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:' Q7 q; V- \7 \. U
请教大家一道微分题，多谢！1 T c/ j( A1 x' B) t# Y

& l9 n& j3 E+ M0 N$ T/ h1 `d [(a+bx) ] /dx = -kc +s
* I t) u. l$ n( ]where: only x and c are unknown, others are all known, requiire c = ?" _0 W7 D* T' o* b! D4 Q9 n
, c. g& ^9 ?& B( N: f9 r
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:6 n4 g: @6 u+ ^' U" O9 N
it is too easy:
) {4 B# o0 b& v/ u5 b& L/ H2 ~# [. a
d(a+bx)/dx = b2 z' C' P# u& h U
# P4 {' X4 F( _" s* b. E5 q
so4 _: U% @6 B; g& M
; U; ~7 q% G/ t% s: ]; z4 `
b=-kc+s
1 e) G" A! a7 P5 o) }
- S9 v* \1 K$ kfinally: 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 }  V- H  l1 i1 z
请教大家一道微分题，多谢！+ r/ Y. ~  [8 q, q5 u

5 {" A0 c6 c- ^# Z8 v( dd [(a+bx) ] /dx = -kc +s
, y3 c* C0 E  g' a1 Cwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?: H; v* p: r0 E1 y
$ x  U+ [) Q0 h3 \. M
多谢了！6 J" x% U/ L3 w
( @3 f* v% I6 F( d3 `
[ 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:
, H9 A, Y+ I+ i% D) Fsorry, did you post another question? your statement is still not clear.
" x9 L, n3 m( L4 O" r/ b6 I. b2 h* l' t& E/ {
a, b, s, k are constants? c means c(x) ?
1 g6 x: ^4 _8 g7 K) k' N. R1 i7 j2 p
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:& `9 s2 b" q( P; v. r
对不起，写错了，忘了变量c,应该是# d+ ]9 z3 V! D, o
d [(a+bx) c] /dx = -kc +s 1 x( e1 a2 W/ D9 d$ r' h. K

8 V o! ~1 x; V) \6 I3 A" 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:7 i: ]( D8 f3 x1 N; B/ I
do you mean it is:
) U, L3 C" P: z. k) m, [) m* _3 n' K I- f8 l4 a" K9 [/ l8 A
d { (a+bx) C(x)}/dx = -k C(x) + s: M& M1 k, s7 r; U: {
4 ]0 N2 |+ h# ~' Y2 s
where a, b, s are constants, you want to know what is C(x) ?: ?/ y. H6 X* v) A$ k

% A, m! s) U/ m& X1 w) Nif 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 14:33:4 `1 r& j% t o/ m
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:
" @: k7 E4 x; e2 Q4 D9楼的答案有点问题吧。
1 ?) t8 Y8 l& P# U! t您说：; O% l- q+ N' ^0 U# ]+ w
. r! w  I) k8 F7 [3 O
d{(a+bx)*C(x)}/dx =-k C(x) + s5 B2 c, @3 u3 [
so:6 f: o% N; O/ z9 h7 B
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s* V9 a5 W1 ]- H5 Q
6 s6 Y& W' E- g# E/ d  M2 K/ R$ l
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx( |% z- S, P. `8 x8 D) a2 S2 K
但这是不正确的
3 l/ r3 A" Q3 f( I/ y7 z& l5 qd{(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 b5 |1 e0 A/ m8 C
....
$ J2 W& ^9 a& z# {+ dd{(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:
, K6 Z- m& {. X* {5 f9 z这样算混淆了微分和导数的概念。) l9 o" b5 `8 S5 Z1 w

. O2 l9 i+ s5 }- Md(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9 O; ?! t2 e8 I! I2 \1 \' k

/ Z( m: T% U2 w$ G) eIf 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:) G' [& h- |0 P
这样算混淆了微分和导数的概念。7 }5 y1 ^0 ?" c- S" G& l1 m

2 `1 M1 k0 n) F" n( zd(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1 v0 l9 V5 j) P

( |. o6 ^( i& c7 w: s* EAh, you made a mistake: it should be:% m/ J0 V! s O8 Q, B
- l: {; j3 }' m4 j( K
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:
+ h# x) ~1 \$ i3 P7 {6 RYes, there should not be " ' " after "f" and "g", it is a typo. :D
& P' T# ]* q/ q! A+ m: X& U2 [9 l/ b6 c& S1 ]5 D
But 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:" c: J' e9 M8 d" D
2 I+ v0 ~( C6 I
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。, v6 [* T }0 r) @( D, |, Y! m: R+ V
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:
# ]8 e, ` b z) V( Y: 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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