数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
( x3 i6 B/ i2 Y) J+ J1 B8 B' E5 U
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?
  \: _0 f; ~2 U8 N
3 v) I( j% q/ G) D* [! ~" T& H多谢了！
* a% p' R- C  b$ P6 u
[ 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:
( i  E& Q8 O5 V. G1 ?/ j7 ]请教大家一道微分题，多谢！
' l' @% p) X" j
; q1 I$ D+ m$ [2 ^( Yd [(a+bx) ] /dx = -kc +s
2 p) Y% P$ T9 v2 j3 Twhere: only x and c are unknown, others are all known,  requiire c = ?

/ W; @; [/ G3 o1 X$ U多谢了！

+ K& q4 ]- [6 h' |# e! \3 rit is too easy:

+ i# g  v8 ^# td(a+bx)/dx = b

so

" \: k: H4 b: H1 w! u0 S7 m" Ib=-kc+s

  b3 Y3 C- t' }- H, \: }# _finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
! F" S, |7 n. ?5 w/ ?% Z( g; zit is too easy:
& p& G7 o1 |+ w+ R+ n, B! C. E
d(a+bx)/dx = b
9 G$ t3 `; K& L  F" K6 b
2 o( W, s7 v+ b- l5 uso
; R" ?3 M& v& v& U4 A3 n1 x$ R
$ a6 M2 ]  W. Y5 `7 Vb=-kc+s

1 k; }+ m/ l+ T' Rfinally:  c= (s-b)/k

; s. V2 p: Q2 s. d3 T$ f; n! G0 V
. @- ^5 D  b0 `. ], jthe 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:
* w2 x9 ~# U' p3 i请教大家一道微分题，多谢！
# J! L; F" r9 E' u1 I5 q1 `
, E4 N2 n9 P' L1 o2 X0 e5 zd [(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?
5 Q4 c& B" N' g
+ w  C, K- T3 {: m+ F3 G8 q多谢了！
( s7 z3 c/ v, e8 x
[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.

( U6 o$ {; n# V9 [1 B0 a* Xa, b, s, k are  constants? c means c(x) ?

please tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
1 }6 a! i; o% C9 s, D. esorry, did you post another question? your statement is still not clear.

( T# K0 o- X7 Ta, b, s, k are  constants? c means c(x) ?
' r+ B; c! z2 h  V% s
please tell me.

9 Y! T9 I2 V. a对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

. h  G& r, J" z  R2 Q6 a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
) _  z$ M7 z) f' e3 @0 r9 ^d [(a+bx) c] /dx = -kc +s
! P7 h) [1 h- F# H: g  U
: R( j) @, o; H' c* @多谢

( V9 Y8 c* u* G; F1 R" E5 v. ]
do you mean it is:
+ {! k% ~3 Z8 z( d7 g
d { (a+bx) C(x)}/dx = -k C(x) + s

, O! B* `% l/ dwhere 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

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:
do you mean it is:
3 z$ y% `  ^' ^4 |; c+ {7 c' y
: n$ O5 L6 G: J. m  U1 B$ H! Pd { (a+bx) C(x)}/dx = -k C(x) + s
, v4 S0 ^. w7 C: l! H. {# ~  J
where a, b, s are constants, you want to know what is C(x) ?
; {- c6 w% A& ~) b
* W. Z9 i/ I" J, Q8 X8 ^: Tif so, this needs to solve the differential equation. please comfirm it fi ...

6 y. q. ~4 K' Z' u/ i+ ~
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:
  p4 k- t( B( h+ u7 E
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.


# P6 b! Y$ J1 Y+ N2 ~2 eprocedure:

, _" f  Y5 F) u/ B1 l. L, BFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
! @/ e" N  @6 Y; M" L% {
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
- S  {" K( |& j: J9 m9 Di.e.
) _9 F" I" v) [% O7 V
(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
; q( x" d1 c$ S& n& p& b. |$ M4 e) Ktherefore:
' T6 b4 r/ t3 G/ D, m" _
{(a+bx)/K} dY(x)/dx=Y(x)
# r* z& P; Y) r: z
from here, we can get:
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, Q& z$ y" E8 |dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

