求解xyy'+x^2+y^2=0

2010-05-17 4:23 am
求解xyy'+x^2+y^2=0 
我算出來的答案怪怪的....
會的大大可以幫我解ㄧ下嗎?
謝謝^^

回答 (1)

2010-05-17 4:38 am
✔ 最佳答案
xyy' + x^2 + y^2 = 0

y' = -(x^2 + y^2) / xy

= -[1 + (y/x)^2] / (y/x)

Let y = ux

y' = u + xu'

So, the differential equation becomes:

u + xu' = -(1 + u^2) / u

xu' = -(1 + 2u^2)/u

Separating variables,

S udu / (1 + 2u^2) = -S (1/x) dx

1/4 S d(1 + 2u^2)/(1 + 2u^2) = -lnx

1/4 ln(1 + 2u^2) = -lnx + C

ln(1 + 2u^2) = -4lnx + C'

ln(1 + 2u^2) + 4lnx = C'

ln(1 + 2u^2) + lnx^4 = C'

ln(1 + 2u^2)x^4 = C'

(1 + 2u^2)x^4 = A

(1 + 2(y/x)^2)x^4 = A

x^4 + 2x^2y^2 = A, where A is a constant
參考: Prof. Physics


收錄日期: 2021-04-19 22:14:13
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100516000010KK07767

檢視 Wayback Machine 備份