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:
; J& `4 W' C* @" N请教大家一道微分题，多谢！' H$ h9 g, J; Y8 [* J4 P- E+ v, U

* T0 ^. h, k% X( b- Id [(a+bx) ] /dx = -kc +s
U# ^2 I; F4 t/ _where: only x and c are unknown, others are all known, requiire c = ?
7 ^: q6 _" ^1 A G) `# r7 B3 F3 F4 a
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:
+ E% L: F# L+ s8 F! W' n' lit is too easy:
. J5 Z) R( G) m4 U5 a3 a/ Q2 S" x5 ?( ~% v& n% ^
d(a+bx)/dx = b
: I( ?# T I$ G: ], \0 y
- d) d5 v# ?( d' e# bso9 Q& L, ~0 g9 a3 z! M

9 e' U6 F9 J: S/ U. c" ]b=-kc+s! d7 R$ T: Y- K. ~: u

% \0 Z3 P1 t) P' l1 mfinally: 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:
3 v1 l' Z: V0 s  K4 l请教大家一道微分题，多谢！: W5 Z5 I& J. J/ ]8 Z' y( v
! [- M! k- B/ C" j% V
d [(a+bx) ] /dx = -kc +s 3 c4 d2 D8 l) U( Y, m" Z& {
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 F5 s/ R# A$ ~' X5 X
; r2 C7 M7 h) |; v- l: Y5 d0 _多谢了！
  q- n& a" v2 k4 x6 L1 S2 j6 j
, G3 W; T. s7 m" Q7 s7 y3 j[ 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:: o6 D6 ]: c# Q2 Z" G. ?; D
sorry, did you post another question? your statement is still not clear.* l5 k5 N! J, L# T
% m8 F5 E$ r7 X! P& W- g
a, b, s, k are constants? c means c(x) ?+ G% y; ?% ~" B* u

) F% G, o# @. u+ D# J7 J+ cplease 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:
; K, c- ~/ V( b& \对不起，写错了，忘了变量c,应该是
; n( p! Q" S+ ~: v/ j6 W, kd [(a+bx) c] /dx = -kc +s # @3 Q6 } G+ g; v% @: n

6 L3 A3 V5 R n" ~- p- [5 A; L' B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:1 f+ i3 A% o+ f ?0 x+ m8 [, p7 x8 R
do you mean it is:# I/ g2 @& h: B$ w

& @+ x) [4 }# I7 h$ ~2 q5 ud { (a+bx) C(x)}/dx = -k C(x) + s
" g/ o/ J9 x5 o% a$ `. ~9 {
( {- @6 h' m2 @7 J' Q9 \/ nwhere a, b, s are constants, you want to know what is C(x) ?
4 Q) T8 }( k* |: C' C% V3 `: s
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:
" [& x9 N9 J9 f" `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楼的答案有点问题吧。
4 H4 O7 P) B( \% \9 k您说：
5 h" v" E4 B/ X- n1 f' x
! w' n: \& k+ a' z( \. Cd{(a+bx)*C(x)}/dx =-k C(x) + s% p3 J1 c; i" k, G
so:7 `/ ]) L; [7 q% }' T
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s8 D% L  K, a2 P4 I7 y) U

. x: S: W0 s% r$ w% X也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx7 h3 S  P$ M3 Z* x- ?' L
但这是不正确的
/ }1 \9 n/ }) J3 x7 ]  Jd{(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 ?' k5 i' e4 O5 u7 p7 t# W8 a
....9 j& p. t4 I" ~* P1 M6 P. c9 Q [/ ~
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:
3 y. v9 F- U: L; p5 c这样算混淆了微分和导数的概念。6 j! v( H$ E g% m+ F. G
8 w4 X6 p8 a3 s) h3 X
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 I: T# f% T z5 F! I0 K4 \* ^0 Y7 T
, g1 D; `, j. f/ O# u3 _
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:
/ C2 m: N; P! T' o6 @1 i2 Q% ^这样算混淆了微分和导数的概念。
2 m$ t& `& A' r' {$ o$ F
; I1 @8 \3 W4 D8 w, Hd(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) d1 B) Z: X$ \# Y

1 f( k1 f- T) }* T7 h0 QAh, you made a mistake: it should be:
8 L1 c) S3 R+ v6 _3 b4 e1 O+ h3 x: [# j R8 A4 e
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:
@# {. \* S! |8 dYes, there should not be " ' " after "f" and "g", it is a typo. :D8 N# V% q ?) W+ w8 e6 @! _( Q6 P
' H* o/ |8 b j* K4 X5 f3 X/ Y
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:
: G9 l$ j3 {5 ~) Y& w: {$ N1 v
& Z/ s8 t! T! k& j9 P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。0 W8 [# r/ Q! y2 A V1 \. I" J
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:
/ y4 F; z; i4 C9 |% ~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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