数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

' M2 f3 w3 D5 W6 N, b! yd [(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?

4 j+ l9 d0 w9 S多谢了！
/ Y3 H* z7 d( Z2 M8 P
6 G  f" C* l" Y- h7 r[ 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:
' |6 Y* n, s0 _5 ?; ]+ q3 }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 = ?

多谢了！

9 a4 A9 G; o4 F. g, w# D% y* U9 R; Qit is too easy:

d(a+bx)/dx = b

so

, s: J1 z9 m4 a2 I2 ?$ fb=-kc+s
7 w5 f+ m2 ]9 e
; S& b0 v$ T/ Y, N" ^& S, Mfinally:  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:
" G8 C: t5 ^0 |$ t) p
% E- y* g  V8 Z: \' sd(a+bx)/dx = b

1 {8 s0 d; `& ^, H9 fso

3 z& x. p, k% k6 L; ~3 gb=-kc+s

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_

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
4 h+ A' ~6 J1 ]where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

3 ]2 {  L; t7 n多谢了！

[ Last edited by 醉酒 ...

' a: T# G( F9 \( ]sorry, did you post another question? your statement is still not clear.
7 Z' w8 b! R9 m; j9 t% y7 t8 m6 G  \
a, 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:
9 w" `! }9 [) b7 ^sorry, did you post another question? your statement is still not clear.
& {3 b8 R9 n0 ^8 s' _
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
, C, d& p& O, S- P/ t5 m) G
1 F" Z1 b. W% X多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
* o7 w9 a( D1 V0 e# Z" W对不起，写错了，忘了变量c,应该是
- s8 l$ z5 J  z' j0 y0 p" xd [(a+bx) c] /dx = -kc +s
4 Y! b, \  c0 X1 J3 X, p
- P3 f5 T$ B+ C多谢

do you mean it is:

0 s0 i  O% M* B% m2 |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

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
5 u8 D5 }7 T1 M
$ @# B' z; K5 J2 k. r: _* I% |where a, b, s are constants, you want to know what is C(x) ?

5 ^: N/ k6 y7 {2 N9 w+ J4 ]if so, this needs to solve the differential equation. please comfirm it fi ...

3 r" x  J4 }6 N; j2 X- x: H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:

C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
# h) E6 S, r/ z& B7 t0 f! I2 }
& d6 B! P2 K, ^0 e
procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
5 [. l- ^5 Z6 ^8 U
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

$ r2 N" q3 M9 I: B* @1 c
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:

2 i! B" H. `+ c/ b1 K{(a+bx)/K} dY(x)/dx=Y(x)

5 p+ i* y. L4 cfrom here, we can get:

# r! F: y7 c8 Q3 k% z2 hdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

7 \. d! h0 S) U/ }0 F! Zso that:   ln Y(x) =( K/b) ln(a+bx)
: p; L; r1 Q4 d8 p# J" w
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
. U9 i3 S/ F1 m2 o% ]1 y6 P+ M! E
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
; x  H& `  R6 a0 {0 D; ~
C(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:
/ f$ ?5 \) i5 l; k- E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  k% _5 |4 m2 t! Q0 m5 o. \
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。
0 M/ o2 c& S+ {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。
, P3 q$ K6 n; j3 \$ t/ L" z您说：
5 d7 Q$ f/ Y6 n$ l0 a
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
7 F+ P4 a& \' l9 ~6 |) w, v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
, L2 r$ z' J+ l4 O& px' 意思是x的导数，y'意思是y的导数。分母是y的平方。
) H+ }  X8 f0 l6 p4 k* l# L8 \& s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。

4 g5 G! R* K: P. ~' z  u[ 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:
( U0 ?/ b' M' }7 [' A; i1 l9楼的答案有点问题吧。
您说：
, i. u+ r* q8 S& P: t
d{(a+bx)*C(x)}/dx =-k C(x) + s
3 z$ f( \/ j0 w0 X  s9 Y  _so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
0 }6 ]8 V" h% w) d& a6 |
3 {' {$ }1 z( p( f6 A# }" w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)* ...

  f! t# E/ E+ f" H
Please note here:
" ^# k# i9 D! d; I, z, Q/ n( H
6 U  n' p7 X1 }& E8 y6 @$ a/ I: sd{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
& T( c# N2 T! I* @5 oit is a multiple, not division. Right ?

十年移民路_

但这是不正确的
6 k& e4 u0 l* T2 n/ y9 k9 E# od{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
- r2 c' ~  \. e! p7 R* E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
....
  J; _7 V3 `# C. A' Z% O( f* p! }/ m- pd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

8 D1 C( }6 ^' y6 K& [( W; k
2 k8 a. V- }7 ]" _1 w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_

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。
# \& i0 H7 D( q6 k5 x
6 H/ L- E& k* H7 e$ ~, j1 B$ }0 o2 ud(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. l( L! K% ?, q2 B, t1 Q9 M
2 i1 }) J7 K3 }; zIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

$ c8 I) \/ M' p8 `: n2 L; I
- l6 T& y2 y8 M) }2 S9 \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
. \1 n* G- e2 Y' g/ U- y" m' c
! H" L( [$ U1 h0 a6 j* gAh, 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
3 C7 F3 Z( c# b' Y" o

' ~+ R: [7 S+ n+ y5 cIf 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)

& h1 y$ a9 n6 C: zfor 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:
8 }1 n" X$ \: Y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)) ...

9 J0 p5 g1 j3 U( c% W# t  ~1 l
) ^6 L% K$ N6 O, vYes, 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:
6 ~; U  M, e$ q) uYes, there should not be " ' " after "f" and "g", it is a typo. :D

- B  {# \3 o5 B. \" ^But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

7 B, F2 k) U0 [' P& nThanks, 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:

/ W+ h! J/ m) S" m" O5 W: Z- ~$ u  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 ...

2 ^. V: w$ ~! D& m0 E
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:
2 A7 u: p0 Y6 j9 k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？

/ Z7 h( b8 s/ q: M$ R* F& r9 N9 {( B
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。