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 20:41:1 h4 E/ G, I$ z. a' V
请教大家一道微分题，多谢！
4 @9 l+ e8 s1 s1 d! m3 S* B8 H8 i7 i5 R+ ^" b3 a, N$ h! V+ A( a! @
d [(a+bx) ] /dx = -kc +s # ~* z+ D2 q! V- E
where: only x and c are unknown, others are all known, requiire c = ?+ T' M8 n4 n4 O( f

7 v% W# v, w1 r- C+ g' f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:
8 [1 Y4 z+ C" G" K- M7 cit is too easy:
- [% R+ s0 {1 S% O5 y
3 R, `1 T1 T/ j8 Pd(a+bx)/dx = b
# E  T0 v7 e6 o
4 |% r* R( g( E7 R4 xso+ v: ]- h+ p% C) v
' \/ R- w  ^6 N
b=-kc+s: ]/ h2 H/ R2 j9 \+ {% @$ z

' r: X7 i5 @) R! yfinally:  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:
+ u  ?" j6 Z( `. }: r9 G% W请教大家一道微分题，多谢！
  [" M. I) b6 D& l! T3 c  B# k0 m
2 ]% I, F6 U" o3 Z; h* w0 jd [(a+bx) ] /dx = -kc +s
9 R& L+ p# @* T; Y: h5 E1 a$ R1 Hwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?2 Z3 M% i- C5 D6 _* q
4 u7 t% i& b/ R6 a) \
多谢了！
* ]) X- J$ R3 J& Z& k' G
4 z! u. ~- G: F6 _0 C' v[ 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:
& z* d0 E" z9 J7 c2 nsorry, did you post another question? your statement is still not clear.% r8 ~9 G- T3 {. ?- C6 y. U/ l5 q

' g; c1 R! H! q6 N4 Qa, b, s, k are constants? c means c(x) ?& y: g* X" h" E; E# f! N
4 V) x; x9 u* B/ ^" D: V5 W( D
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 l; J+ }) N z对不起，写错了，忘了变量c,应该是 ~5 l `# H/ ]5 @
d [(a+bx) c] /dx = -kc +s # t. q! V3 X/ A& J. a. X2 C
; T6 D0 L& y1 V) M/ t
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 a! z" Z' ^; h7 I3 vdo you mean it is:: B( b- @( e0 H: j# q
$ ^- T9 m, x# n0 |9 n. Q
d { (a+bx) C(x)}/dx = -k C(x) + s- N! F1 a1 t1 @0 }

, S0 v( L2 {% G" B K2 C: m9 Z) gwhere a, b, s are constants, you want to know what is C(x) ?
% `" a0 m% H3 L& s2 V
' L& A: d$ P5 T2 J. ^# Xif 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:
1 y& h/ Y w/ A( K% C( B- p/ k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:! ^* n3 g# P2 ]
9楼的答案有点问题吧。1 X3 o$ I% P) @1 a9 C$ c" s
您说：
3 V0 x" R, Z7 d* z% e% \5 T6 y: l6 \
8 l8 g3 z1 x) W( N- P5 {; G$ bd{(a+bx)*C(x)}/dx =-k C(x) + s9 C: v. G! e& v" W
so:* }" u5 C8 J6 x3 E8 W. ?
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
& i8 W5 {* X0 `! Q0 h9 q, K2 {" l
0 Z# ~, O5 C8 V4 I5 S也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx0 L4 I# X2 g: y* r3 ]. X, w% h7 F
但这是不正确的3 U$ S; v ]7 p$ ?
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% ~& k# k+ v5 J1 }5 w. r
....
/ ]/ p& f; ~) ~/ J& J# Gd{(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:- h- S7 B. e( r* T: \1 h
这样算混淆了微分和导数的概念。
4 I, @8 j* \ d" a) z: d8 t- a& ?6 ?$ D. D4 C ~1 O
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$ H4 N/ @* y& v: |. D6 j
& D6 x2 y! j0 b; T/ I; l
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:
& D) a$ U" R9 e0 \) \; k这样算混淆了微分和导数的概念。
6 Z" d4 ] d7 `2 [: `& `, k; @2 f- 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
' K- Q9 s9 Y1 T' C' ]+ _% z" ^
6 ~8 ? h0 X1 ?Ah, you made a mistake: it should be:( `+ o8 T0 |" X+ E
# o0 w1 v) r9 n5 j0 i% J" }0 F0 ?
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:! P( ^8 Y$ o. m$ V
Yes, there should not be " ' " after "f" and "g", it is a typo. :D5 A2 z2 J, I& l3 T' Q; T

6 H% b$ }) E1 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:
. p+ p" K+ J. i4 B' I C) U; @: E# L: ^5 B# ?9 B% y" y
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。 B p# W5 p* h) z' |
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:; H0 Z% C/ P/ q8 L- m
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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