数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
1 T7 j0 K7 d8 g0 a' n" O
0 ^* Q* x! r4 ~0 ]d [(a+bx) ] /dx = -kc +s
1 e* t# i' l2 |/ Cwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
2 u. \- y- E8 P
[ 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！

d [(a+bx) ] /dx = -kc +s
9 |$ \% l5 y, T7 {% dwhere: only x and c are unknown, others are all known,  requiire c = ?

$ G! A* o( i/ M, H+ @" l( A$ x: K0 b多谢了！

it is too easy:
3 n. o5 `/ v8 g" x6 Y
d(a+bx)/dx = b

! E! B0 Z" q1 C, k6 n- K. Bso
  B8 E, ?. a$ \
$ t# ~7 a4 {! _) b& u8 Ab=-kc+s
( t' X. h* G2 V- X: f) c. C) C
# q6 j* `, v3 C2 Ffinally:  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:
4 e% P5 ~4 _) w  k
d(a+bx)/dx = b
2 r7 r  u, k8 S% }$ W0 O: L' @- j
so

b=-kc+s
3 P/ y5 G: u. }6 ?% Y
finally:  c= (s-b)/k

3 J1 A" L- Y3 s! _
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！
0 A! |3 x6 ?) R, @/ P
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！

8 L* O9 I8 E. Z" a, j[ Last edited by 醉酒 ...

# }% M3 i2 T  k  \) |5 Y( K/ {
+ v: V# `) O7 t4 A6 Hsorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
9 j' q! L1 D$ ^- @+ u" R2 K
7 s# D) n- i1 {' Qplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
  P, Y* \+ T- ^% u) I' W1 |2 `9 l+ bsorry, did you post another question? your statement is still not clear.
1 t8 k+ F( n0 V
% t0 l! G* q5 c+ t% w* p% t" xa, 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
* L, t: C$ |+ z  U' Kd [(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:
% I& q2 V$ N" g1 o# l0 O对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
( ~. t$ g' c6 q$ F  z7 Y: q- ^
多谢

/ b# l. Q0 w  k# d! {, R0 w6 K
do you mean it is:
& k$ J$ ?3 \  @7 d
3 M/ {1 }3 F& _0 ?8 fd { (a+bx) C(x)}/dx = -k C(x) + s
: y4 ?: L8 T' N; Y! ^
9 X' N$ T, Y( kwhere a, b, s are constants, you want to know what is C(x) ?
! f: u) z9 |% X2 n) v! m, j
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:

d { (a+bx) C(x)}/dx = -k C(x) + s
9 v2 E! _8 f- j8 T+ x" q: t7 w7 I
5 ^( q3 o9 M- E# `6 bwhere a, b, s are constants, you want to know what is C(x) ?
0 [5 _) s6 w1 x& i
6 P3 a( C9 V4 Mif so, this needs to solve the differential equation. please comfirm it fi ...

$ v' y* {+ [* {. q8 s  a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:
, H) b% V4 Q6 T7 a; I
) V/ y. Q5 k5 X6 \; nC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
1 [. x) e# M6 A( y6 ^/ M9 a

procedure:
, E4 j1 _0 p# x6 ~/ t
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
9 n4 A& C: ^$ h2 ?
+ a5 o( I8 @* tbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 |+ }# X3 t: U9 [% \i.e.

  S7 J+ J1 M, F4 J9 a* ~' A! a0 u5 u(a+bx) dC(x)/dx  = -(k+b)C(x) +s
* R7 w( T7 W3 J! D. m

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:

/ V0 X) X' D7 z" I  p- E* D* SdY(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)
) l$ S  b; O$ {% O3 U! Q
this means:  Y(x) = (a+bx)^(K/b)
' Z; p# `1 T7 R8 z) O, ?by using early transform, we can have:

  _( v$ k& K. b8 ~4 n9 M; u-(k+b)C(x)+s = (a+bx)^(k/b+1)
( w$ e2 k: h5 P. H: G8 t& R, B
" s1 j, V$ X& d5 n; d  t* mfinally:

# w  y! I- u; g* N5 b. _+ v1 aC(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:
$ A) I1 d+ ^' [0 n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！

: X" k3 c) l! N- c; ]+ Q1 ^4 w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。
' _$ H0 C- Y% m* ]5 w# r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
so:
7 n# \) l  M* J/ i& b& pbC(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
* T4 y' j9 x& K8 D4 G7 Fd{(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
; E. N, j* j# s) V& j( dx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
; ?7 B, |0 B/ a: j4 I" s  I因此：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
+ Q3 V8 q3 Z/ u9 a1 R7 ], H最右边是x的平方。
0 H# f* ?8 X  G3 `
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。
. }  a! E: b  u3 b( t) C您说：
" k( s. y; f8 S! @
- D0 Y) {  P# d- b) {d{(a+bx)*C(x)}/dx =-k C(x) + s
, `9 r4 t6 P7 W' v; d) ^9 xso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
1 d" b: d/ @) b  z
. ~6 t# v& Z' e9 l/ s+ f0 h' Z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)* ...

; H/ ~  |& l$ e  e, [! RPlease note here:

/ n5 H% }9 S9 }/ j7 Od{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
4 @) T- J! H+ q" p: s* Lit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
1 X+ V9 i  z! |. yd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
5 E& T  P/ P" o9 E/ [7 ^9 {并不是   (a+bx)*C(x)   的导数除以   x   的导数。
3 t. ^- _! w+ u+ P
8 x4 C9 r: D6 Ano "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

这样算混淆了微分和导数的概念。
3 k6 U* k  c0 H+ z3 K2 {2 s8 C( t
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)
6 ~* S. B- q9 j8 L
6 O5 d% p* J- p6 {: s5 L5 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:
这样算混淆了微分和导数的概念。
+ E' M: c& Q" x* s1 t2 r) z' h
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
8 G) }  o3 t" X* k6 D8 j
) {0 {2 X& L: d; I( gIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

) k4 A- E+ o2 X5 M& 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
* a! i) L. X, B  m- ^, U3 b
Ah, you made a mistake: it should be:

, W- y+ ?+ z/ H( i- sd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
/ N2 x  I7 N$ O2 z7 f) m2 Y

- Y. W5 \7 h& GIf 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)
" j) q. E; Z2 J: W
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

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
这样算混淆了微分和导数的概念。

% T3 L! K( l5 n  r5 V( ], A, dd(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
. d7 S3 H  b9 X
: S* i8 K$ g. s# ~- xAh, you made a mistake: it should be:
  t! Z, n% q7 W, k/ M
d(f(x)g(x)) ...

2 v( N/ X; a: P- v4 KYes, there should not be " ' " after "f" and "g", it is a typo. :D

9 O- l- l# z# N. q) cBut 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:
9 d: r" p/ N: }2 S! J$ xYes, there should not be " ' " after "f" and "g", it is a typo. :D
; E9 E, i$ A3 _
But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

+ P) W# w0 D+ R0 k% X2 S
. k# s' ?6 v. F6 l- H5 F+ i6 ]Thanks, 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:
1 |& W( l3 \+ l
" Y7 G2 W: a3 k& V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 ...

9 C- c, j; e# j3 u2 c
% Z. B; {4 i( {; a" ]0 [3 `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:
7 D: D$ W4 j5 ~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？

7 ~2 b" P( ]: ?/ U$ k1 d
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。