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 {+ J+ v+ U' J( C5 Q# w请教大家一道微分题，多谢！- {2 I# u  _$ ?" c% F
- u; p; U+ Y1 a1 ]- Z
d [(a+bx) ] /dx = -kc +s 6 q$ w" M5 h& Z* t% N
where: only x and c are unknown, others are all known,  requiire c = ?1 H) U6 }9 W# h7 A5 S

  {& Z6 O; x4 @! t- q  y: ]3 b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:8 a0 ^5 z) a5 O) d- p) ~1 n# u
it is too easy:* M8 M7 f* v9 c/ |: V' H5 O
+ P# G( ~  A9 W
d(a+bx)/dx = b
. \/ `) A& I. J, Q# K7 M  k* n3 P
* F1 [2 g, x5 b* qso
8 e, y8 i$ ]0 s& j- E2 F7 F
* B# s$ c4 T4 `7 N, A' b; ^; Xb=-kc+s
$ j/ G! O  U- d; I( O0 z+ V, e
1 H5 t7 ?0 v6 e$ G0 t* D' x5 ~$ afinally:  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:/ n0 `% Y+ B+ h, Z. n
请教大家一道微分题，多谢！
$ K- g( C4 _, Q5 }9 ?9 M+ a% n0 y+ Y* U3 H& K: N2 i
d [(a+bx) ] /dx = -kc +s
3 E6 C/ v. ]3 f$ o* }$ wwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
: ^: n% O2 ^# g1 I8 `0 ^( W. K: |, T6 D' P1 y1 q
多谢了！0 B; A' X/ {- p) ~

7 M5 C8 ]" l- o$ x' u3 M6 V[ 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:
1 \3 c6 r9 D8 y7 H& I4 H2 ]sorry, did you post another question? your statement is still not clear.
0 I8 E- s4 K, Z& h5 U. r n( c) ^9 b; {0 @, U# m2 h
a, b, s, k are constants? c means c(x) ?/ t4 F( X' h% g
3 r# \8 z! k+ q( s9 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:
5 ]' ] v7 N3 [5 [! q# p对不起，写错了，忘了变量c,应该是! l) q: @8 z# |3 _
d [(a+bx) c] /dx = -kc +s / m6 [5 f/ e) X# V
7 M+ Y) w) K r% i$ a
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:
( W; u( ~( p3 G. d# h' [) Bdo you mean it is:# t( Y1 F; ^3 s  n) N" v# H
1 g2 L7 z! K' B, o# r" d. v
d { (a+bx) C(x)}/dx = -k C(x) + s
5 Z3 `1 N. S4 W. Z/ I4 C9 |4 w9 f. y/ K
where a, b, s are constants, you want to know what is C(x) ?0 D. \/ _* J9 X. i  Z) I

6 l9 _* S" L$ G  G5 F% V# X2 ?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:# C8 Q+ k$ @! u6 |
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 19:34:4 j9 s& o% U, C
9楼的答案有点问题吧。7 c. g# d- f5 O% q. |+ n- f1 U3 h
您说：
( E2 k. C7 X5 x6 l( m5 s5 R: w# d+ H3 ^5 T7 e
d{(a+bx)*C(x)}/dx =-k C(x) + s
3 b/ d3 V- h* E3 ?& l+ cso:4 J0 G$ H' L8 z; ^
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
1 W( k2 q4 C3 q) a
  O& `, [( k- B# G也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
! q$ T6 E) R0 _. s' R% q但这是不正确的% B1 H2 Z9 M8 S* i  R6 L) x
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
! a) o# C& y# F! z1 j4 P....+ o @5 _, D3 W$ g
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:
, E; ]" {0 z/ |这样算混淆了微分和导数的概念。; [4 y& z$ |3 m# U3 u- F

+ `- {; H7 z3 f- w2 D/ s5 i9 Hd(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
# f3 J$ `; X, j' P+ [: F" Y& J! x, @- z' u1 f* `
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:, [7 r$ O" A" F3 U. E9 `
这样算混淆了微分和导数的概念。4 g1 K9 v% F0 C, g1 o8 W1 q

& I# t8 D. i* k( r: \; |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, r: O6 I/ ]3 B7 r

1 J2 e6 J0 S! ?5 _# qAh, you made a mistake: it should be:7 I; N, T. W7 e( n t" O
+ |+ n3 A9 q' H: J" a4 L' S
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:* y! j9 n$ Y! y. ^/ R
Yes, there should not be " ' " after "f" and "g", it is a typo. :D. E/ r7 M0 r- }

9 D8 Q8 y' L. f( ?% s, I! I; dBut 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:. M! z- z' ], x. E1 }

7 V. _, P2 o/ G& @; M+ w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。% k, X- w3 [: g0 C: O w2 [
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:+ N& W P, O) N/ v2 j+ g
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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