]4 _& N: id(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; d4 z. o+ _8 F/ d' p/ K+ e
) ]$ L) R! ?" ^% w( U
Ah, you made a mistake: it should be:
' U8 \7 [ E* n% ~( M- j8 m$ N4 L* C4 L* C$ m1 s8 |5 @
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:. k- }. E$ \) N" n: K" B( A3 ]# E
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
/ }6 r. S$ N9 m1 ^; n: c1 Y* h# m8 m0 ]+ ]" p7 m0 H8 ~* J& C4 c
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:
: v" b$ E6 z8 u$ ~. y4 P( m+ Q: P# }) ]( l7 y
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。
* _5 X1 Q+ z& ^7 s! ~( q9 z \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:
, }2 I0 W1 _' \: v* O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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