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:! l* s/ _( A) e# n: M7 p: [
请教大家一道微分题，多谢！; I, I; D; r& o1 p9 q4 v- b$ H& ^: J
3 @4 F% a( y3 U& ]* `
d [(a+bx) ] /dx = -kc +s * w0 t; U* e( \8 Q1 S: o2 l. d. ?  [. `
where: only x and c are unknown, others are all known,  requiire c = ?
. A# J! `. R8 q; j% @  }9 k2 ]8 c2 C" n1 P
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:; }6 u( ]5 G3 ^$ y
it is too easy:
$ T, }) d& x" `4 P0 V: q4 ]8 N+ ~; `5 @4 J# d2 k3 k+ Q
d(a+bx)/dx = b' |0 K# b0 V' E4 A; K; I
8 r% G9 s& E& d" C5 @1 k5 O
so
6 s# z( ^% | {% Z" \ O0 W2 L6 [) l! u) S
b=-kc+s! c' V6 @9 j+ h% j3 V1 j# p
3 f3 a% @9 `- _* n
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:
2 t0 N3 W8 E' L, S5 W- Y. @5 C/ W请教大家一道微分题，多谢！3 ]. t. {# a, A; k0 `3 V+ ]
. \+ D9 {2 T- J! [5 t
d [(a+bx) ] /dx = -kc +s " a" ?$ Q5 {, z, M+ z8 K
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?$ \5 F/ x; t) g3 E) U
* c# \# B. y) D+ k3 U2 G
多谢了！
6 p( X4 ]" ^) G& v* O# s0 @
: f- y5 H$ F) 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:
$ g; ?/ w5 L" n- Nsorry, did you post another question? your statement is still not clear.3 g s, H: u8 d( u
, Q- @4 |, O- g5 _* \
a, b, s, k are constants? c means c(x) ?
4 S$ `$ A' N, W) F3 |
# \/ o, Y+ l- wplease 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:% K1 r6 j+ T) V( {# u
对不起，写错了，忘了变量c,应该是. ?4 U% g; O4 D" y, U
d [(a+bx) c] /dx = -kc +s & N) Q+ s3 P$ w2 x

! C! b: t9 }5 I' a' x8 k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:% q6 I; J" t6 q; s: F
do you mean it is:" P H: s( P* i: v# r$ ?7 z( }
5 ], k! n& c3 x" U' e! }' x, f
d { (a+bx) C(x)}/dx = -k C(x) + s7 o. C+ s- G$ N. {% N7 Y+ q
, [2 ?! s: m2 N f3 R0 ?
where a, b, s are constants, you want to know what is C(x) ?5 _8 D# M8 R2 _8 B* v; E

8 ^3 V8 i- I# ^' qif 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:: ?$ X1 m1 i- B' 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 19:34:2 `' x' n1 ]; G0 P  Z% R- n) t* h' I
9楼的答案有点问题吧。! o# a8 r% H/ v
您说：3 G% A/ e) \0 g* y8 Y
, z/ G: S; F- l$ j3 ^* U: H/ ~( N- O& m
d{(a+bx)*C(x)}/dx =-k C(x) + s' r# }% g/ S! J1 [3 W4 M
so:' z  |, t% `& q" U% H  z6 f
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s# M9 y& v2 w9 z9 {3 J( W" e
4 N7 F+ M) P9 r7 S
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
2 L6 M" a% O: Y0 I但这是不正确的
: }7 B( ]- a5 B7 G: m$ n/ `8 Id{(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+ r* w; [8 `5 W- {$ p% j! B
....
w( j' I; Q" @2 _- C& x {( cd{(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:
( F( ^( \" w/ ?% t" e6 j这样算混淆了微分和导数的概念。
" }6 e! K4 `7 {+ \4 m2 m) j
' L$ g) w0 N. A _+ K- I5 {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
. j* Q2 z+ w$ W7 _$ M
, C9 c7 M2 z, A8 AIf 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:
5 b( j1 m6 k0 G+ r4 |' d这样算混淆了微分和导数的概念。
+ u- V( W# g! ?/ g; ?" z! T5 H" N1 Q; L7 K: T* _9 e& {
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( ` e2 D: v7 p

]9 f8 A: Y/ |1 e& YAh, you made a mistake: it should be:
, ~; ?/ n6 X4 `* N0 ]9 d/ V2 T* l0 S. V) y$ {, G, w% D
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 20:51:6 u; h8 I1 Q# {/ R! `6 P; M
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
! J0 D0 F, i; v/ h6 G7 N$ m
8 ?* A, D$ C/ K4 [) _/ IBut 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:
. E+ f9 N( n# x! s0 m/ ]0 j8 P- A) s7 `: {+ X4 \% {* D. J
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。
# m/ M$ u& b4 C1 z+ a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:
9 X: N! P2 c; M6 c7 h3 i* `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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