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:
6 g( @: ]# R+ g1 [1 c) A1 A请教大家一道微分题，多谢！: N  s" O9 ]9 h" L5 I
3 D; }& \% b$ E5 d* n! p: a
d [(a+bx) ] /dx = -kc +s , D* D  k& K, J) B' I5 T5 g
where: only x and c are unknown, others are all known,  requiire c = ?
  y& ]  ?" n, I& a5 j+ P# F0 O, U( m9 W+ E: A- G4 u& Y
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:$ k0 M( i) J" z
it is too easy:9 A8 j7 R; Q# q9 i1 T
* Z6 [2 @* U. r- N8 r; G
d(a+bx)/dx = b
# W/ K7 J9 |* A; p/ S( j2 ]0 K" F$ c* L* o6 E5 d  R, x
so
3 P' ~( R7 Z# e9 E( v0 D  j
7 b8 S% P, l  Y0 |b=-kc+s7 L6 [: A& u& p  C8 P
6 I0 t$ [+ ]# v* O: o
finally:  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 20:41:9 \% t# @7 h3 z/ C* T& m1 \
请教大家一道微分题，多谢！1 u+ G4 v( E" O8 U. p
' M- K' t* j, x& w$ P. _3 f
d [(a+bx) ] /dx = -kc +s - ]. U, o$ p" s+ W* B- I& A
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
. M2 V& ?/ f. M& P$ h- Q" W' ?7 u6 `: T0 e: ~3 v0 a
多谢了！
3 ?0 _, ]% u: y4 Q+ \$ e& P0 v% p2 ?. b7 l, o0 \- p
[ 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:' d! S! v! q1 K* x) Q
sorry, did you post another question? your statement is still not clear.
8 U# z: H$ E5 k1 @6 e
6 {" Y; X! I! b: E0 u8 k4 K8 }% ia, b, s, k are constants? c means c(x) ?- L- I9 N6 E1 S0 }% g8 [7 ~

+ Y# w9 E, ]! ~' t2 Iplease 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:
, M2 B0 E' n. V& ?: w# B对不起，写错了，忘了变量c,应该是
+ C* D' d' h# `- ]# a# ?d [(a+bx) c] /dx = -kc +s P2 }# ] u! a8 L( O/ k" ^
3 n D: w% h, z( ]" X& [
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:% T2 s6 `0 b3 A! y
do you mean it is:
3 S0 ~! G9 k) u; ~! Y3 K
# L; S3 _/ B2 Xd { (a+bx) C(x)}/dx = -k C(x) + s
! J' ]# ?, r0 n+ w. `! t/ t+ K" E( r& P) ^% p+ ^' E; v: L
where a, b, s are constants, you want to know what is C(x) ?
# T4 {5 m" S& }2 o+ b6 o- n$ i2 ~4 k
& C1 p; B+ z9 |5 C, D3 Lif 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" i# g! F* D' J" x
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:
" X( u+ e9 y3 e, l9楼的答案有点问题吧。3 F5 ~8 Q" h1 D$ I. D
您说：7 L7 I' t) N3 Y8 h$ d8 q) S
# Z1 E% Z& w* @
d{(a+bx)*C(x)}/dx =-k C(x) + s
+ O, @3 o% T/ _7 Q$ x, E) yso:
! {& M- t+ D4 \8 g  G+ kbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
. A( ]) D  J# @+ n7 N$ I8 @) X+ B& j" x- n" V$ }& C0 x* ]
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
2 z: a4 \! K# W. x6 i1 I但这是不正确的
3 P  x& M5 ^8 ^. ?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. E# r. c7 F/ U
....' s8 E- v% v/ `1 _( y C( j
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:
: O2 Z( g" o! y' o& I+ P: Q D3 W4 K这样算混淆了微分和导数的概念。( z' y) Q5 w3 ]4 g
& t) s8 s/ `8 |- l$ l) r
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7 B: `" f; E# c

! g! g) \& i) _8 H$ Q3 v: lIf 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( ?! G! ^9 O: K
这样算混淆了微分和导数的概念。
7 R `; \1 O6 x: @4 y2 G8 w8 D
( C' R9 K6 I$ X3 P2 xd(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( ?! C5 @* L, e$ P
3 G4 S! N( L6 W; N% @1 u& h, c* o
Ah, you made a mistake: it should be:4 a' v# _% s& ^- A

! K$ x+ A; K+ Q3 j% id(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:
7 T% E$ Q. j& k* \2 p `: CYes, there should not be " ' " after "f" and "g", it is a typo. :D3 Q) ^$ \* s4 `
/ M6 D' V; }- c8 Q8 o! 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:
7 @4 }8 C7 X7 F3 a, w% E# p7 W6 [( a- n$ [
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。
$ o6 b, G( N0 k" }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:
" @( U1 K j3 x6 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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