Question

求数学通解？

Answer 1

let
u= y/x
du/dx = (x dy/dx - y) /x^2
xdy/dx -y = x^2. du/dx
dy/dx
=( x^2. du/dx +y ) /x
= x. du/dx + u
//
(x^3+y^3)dx -3xy^2dy =0
dy/dx -(x^3+y^3)/(3xy^2) =0
dy/dx -(1/3) (x/y)^2 -(1/3)(y/x) =0
x. du/dx + u - 1/(3u^2) - (1/3)u =0
x. du/dx - 1/(3u^2) + (2/3)u =0
3x. du/dx = 1/u^2 -2u
= (1- 2u^3)/u^2
3∫u^2/(1-2u^3) du = ∫dx/x
-(1/2)∫d(1-2u^3)/(1-2u^3) = ∫dx/x
-(1/2)ln|1-2u^3| = lnx +C'
1/√(1-2u^2) = C''x
√(1-2u^2) = 1/(C''x)
1-2u^2 =C/x^2
u^2 = (x^2 -C)/(2x^2)
(y/x)^2 =(x^2 -C)/(2x^2)
y^2 = (x^2 -C)/2