Question

xy'=yln(x/y)通解

Answer 1

xy'=yln(x/y)通解
设u=x/y
则y=x/u
y'=1/u-x（u'/u^2)
代入原方程：1-xu'/u=lnu
u-xu'=ulnu
u-x(du/dx)=ulnu
u-ulnu=x(du/dx)
du/[u-ulnu]=dx/x
d(lnu)/(1-lnu)=dx/x
积分：ln|1-lnu|=-ln|x|+C1
x-xlnu=C

Answer 2

let
u= x/y
uy = x
y.du/dx + u.dy/dx =1
(x/u).du/dx + u.dy/dx =1
dy/dx = [1- (x/u).du/dx]/u
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xy'=yln(x/y)
x{ [1- (x/u).du/dx] /u } = (x/u)lnu
1- (x/u).du/dx = lnu
(x/u).du/dx = 1-lnu
∫dx/x = ∫ du/[u(1-lnu)]
ln|x| = -∫ dln(1-lnu)
= -ln(1-lnu) + C'
x = C/(1-lnu)
= C/[ 1- ln(x/y) ]
x[ 1- ln(x/y) ] = C
x(1- lnx) + xlny = C
y = e^{ [C-x( 1- lnx) ]/x }

Answer 3

Answer 4