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:
" d+ K4 H$ Y+ {0 p请教大家一道微分题，多谢！0 w2 G7 |6 [  B

$ M6 u7 b! }- h; |d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
* W  ^8 h; s3 r% f, g, N; l/ x
, q0 R- J3 A: z/ z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:
5 i8 q$ C& x" Jit is too easy:
) C" Z* D( J2 _; g0 q1 o: B, K8 w) U! ~9 U8 e( F1 o
d(a+bx)/dx = b1 u2 t% n& O* |5 d8 c! Z5 i
/ T. B$ d( q& J  W; }4 n
so
! s1 R! n$ r, r; K, I9 A& ]/ i8 \# m
6 e+ B6 M/ X# |. y, B& W- f  wb=-kc+s
+ l; s; y9 N: R( r% \4 i4 T
2 j/ f% X, j& _7 ?8 h* m. S3 x, 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:
: v* E$ L2 B6 }+ }: u% E* b请教大家一道微分题，多谢！# t/ r& |. I$ K% R
7 {0 l+ |+ H5 Y+ V. B: ~# O
d [(a+bx) ] /dx = -kc +s
1 i+ k+ M. @7 Gwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
; J. ?# F$ h* _& h5 c6 Q
* n2 F# o& ?. F8 P2 l" L多谢了！0 _) Y6 ?9 o2 q( |7 W8 x
, w5 j) r1 P4 `6 |
[ 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:3 o2 r; ^& {! k" x
sorry, did you post another question? your statement is still not clear.
' f' _6 Y& y; k P+ l% X0 Z v: T
: k$ b7 W4 V5 C# y+ [# i7 i$ _a, b, s, k are constants? c means c(x) ?' R* ~$ T0 x& ]+ m! ]/ j5 H

' K, k% S8 ~$ s0 q: N/ O9 Mplease 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:
! N) u% [3 t: T" P对不起，写错了，忘了变量c,应该是" _& q% {' \& W2 @
d [(a+bx) c] /dx = -kc +s
. V6 y$ {% |. G- L" U( p! p
: L% p5 k3 i. U- y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:
% W9 g0 |7 x: m  n8 i9 E0 M( J& tdo you mean it is:
4 ^4 J9 i: U) j/ t1 n1 Z2 e3 B' R
8 ^. j7 Y$ D. \( _, ]' [0 ]d { (a+bx) C(x)}/dx = -k C(x) + s
9 u) o% n2 _  d1 s
. v; r' s8 n& ?1 V6 E3 G- awhere a, b, s are constants, you want to know what is C(x) ?! d4 b0 I7 `. L& v- n( G& ^8 k
' H6 O6 H+ p1 d  x+ q9 R
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: Y& K; `; J# _$ p0 ?1 e: u
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:) s; p& x; \5 t# S' v1 q0 H  R
9楼的答案有点问题吧。; M8 t, ~: ^( v" F$ f
您说：8 L; p% \! e. B) v0 H% K
# o. X  P7 L( U
d{(a+bx)*C(x)}/dx =-k C(x) + s
5 x3 R6 j4 {3 d) y* b, W; Jso:
3 f: e& v8 w# y, d& ]bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
) ]1 j: D+ U& `4 W) n& g4 T, K3 X* {* `$ V% b* A
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx' Z2 L! A% S+ L! n" [
但这是不正确的
5 W6 m& `( J6 I! Dd{(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: @2 r! v, y4 k; H* P* B9 ~
....
' _% |! a& J3 p a- [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:
/ t: j6 y: X+ W% @6 L2 w/ o9 c! C这样算混淆了微分和导数的概念。
! a4 v: q$ Z: W' [& K$ A7 m& U
. W. M0 A- {4 S, d0 Q3 kd(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
, V. d- `. j# H# ^' z3 e4 i+ _; V, t- U9 A5 a' k
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:
" l+ `+ g' c- }, v* N这样算混淆了微分和导数的概念。
, ^8 k- K9 ?! G" T2 q4 n( G' Z1 g
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
2 h2 f# K, L/ Q2 X( D) M2 w& I$ t$ b/ W$ A
Ah, you made a mistake: it should be:6 G1 g+ r3 Y/ |. ~7 V

& ^/ D! x( s, n D: r& @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:
0 ]1 L9 y' U) o& h$ X KYes, there should not be " ' " after "f" and "g", it is a typo. :D
3 M: }3 i) e# }) C9 A4 F/ L% Z8 J! l4 d/ ^2 B. j
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:
" o+ ~- [& N" ?- M' R( _* L9 O: x r! u/ f6 t F9 E. `
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。4 B, g- n3 Q. U" G
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 23:24:5 [; H; e9 F( c# _ \# ?0 U. F; a
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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