数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
1 C8 a, D7 d- s3 n- z1 T9 ?" G
9 J3 Q  p1 v1 w% m0 od [(a+bx) ] /dx = -kc +s
' d% v3 M- I) |5 kwhere: 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_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
请教大家一道微分题，多谢！

$ A7 M5 m5 |- A' Od [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

. U1 i3 l# g- T2 S" R4 c, T+ }多谢了！

2 r. g# c6 v4 A' e6 G& u
it is too easy:

d(a+bx)/dx = b

so
! n3 R2 e1 q& d& X# {
2 L1 E3 W2 k' Y7 m' f7 Q( Lb=-kc+s
: y+ o% h4 [7 ?$ S$ T- t9 J
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
it is too easy:

d(a+bx)/dx = b

so

& X5 |( v; h& y9 X$ l( K1 E( l: ub=-kc+s

+ t( l4 f& Z  b, K5 e7 ?. k1 k0 r: Yfinally:  c= (s-b)/k

8 Y2 o+ U' F0 i3 H
the right side cannnot contain c, but b contains c, so it is still not solved.

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
请教大家一道微分题，多谢！

# l: b7 g4 m/ l2 R1 d( I" {3 s; _3 id [(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?

) x5 r, S; o5 M; n) ]" ^4 F多谢了！
* d& f, o; K% K* Y4 b
* k9 V" L' V3 L( r8 k! g[ Last edited by 醉酒 ...

2 @) l6 j2 L$ B& |) |8 C
( i2 u+ P" _; t1 D$ X" bsorry, did you post another question? your statement is still not clear.
! y+ w' ]  I0 o# M( ^4 a6 c4 ?
a, b, s, k are  constants? c means c(x) ?
( o% M% c7 q) E  z
please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
. b# ~+ c/ |4 J( gsorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?

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_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

. w+ {: y, ~. @) e6 e
" p5 b$ ?( ^. W7 j  t. kdo you mean it is:

7 J( ?. m) D5 {" h) Ed { (a+bx) C(x)}/dx = -k C(x) + s

) H5 X* s# F+ gwhere a, b, s are constants, you want to know what is C(x) ?

( J1 i0 [* Y. H: L: @if so, this needs to solve the differential equation. please comfirm it first.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:
0 R% R. _) R0 B" w" b5 Ldo you mean it is:
4 c9 g9 F1 _8 }0 Z$ F/ M; \+ j
' r5 g! o* Q, \3 y. K" O  [1 T, Q4 Id { (a+bx) C(x)}/dx = -k C(x) + s
: P2 `8 \8 ^2 Y5 \1 _
where a, b, s are constants, you want to know what is C(x) ?
' q0 b- q5 Z0 @/ x  X4 c# q
if so, this needs to solve the differential equation. please comfirm it fi ...

4 n1 x: E0 ~1 G) c) {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_

answer:
) _2 M6 A, Q' A7 K
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
2 y3 f/ j+ w. G( V; U6 |( M

procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
$ q: q( k1 G2 I2 j7 Z- hso:
. P" C4 u/ T) o" |1 q
) H+ k; ?0 e4 A  H7 \" I9 ~4 ]bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
7 \: g  R8 v1 q0 k8 Di.e.

4 F& W* O1 y5 j6 i1 x) c4 w  z1 D(a+bx) dC(x)/dx  = -(k+b)C(x) +s

! j( J; Q" k) k- [) K( ?+ s- o
' H+ S  O- X! i0 R" O0 s! N; e, Rintroduce 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:

3 N9 l4 \$ n; o. z2 ^, q1 Q$ H{(a+bx)/K} dY(x)/dx=Y(x)
8 _4 u" c- d3 u& a
from here, we can get:

& y* I! U; @" ?0 Q6 rdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
/ s/ c. r  k& ?0 b: r
so that:   ln Y(x) =( K/b) ln(a+bx)

4 Q- d2 Q, p) H; Z2 vthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
* c3 I" `& C% h0 O; ]( u( \3 {; p
# @# H  f/ K( g! m-(k+b)C(x)+s = (a+bx)^(k/b+1)

