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:
. J; Q  x. r! n请教大家一道微分题，多谢！! Q" L0 ]/ O- @) m2 ]! F' U
* K! V6 S! b0 d2 k0 c: D7 R
d [(a+bx) ] /dx = -kc +s " h' w1 C8 `. V  r' _
where: only x and c are unknown, others are all known,  requiire c = ?9 i6 l7 _0 P. @
2 x" k9 i( ~+ Q' t
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:
7 w6 d+ W# p4 f1 jit is too easy:5 `: m4 p. y& [( w$ k+ g

) g9 E) |- ]; L& U1 D" K5 `' e, od(a+bx)/dx = b
" N4 Y9 }6 X3 ]: j. m, r- C4 i% {$ J( {
so0 k/ _2 x- b7 R9 [0 B
: U6 _3 V* y4 D  O" C# [
b=-kc+s
( [( l; V  X" j' a6 r. K1 U( Y  o3 J- ^
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:
" E* F9 t+ h" {" Y- f: x7 k请教大家一道微分题，多谢！
) n9 d4 f4 x5 e* E5 a
8 r5 `+ n$ n4 X6 ]d [(a+bx) ] /dx = -kc +s : z- m0 [2 |7 a+ w" ^4 Y: X% w
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?! I0 w) o2 a/ `( t# K% {$ I
3 B* e7 C1 {: G' N# ?0 S
多谢了！; U9 Z9 A1 F; w2 e2 g0 Z$ |3 N
/ ?0 t( r0 h5 r( Y. U7 h
[ 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:0 [9 Y  s, i; ~  ~- o% d/ W# N! S
sorry, did you post another question? your statement is still not clear.
# k. o" q$ i7 [- D/ W3 x' v" n, _( E% Q( ]  J
a, b, s, k are  constants? c means c(x) ?. ^; {1 ]$ B" N& y+ @+ {) u. l
. K) q1 _; n+ 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 12:30:5 M6 U9 [! f) K1 b2 A7 F5 z# }* M
对不起，写错了，忘了变量c,应该是
5 t4 Q1 h M6 Yd [(a+bx) c] /dx = -kc +s * G i% O! @. @/ x$ X% s2 v
5 h: c2 V2 W& L
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:" j0 b9 w9 t( R6 A- C
do you mean it is:
* I8 D3 A+ t4 S2 d6 U+ F
1 \' F/ O6 F' f* j1 L' Ud { (a+bx) C(x)}/dx = -k C(x) + s$ i r k1 W$ k' X/ N7 |2 H' h5 f
2 g# Z/ E0 L x, f& U5 m% L0 l
where a, b, s are constants, you want to know what is C(x) ?( X- ]9 U) y2 n
: `" t. ]0 J+ T7 E- R( q/ J
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:$ X: T6 {$ D) i. _( E# y( Z
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:" q2 g0 B! L$ m1 J' |
9楼的答案有点问题吧。- W  V( U9 ]! j( J0 V9 u' B) J
您说：
+ j& K3 Y! R) H! Y* \4 L0 U! l1 e, Y0 [/ `
d{(a+bx)*C(x)}/dx =-k C(x) + s: L: u$ I3 l& z# h% G
so:
4 Y+ N% x; @, \/ e$ ObC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
' A& s: e% i' R  M0 y6 E1 V/ R; _* K6 o( i( E- [3 @5 `
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx- D) G( V2 K4 ~
但这是不正确的  H% O; O' x+ {% t+ z7 D
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
6 C6 q( n! ]% T: r) b9 x2 g....
( ?/ B5 C3 ]: @, c, J) g! Md{(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 20:30:9 \% P9 K1 j& H& H. F
这样算混淆了微分和导数的概念。: e" Z, ?; m5 |2 Y

( ^5 E( m& ~2 e2 r$ {5 t( ] Nd(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4 M1 f5 r9 }% h
5 R- d$ Q; [# i' {, }% u* S4 D& z
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:
0 z/ v4 H) i! y% ~! W5 c: T3 ~; [这样算混淆了微分和导数的概念。
& x( ]0 U7 d2 z7 T* r& |/ ^/ F- B; r( U
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* n. Q% u$ m% s- X& o0 T
1 E$ g2 q8 P x1 I7 ^
Ah, you made a mistake: it should be:
7 p* A+ J/ S {+ u8 `
5 b* N/ m0 b' ld(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; ]' @( n7 C. I
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
& D) J% e# N! L: `0 Y
" [& `- z! M$ e9 ~+ W; W0 NBut 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:$ T: g5 a1 H p: y5 q3 z. [

2 d8 r& V) B! R! j: F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。" J4 p a& G1 H
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:; S, k0 q L! R- h3 J$ {1 f' V: \' K" d
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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