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:
, Z2 l$ X! m, _# }- _' \4 `请教大家一道微分题，多谢！0 |$ e% k% ]2 l8 E8 r
; g3 {! }7 t0 h* K& t- d0 z
d [(a+bx) ] /dx = -kc +s ! ?6 E; Q9 z; a0 r+ g
where: only x and c are unknown, others are all known, requiire c = ?' a5 w: }3 m# \2 y
% O2 P( A8 y' c3 ?, H
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:3 \  i; F4 z2 @5 Q' [$ o# }
it is too easy:3 j+ x+ A9 R1 b

' m, z' B& k# P' w3 t9 Z& E; ^d(a+bx)/dx = b
3 j: }0 S) v$ V% {5 X. i
& n& I: N* j# V1 z6 Oso1 {( W& M4 s- G4 C. S7 A  l

9 m  ^) u! c' d- L- ^  Hb=-kc+s
; {/ O5 n5 j  p# c9 R* ]" ~7 J- R9 V1 P
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:
) I2 f  ?) N/ J请教大家一道微分题，多谢！
7 H( p$ s0 a9 L5 j4 c- Q
% ?0 F3 F" V8 u! {5 t. v  q) gd [(a+bx) ] /dx = -kc +s 8 j& k5 g+ t* K$ C) d
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
8 \8 f: ?: l" ]$ x2 ?1 d
# F% q) ]/ J& p. o4 |6 ?多谢了！
% J2 D+ I, a# W# ?% ~: P
1 J( p8 t; d+ W" {4 ~' @[ 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:
" c! Y3 y6 \; S6 T/ L+ n/ Vsorry, did you post another question? your statement is still not clear., Z8 F" g$ d. K6 u U

' D% g$ |7 |( @3 B( v' sa, b, s, k are constants? c means c(x) ?) G# N( E7 V( {3 L+ y1 w, j
9 n. Q3 p1 U# w+ E. O: L
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 12:30:7 q( [& K" ?8 q/ Y& d- C
对不起，写错了，忘了变量c,应该是, b# h; p. A; \% k% F; C9 E
d [(a+bx) c] /dx = -kc +s ' {: J' O: |2 R# {( ^6 p9 s( @- g! I

1 \3 A/ z+ s3 ^4 ~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:
$ p4 J Y6 E1 Q0 k8 |) Ndo you mean it is:0 q' r5 f! s- ^9 D6 L
* c" z; _# o# G! {; c
d { (a+bx) C(x)}/dx = -k C(x) + s! O5 s0 W, I' |/ B6 s3 r$ }5 U

% p4 C/ h% |4 |where a, b, s are constants, you want to know what is C(x) ?: E5 D& I& V- c( {0 N+ {& f
1 J& Q2 e5 @% ?! V
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: V7 T, H7 ~( U4 l/ K* e
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:
) T4 B3 L, b* p" G; E9 q9楼的答案有点问题吧。+ s8 c' q( S0 l( I
您说：
4 a- s4 o% Q7 F; s0 f$ B, e
8 k: {! e$ ]! Yd{(a+bx)*C(x)}/dx =-k C(x) + s- N* N) z% B6 Z4 v
so:
m+ @6 t/ a/ A, pbC(x) + (a+bx) dC(x)/dx = -kC(x) +s* n d# W1 h& z- ]! r8 r4 @, r0 E

3 K0 i/ D% M7 l1 S" A也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx9 _; l K, Y3 ~* a
但这是不正确的
: S; W. t3 y. T9 ^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' V4 o5 e: I% Y6 }6 E7 j' L# l0 F1 T# F
....
& {. e# E6 `! y7 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:
8 O6 H- K' T% t+ I, F- t% [4 ~这样算混淆了微分和导数的概念。
/ q5 C; }5 C: J* b+ l( G& ~- K l- f* h; a% 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7 ?9 C7 U: H; G, ^
: h9 ?$ t5 K* x: c, v' B
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:
! P. i$ _# B5 p! L7 q0 n. U这样算混淆了微分和导数的概念。
5 U" y# b' F1 W' d( ^7 A' R) W- {5 `. 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
3 W4 L) r4 k, U# T( d6 [8 C. N. r7 X4 Z1 u, X, F( R
Ah, you made a mistake: it should be:# B" B0 }1 B$ k$ v$ X; A

: S1 \! m4 u6 D- \& v" q, ed(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:
" C6 _0 m9 K4 ^$ Q& Y+ F; NYes, there should not be " ' " after "f" and "g", it is a typo. :D8 n4 a R* m9 l$ q

8 E1 x7 z* j0 D# x. Y! f! i' pBut 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$ V( o- g3 V5 E' ]4 [9 u
: `7 k+ p. u% _8 v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。
R) F# v; a) s( @; _: z0 G3 j" O4 R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:
$ T2 g! h! T* A+ u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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