$ ?( k" R# c" x/ afinally:

4 A! v% L% Y" WC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)

醉酒当歌

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_

Originally posted by 醉酒当歌 at 2005-4-5 14:33:
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！

* e( [: d2 d  Y. N. h
/ |' w1 {) R6 l7 n+ K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。
+ q- T) I( a* a% u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

9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
5 k" f' `, B3 y" {2 I- Q7 Pso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
- `  B. T: R  t/ i
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
/ ~$ l/ ~+ W( M" Z7 g但这是不正确的
# R! _* k8 F7 {- ?* {d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
. U: m) H' V4 t* L并不是   (a+bx)*C(x)   的导数除以   x   的导数。
  K; V* e7 D0 `+ O& }比如说   dx/dy = (x'y-xy')/y2
4 |+ v5 Z7 e- T2 Y  }: i0 b9 _+ p: A5 Lx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：

4 \. J- G- U, \& l" S! ]d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
- o8 F0 {  d3 H. m3 Q. [, {最右边是x的平方。
' ~" ^* z; b9 x( g4 z/ _5 k/ n# J4 _4 X
所以您解的有问题。

: `- K5 [$ o" f; x3 H: Q[ Last edited by 龙舞九天 on 2005-4-6 at 07:35 PM ]

十年移民路_

Originally posted by 龙舞九天 at 2005-4-6 19:34:
9楼的答案有点问题吧。
: q- F" K9 I/ o$ E您说：

1 _1 Q2 D' ^& F/ c& F0 B6 yd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 V4 S0 l8 a& h0 E* p3 D. v6 k
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
) b' a) O: z- b1 [0 S. kd{(a+bx)* ...

Please note here:

d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
$ D5 G% U5 ?1 Pit is a multiple, not division. Right ?

十年移民路_

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
) i! W1 I1 Z8 ^# U并不是   (a+bx)*C(x)   的导数除以   x   的导数。

no "x" in that position.

gnome-kde

d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

1 |0 v: N% w2 j' m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

" |  P) ^7 E, e. d7 _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)
; V2 C" r% ]8 O2 N9 x
[ Last edited by gnome-kde on 2005-4-6 at 08:31 PM ]

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:30:
8 D9 V. _+ g$ ?5 b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
5 y* z6 J" w( q7 q. h# S# x
& n8 p$ ^  l' ^* Y% b9 j% \If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

4 T* H$ z  o4 o: c6 N这样算混淆了微分和导数的概念。
2 _& Y6 G+ ^' r+ x. A3 z
. ^5 F0 V! E% m7 h' 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
, r8 v8 x( ]' l
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
$ ]& l( w! l" a4 l" [7 I  l6 t4 e

7 S- d. A. h$ ?6 d5 r0 WIf 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:
+ x  T" R. o4 }; }5 f* K
h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
5 C0 A5 [* o0 S6 E; Z9 X这样算混淆了微分和导数的概念。
. I$ C1 o" ?2 T5 G" p! \
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
* D! t3 F' I+ Y5 ?7 L; t
3 D: G" b+ ]4 L( X- a' gAh, you made a mistake: it should be:

d(f(x)g(x)) ...

7 i" x3 r1 X, X6 X* S
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_

Originally posted by gnome-kde at 2005-4-6 20:51:
! S' {" P2 X) c. ^; J7 d! u$ mYes, there should not be " ' " after "f" and "g", it is a typo. :D

& l; n2 h# m1 X* FBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

4 x  e% n& E2 s9 [+ b2 m. S
) g7 e7 v2 M0 EThanks, I am happy here to discuss math.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:

3 n7 \, F$ i) C2 x) z- F$ {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。
' \1 \' a& Q2 H7 c7 n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 ...

3 I; X# r: f6 B1 R2 k& _6 G  a1 f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_

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
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？

6 B9 i$ s7 W8 F3 p
; n; x' e# @# t( {& U. S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。