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:
/ k6 t l' H% \ B8 Z请教大家一道微分题，多谢！
5 j& m* W1 S# {8 O/ t& W; J7 T# W5 p3 U: }& A+ u. S
d [(a+bx) ] /dx = -kc +s
5 t' _0 t. {( ]' P# vwhere: only x and c are unknown, others are all known, requiire c = ?1 T/ J7 [; y( f9 v) E+ D: X0 z

& W" Z$ l) z% m7 @& 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* l$ l- c+ G. o1 o& s4 i" h$ d% m
it is too easy:' I6 ?5 g) p: J1 m
. {, F/ l7 B+ y0 B9 R
d(a+bx)/dx = b; d3 a9 O5 i6 x

$ f5 P$ n- U/ n+ _so, A+ g2 _: T2 m$ B$ }/ e

" y/ E$ f6 B/ t: Lb=-kc+s. n9 w! g6 r* U4 @! B4 P6 P
+ }4 {7 A* y% E% ?1 @# A, I7 {
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 20:41:4 k* J& @7 ^( ^7 v
请教大家一道微分题，多谢！2 R+ x( f2 ]% z% z. A

, y* r7 _, F! h( Wd [(a+bx) ] /dx = -kc +s 8 p2 k( I2 d5 I# ?. Y9 D
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
4 b. l# B8 y' y: X0 T l3 [6 b& Z C' ^+ S, d% j! f$ w5 ]
多谢了！
' L1 Z# V* X' h( H: v4 _" [# T1 R$ o" `
[ 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:
; U2 C( h5 K+ z! v; V6 F2 qsorry, did you post another question? your statement is still not clear.% L! t3 F' g! Z8 y

- h+ G8 @% H- b* Y) _a, b, s, k are constants? c means c(x) ?$ [. ?+ ?2 c3 W7 Y

& }# |+ S |3 Splease 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:
. e5 a6 |* m! k! }3 h# v! r2 A! W对不起，写错了，忘了变量c,应该是' M6 w+ D7 a) q0 c8 M
d [(a+bx) c] /dx = -kc +s " |3 G2 F9 ^5 S& f+ i; n
8 a" H1 Y3 k! g! a6 f, N0 X
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:
6 j% @- z$ c1 m- y7 ^0 ]do you mean it is:6 D( h, u3 @1 O
2 E% e5 e; k H" Y
d { (a+bx) C(x)}/dx = -k C(x) + s
) C. V; j7 {, ]1 U3 r& `* ~6 F, a& F; r2 p
where a, b, s are constants, you want to know what is C(x) ?
) ?$ H3 p; R z1 Y# N7 \; u$ E/ e7 H9 k8 }1 r! x
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:
2 e) z3 W( b% n, W4 S/ S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:
9 q2 s7 ^6 p- R9楼的答案有点问题吧。) |3 C. A" r* O1 `  o
您说：7 H4 x' p- W# O" b

, a. b# u" `  I" G4 b/ O' dd{(a+bx)*C(x)}/dx =-k C(x) + s
1 m3 }& ?2 T$ y3 v% Cso:
) T& o& E4 ?- J( zbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s0 r$ p: N  j2 b5 h

8 p+ C- x3 s7 Z  y/ @% R% J* P1 P也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
) o" @- w, R' V9 Z" {9 r但这是不正确的
; f7 f0 `! k: ~7 ad{(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 R/ z8 X: D5 m
...., J: ]5 ~$ s- i# }9 r
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:
. @0 I; J) u& E& k/ x# b这样算混淆了微分和导数的概念。
, s( f$ v0 t) N2 e$ x9 l1 m9 D2 m+ a" j* J
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( Y+ S7 E7 {! \* U
/ P- A g( f6 X& L( p6 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:
1 U' \9 ^8 `) S6 U" R这样算混淆了微分和导数的概念。
0 S& [+ `% H& ~- r, P+ A
4 N% A# K9 z Kd(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% V6 Q; ~$ |+ s

! N$ a( {5 Z7 z3 N$ M" ]Ah, you made a mistake: it should be:
" \9 k# B1 J7 X% g! d$ M, p7 a3 T( K7 F3 p' Q- e, h5 A" Q
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:
7 a# B# Q/ v) _Yes, there should not be " ' " after "f" and "g", it is a typo. :D3 g0 A0 g6 e" {1 D6 v

& R/ u4 x$ B6 Y% F4 YBut 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:9 r {8 r, x h* n0 w, q9 k
+ s7 u/ M2 R- s4 z5 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。
, s( Z# P8 n. G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 23:24:9 p9 N; I+ N7 [" B0 a1 X. v
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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