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:
) M# O" p2 K  V* i+ G+ m! |0 E  h请教大家一道微分题，多谢！
) _9 F$ d% b7 ~! T  A
8 G; k5 J$ i$ t& d: s! o. Fd [(a+bx) ] /dx = -kc +s 7 s1 m0 ?: C) D* o
where: only x and c are unknown, others are all known,  requiire c = ?% j6 q: ~# ^  b, j
3 ^' Q  H' E; t  B" m
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:8 e4 O9 v! w! v/ L( g
it is too easy:
  s4 u! s* R- s7 u$ V. a& V
( ^/ e2 ]5 H4 h1 x. `; Sd(a+bx)/dx = b
$ W2 M) p& V' }5 q  @6 i% g
8 _6 w0 M& I' rso
1 N  l2 j  `# o# V/ T) L
& f" q- F3 D' P  _b=-kc+s
+ K' d$ A- _7 F0 ]6 ]2 s# W# N9 C) c. r% `* G
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:
: J. o7 I/ D' z. w3 _. Y% Y S请教大家一道微分题，多谢！
7 F/ z# p$ X9 s5 `+ x! n! {% k6 k6 ~+ n% s
d [(a+bx) ] /dx = -kc +s
5 m }$ H' I5 [' g4 G8 Mwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
; A7 B1 l9 Z. C' }8 E
' _1 Y2 Y. ^1 j; B! e多谢了！2 k# K" w+ ~- d7 Q& u

& ~, v8 w" o: F$ ]3 s[ 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:1 W8 d4 _# t5 w$ v$ l1 `9 b( m
sorry, did you post another question? your statement is still not clear.
# B+ a; F3 D$ ~# z3 N" f3 K
3 G+ w( Z5 Z5 K$ z, `5 {a, b, s, k are constants? c means c(x) ?* i+ b7 U9 r* ^: ^8 {/ G- V

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:' y% w7 m$ i& C- [
对不起，写错了，忘了变量c,应该是, i2 A# O% |! m, T/ o5 X& `
d [(a+bx) c] /dx = -kc +s : `& I, E6 T) B* Y

& Z' j; x- l ~! I# W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:' A) I! W- ]. U) z4 R
do you mean it is:
; Y! c8 |- [/ @* w% a$ V% P" X+ Y' T7 U V
d { (a+bx) C(x)}/dx = -k C(x) + s2 M( }! \6 W; E) h9 b

; B0 o: |# I: e) \4 Ywhere a, b, s are constants, you want to know what is C(x) ?6 d0 {3 w% P y" h+ N; t

7 o* n$ n! a/ u5 {# Xif 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 14:33:5 i* r5 s( J! h- @
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:% I$ u. ?4 o- _7 o8 l. y# e" {
9楼的答案有点问题吧。' a% ?, ?5 u# J5 Q) P
您说：/ n( f1 u! U+ w% l/ t
) f; j' d) C4 r6 R0 |9 B
d{(a+bx)*C(x)}/dx =-k C(x) + s
4 ?8 k. J7 ]; r: S$ tso:# V. h4 r3 S, I: V# H& Y7 Q4 ?& R. C
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# u# h' h+ }6 T S# a; z# G. U% e
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
" x; ]! m. D0 K* a: i: E, N" `但这是不正确的* d1 d% u( D% l& X/ o0 r! C
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. p" T; g* ]7 _5 u" M
....
/ B8 @' H9 d+ @8 rd{(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:
+ b) s4 r- l& c: U2 K这样算混淆了微分和导数的概念。- l$ p/ P) Q+ G b3 F
. Y+ l9 m. @* `, w/ Q& g5 f0 w
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) H& V5 f3 F* D% ^- z8 }0 Y% @% j

( Q/ D3 Q9 x! Z, l) ?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:
+ ?7 N% W$ Q, l+ _7 {这样算混淆了微分和导数的概念。/ y, `' X' r" u2 o. W  Z$ q- J. n

3 _. t, ~# T7 l" |$ Z  h. vd(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! [# |" m2 G+ l6 u$ q$ |
+ M: y1 W2 t2 ~, ~2 E" G, y3 D+ PAh, you made a mistake: it should be:0 k4 H" Q1 h2 r

  A! L! D( z' B- v, nd(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:, q3 x1 s3 V1 V
Yes, there should not be " ' " after "f" and "g", it is a typo. :D6 l* q5 c- l4 j x; E% N

3 i% [$ K' m6 T4 j% qBut 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: l9 ~+ [) l: f0 ?9 m- I
! U, m/ p0 C2 Q
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。8 s! d8 }5 P- G: W2 k3 \6 H
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 23:24:0 f6 t3 C5 }) [5 S- w# w
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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