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 20:41:0 c) O* I4 O7 |/ F. E- t- j
请教大家一道微分题，多谢！$ t1 G/ Z& C& }. @, d
& N& y! D) O, z3 I/ Q% [" G
d [(a+bx) ] /dx = -kc +s ; c+ s8 L5 A" x/ z f1 {
where: only x and c are unknown, others are all known, requiire c = ?; g; H1 H+ j% B+ N0 N, `
# c) T1 B, V. ^6 F# Z( w% s4 k
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:
" u# t( S0 r/ l7 h4 k8 ]it is too easy:
2 c0 h1 ?# ^6 R$ V0 A" C9 Q, e1 q8 {. l! R" j; l( r/ A% F, S$ r
d(a+bx)/dx = b! C0 [7 p S5 u Q# i

2 S5 d4 N3 N% D1 [/ s* bso: X* m. o0 r( f
8 x1 @9 X5 u9 t, S7 k/ V
b=-kc+s8 ]5 m2 f5 a1 @2 G6 ]2 u0 g7 b
) O* a2 \& ~0 i, C
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 20:41:2 E8 ^1 i) I9 J4 s3 Y- j+ J
请教大家一道微分题，多谢！# w6 M, ~7 `. _8 C) y2 i% G

% S, A" u+ ]& r: Dd [(a+bx) ] /dx = -kc +s . d; \2 w4 M: Y
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
5 X; Y! r; E5 v% J8 v2 W8 s/ Q# b/ S. t- L* [3 x
多谢了！9 L( H+ |1 j- @4 q* I, R' [- ?0 t

( [8 f8 X) g5 D* Y5 q1 n+ D: R8 b+ j9 x[ 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 q/ g, a; m/ W% i, j2 Y; ?
sorry, did you post another question? your statement is still not clear.
" s* e+ l& D! c
4 N, t( p2 o" Y" Z1 {( Ea, b, s, k are constants? c means c(x) ?
. t- W' p1 t, J- f2 t' e
8 L9 \" D. |! O p; I" J: Eplease 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:% Z1 Z; I' K# o9 r( ?& }/ S
对不起，写错了，忘了变量c,应该是. J; i+ N5 v' q5 ~# e
d [(a+bx) c] /dx = -kc +s " S2 q$ C% v& @! m6 G2 h
- O+ i3 f: B& E- u3 m
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:3 S" N7 j  [8 S
do you mean it is:! O% f0 l: [2 ]( A5 J3 S

  s; _+ ^" K% q% o7 ]d { (a+bx) C(x)}/dx = -k C(x) + s
! _* @7 o% ?0 y* g
/ L" e2 X* }1 f( K- z  awhere a, b, s are constants, you want to know what is C(x) ?0 [% w( Y4 P$ }, K* B2 a' H5 K

9 @# i7 U7 m" Y* F" Iif 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:
* o1 v5 Y! a5 F/ \. J# h" T+ 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:
0 ~# I' d3 T# p) t$ }( ?' o; v9楼的答案有点问题吧。; b- @$ B" ~" l9 x' `
您说：
" ^: Z6 M/ T$ _5 E) O* r% M3 I6 {2 N, B# B( ]0 f$ F) {" M+ ~% f
d{(a+bx)*C(x)}/dx =-k C(x) + s
0 t* r5 ]% P, Y' {& q3 Sso:% y: Z' m4 P, G) l' Z
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s$ I2 r7 j4 q9 K& D( ]( G, v

$ D6 h6 w9 t. ]0 }4 L4 m6 U1 S也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx$ N# {7 {- Z7 Z0 w4 O1 [) F4 W- o
但这是不正确的
6 E* s# P2 w: P9 \" i8 B; Gd{(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! S. g9 F8 A" a' e2 e/ L
....; y7 @* F9 h! n5 k4 O# n9 x. ~. u
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:: u& A( o/ P) C4 y
这样算混淆了微分和导数的概念。
/ ?* e# ]' q. s* P" n- z7 {2 Z, F* o# y2 h2 u9 u# t9 z
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! q8 w/ s; |& s
1 m6 U! O' Z y( W+ I4 ]/ L6 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:; q7 o: G3 o- p' |: @8 m/ E
这样算混淆了微分和导数的概念。) `* U4 k$ y0 w4 l6 N" u

4 V6 Q. S) I1 |0 a; V9 |6 E1 td(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* h# l1 ^$ m) J3 Z

/ Y9 R' Q* \% w, e( D9 E/ A9 ^Ah, you made a mistake: it should be:
4 q! {3 I, j$ H6 s7 C0 O
4 x; N. p9 S: D; F" |8 p) e$ Od(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:
6 C u! a2 v- d5 i# XYes, there should not be " ' " after "f" and "g", it is a typo. :D; F& A& {0 D; y. b% ` H; O

# o+ V8 }# ?" X" BBut 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:
/ p0 u" \7 o$ j
' _& n$ J9 Q3 {* i# z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。
7 T8 E% X/ d& ?7 F$ A2 F% 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:
( ~& ^$ O4 m8 N0 ]8 o% S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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