, C# ]# }2 M" c" F# n5 [. {so that:   ln Y(x) =( K/b) ln(a+bx)

& Z9 w$ b) @" `& dthis means:  Y(x) = (a+bx)^(K/b)
0 E: Y$ j& w$ z3 u  r2 R! p" e: Tby using early transform, we can have:
! |8 m- H5 c8 v8 Q
; g" a$ |# ^3 ~8 j0 M$ v2 y-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
* X7 Q" n8 q8 e& ~. y3 E9 w
7 ]1 u5 u9 f; }- x9 G0 B3 mC(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！

) s- ~: S! @) d5 ^# k9 t% 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。
6 a( n3 R& {- m5 T* k) P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。
( N: x2 J: a8 h. U, c' O1 F您说：
: y0 F- \0 K" r' M
( u( ?4 s0 ?! Td{(a+bx)*C(x)}/dx =-k C(x) + s
2 a8 t" G4 s+ D: j+ T4 Cso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
  d9 w7 e$ X5 N1 B) q
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
4 r: j" p; ]& T* C/ [; r1 z: a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
% [1 W* k/ ]% W& D9 C8 M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
$ a9 m& J/ F9 \$ ix' 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+ b- \2 O7 ^. r) r
9 d( `5 p+ h9 d6 Cd{(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。
9 Q# r9 F9 O- ^
! l4 N* I/ B  f: v  x1 g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_

Originally posted by 龙舞九天 at 2005-4-6 19:34:
9楼的答案有点问题吧。
您说：
/ f3 I( y1 j3 |" M4 z  {
d{(a+bx)*C(x)}/dx =-k C(x) + s
! |* D8 M3 d8 I9 Sso:
  V* h% e, P/ }bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
& T1 Z  d! a. r4 v1 W
5 S$ f; N3 n& G+ Q8 Y8 y也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
( S4 u- D5 g, s# r但这是不正确的
d{(a+bx)* ...

Please note here:

6 b' f6 C' H- S0 O5 ld{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
. ~. Q; }: c3 w2 u, sit 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
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
6 B! v3 z0 |9 S% D" n2 D....
+ r9 H! c5 Y. h/ T9 G- n! _d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

3 \6 y' R  d. k# \! p/ U$ s0 p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
$ ]9 E5 i% [3 [0 D; v
* ?) L3 Q2 g' kIf 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_

Originally posted by gnome-kde at 2005-4-6 20:30:
这样算混淆了微分和导数的概念。
0 Q! C; b% R$ j! S& Q0 Z9 Q, s
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
/ ~: b  y0 w9 y8 C; {
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

# X+ e1 r$ z& v* ^. ~0 A
" ?# F. W0 U) p1 I' t: e: w1 K这样算混淆了微分和导数的概念。
, v! n, z: U7 z; l, O. x3 C
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
+ ~: z1 C# x7 }( z# H. m
# J/ \4 g$ d8 V, ]; V6 h) @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
5 H' _9 f; `& }- h8 j! t7 M$ H1 g. j& T8 P
6 z$ n) X8 A3 |* y) J8 O
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)
1 U$ g  e; F8 e( M' L/ Y
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

7 r& m$ `  N, M- J7 Fh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
6 ?3 a" e+ V: k2 @( S/ |2 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
9 t3 h# a! w0 Z' f% p
Ah, you made a mistake: it should be:

& z1 H( U3 b) |+ W9 H. ]1 r& u  ^d(f(x)g(x)) ...

1 k6 x- ^1 |. [2 h8 `4 Z6 F3 IYes, 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:
+ `. D! W& U8 R6 @' U& L8 {) ZYes, there should not be " ' " after "f" and "g", it is a typo. :D
" b/ P2 }0 h% x- J: [9 \
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

) [  _& z/ E5 k( i8 z7 qThanks, 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:
' {4 \, I% a) p7 y6 p6 ?1 k0 L
' [: @/ b- a; r, ~  v5 U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 ...

; o  {3 n) s% \7 D' Z' F; y
/ f0 O9 z. X" z2 c" C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:
( @, E) T; Y! S" n1 l( A1 h/ `1 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？

3 d, s, M6 a) T6 |! s8 x  ~
$ W& d" R4 ?3 d6 {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。