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:
% h- E( m7 D) W# W; {5 P: y7 _请教大家一道微分题，多谢！
: Y7 f2 T) I# j# f
/ n4 ]; H2 a: _, ], h3 n: Bd [(a+bx) ] /dx = -kc +s
+ O: X: h$ X1 Ywhere: only x and c are unknown, others are all known, requiire c = ?7 _( m8 u9 ^1 w! y
: k' S5 K- C; }2 _5 ^
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:' O& o, S7 s/ W* P5 P/ @
it is too easy:
0 @! ?& ]  ~3 N: X9 Z
& g- ~5 N6 n. ~$ a, P# Yd(a+bx)/dx = b
; \( N1 u- m* }1 {- ?4 y7 ]3 m# b7 @- P. B# e
so/ x% \1 \' y8 O$ K; ^8 @3 u* [
+ d( I/ E) b4 G9 I' T& z: S
b=-kc+s
3 j: B. E  L3 N( }6 c
' a9 o4 U" ?- y/ ?2 Tfinally:  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:
' a" g1 _, T' I5 E. @请教大家一道微分题，多谢！. @! C# }0 U0 ?( O' @

) w3 B$ c  y* n0 x' m! {2 v' j5 cd [(a+bx) ] /dx = -kc +s
) A$ x" s2 A$ H: i: U2 M3 Kwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?$ Y9 i3 p/ H: Y
1 B! ?6 w7 c* h) P- D- `
多谢了！
% a$ \2 F' r6 k  ]
% i; V2 Z* h- B3 \1 c5 L[ 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:8 X  i' _2 I0 P, A4 J
sorry, did you post another question? your statement is still not clear.8 ]* S$ s  w, ^( Z2 f; f
% i1 |$ ^  M1 W' g$ k5 @2 |
a, b, s, k are  constants? c means c(x) ?$ ^1 f) R# l: T/ }
2 q* h/ t1 b) P. ^  E6 ^: a
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:
7 ]( h0 h8 r4 X1 |4 t+ ^ p对不起，写错了，忘了变量c,应该是' b+ \- C4 V: g5 k! w/ w4 F* h
d [(a+bx) c] /dx = -kc +s

1 ]6 k9 ]# L0 |: S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 K; z  k5 O* U' G! I  A" B* M  cdo you mean it is:( c" j, R* O9 z" G, t7 ?+ K

/ i( [) I* N  s! m6 r; N4 |d { (a+bx) C(x)}/dx = -k C(x) + s% J; N5 x$ o( P# O
: c* x' L" D7 N% C" _- r
where a, b, s are constants, you want to know what is C(x) ?, Y* q- e3 Z  U/ [  R- t4 T' r
7 d# q4 e3 }; W  Y9 c  v
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:
4 `9 S, ? P1 g" U( N: ?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:
3 r- z0 q" ]8 t1 t+ \; S9楼的答案有点问题吧。
% e$ H2 t; ~2 |* `9 d% t( a您说：7 K" b9 W# L) p/ W

6 s1 c! D7 g! l5 ud{(a+bx)*C(x)}/dx =-k C(x) + s
0 J8 b! y) L2 ~1 C5 U0 j8 dso:7 C+ E& Y1 X. V m9 o V
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s \- Z, R) z3 x# o! b
( `6 A" @/ q$ @# e
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
, ?* L9 {6 c6 X' q但这是不正确的
1 X. A! W1 v* K+ q9 T5 |" Ad{(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 i L% R, i9 @....8 F: s! t d) G( H$ }, D4 z0 P5 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 20:30:9 z) r* {9 @# E
这样算混淆了微分和导数的概念。3 A f* a, v. c! q: L: V

; ~* R3 U8 Q$ C6 ?* S7 }7 Ad(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- I& e! D( s0 i' Z+ s

" O. J. n% J* k% \* Y: s+ ]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:0 c8 L  U, ~5 a, {) i
这样算混淆了微分和导数的概念。
6 O0 W' F: }5 U( D
( ~) S# \7 u# R6 q* _- ]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
. ^+ |# i5 J6 V# P; G' D/ O/ e. U
Ah, you made a mistake: it should be:2 h$ E. M0 g  z6 [# I

: s. M( A. Z8 C) j  b- ed(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:
% W2 d U# F% i2 Y$ R/ `$ DYes, there should not be " ' " after "f" and "g", it is a typo. :D1 N) r% J5 N1 \. y
$ x& z5 K4 F; J$ B L4 |# E! q
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:
' l" Y7 m% d, K) W* I: n% R4 y! L e6 t7 L! R; s$ S
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。9 E' T4 l8 k9 d' {+ o) Z# E
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:
3 f! M( V3 x7 B" L# d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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