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:/ x9 C: y3 W1 ^) K2 r
请教大家一道微分题，多谢！" ]! I" M! s) z- O
7 D$ R$ L& X' O% [- z8 t! [8 n
d [(a+bx) ] /dx = -kc +s
4 o, `3 b% ]5 w8 w9 @5 rwhere: only x and c are unknown, others are all known, requiire c = ?; {) A# x5 y! X O4 g

5 i% H* U; g1 K' n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:
& _ v2 Q- i0 A( git is too easy:) w0 t0 C. o5 u6 j

9 W; q% [' w; b! d5 S3 yd(a+bx)/dx = b
1 D6 Z$ H2 |2 y& ]
, Z! P- G! f! Qso
; j+ v5 @8 {( E: q
) I2 [5 b& S( C0 Q! Xb=-kc+s
' D3 b* v! M5 }: z
; z; X4 s0 i7 W1 E% R3 ^" \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 20:41:2 z# m4 o* M) Z$ g9 I+ ?2 V5 s
请教大家一道微分题，多谢！
" v5 R4 f8 ~- R  D3 |2 R) ]! m9 a5 {& P* J% U* h% g
d [(a+bx) ] /dx = -kc +s
+ E! |7 F; a7 J' V* l+ I8 Zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?: d6 P8 ^: ~0 N7 Z- k0 T9 i. g: w3 w
" R7 r! ^: m2 g' w9 t, s2 _
多谢了！
2 z! n' S8 z/ ~' q3 I
$ ^9 L  a. ]* k1 ^& p  C3 P[ 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:) W" S! X' s* s1 J/ u! g
sorry, did you post another question? your statement is still not clear., }1 g" X9 f# I$ K' Q+ k

1 K# U9 w2 G Y" S0 U- Ka, b, s, k are constants? c means c(x) ?
4 Y8 S1 y) w/ r% E6 d! x
, j I; z" a9 vplease 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:
/ s6 C& Y' i1 [7 S* f7 g对不起，写错了，忘了变量c,应该是0 O- P1 q2 r( o/ u! ` B
d [(a+bx) c] /dx = -kc +s + i* s- Q0 q6 F% @8 ?9 |

" v9 H8 o$ B- i- z2 t% l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:
( E2 W8 P# ^* c; v! y Gdo you mean it is:
- K/ N9 q( ]7 ^7 V6 E0 F: o: P1 n0 W/ n) e# c
d { (a+bx) C(x)}/dx = -k C(x) + s3 n% k8 G! o. t: o" [/ C9 j. v1 c$ F

& K$ e: ^# t# P1 J3 h8 fwhere a, b, s are constants, you want to know what is C(x) ?; k+ i6 U) F% d6 |' B, B2 |# o

! b( V: P5 |' E2 c1 R+ G: Iif 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:
4 F7 w/ \7 U# h4 d- p, 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！

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:' f/ w- a2 {& ^1 ^$ |
9楼的答案有点问题吧。
9 T2 `3 m; s5 ?& S您说：/ b9 b( }4 j9 M* B

d* Q1 h5 Q) T8 [7 {+ @, zd{(a+bx)*C(x)}/dx =-k C(x) + s
" f0 R6 l1 m/ f0 {so:* s6 s6 f7 _" j3 D6 U% j
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# O4 T$ D" b6 b5 l' W+ N, K$ g! ~! q3 {
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx% w: Y! v( u+ x( o: ^) C( q2 u) n8 r
但这是不正确的; a# p& H: {1 v0 S7 ~
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
8 z9 \' f- P( V& w/ X....
7 d2 r! M) L% q- |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。

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:' F/ D- L, h. j" g1 X" }1 M/ g
这样算混淆了微分和导数的概念。. b2 t6 G: T2 p' |: c5 M! ~

' a/ U) l. }& {, h4 h4 nd(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3 D: ~! n/ {* B7 [# U6 O2 C
# p: j$ W- N, m0 G4 g
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:) K) E/ b M2 P$ J5 h9 k7 C
这样算混淆了微分和导数的概念。- S/ H4 z- Q1 p9 K, S, `
$ _2 b/ d' y! z
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
& O% e& N D- a8 B4 s. t0 h* j6 s9 {7 X, e# J+ d
Ah, you made a mistake: it should be:
6 h% ]) H& b) C+ [! @1 p0 P* ?" C* W" w9 J( Z" C
d(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:% w4 K4 {( I3 e# j( S
Yes, there should not be " ' " after "f" and "g", it is a typo. :D- I; u e' n8 `; J8 T- w

9 L: L! S3 ]$ W0 |9 H; [) gBut 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:: Y5 o6 M0 R+ W J; k/ C
5 a0 R/ k6 L- d- O+ x
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。
7 W0 [& P( P7 o+ ?/ t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:" m( e$ R, ~" {! d R
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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