Линейное уравнение первого порядка 2y'-xy=xyx2-1

2y'∙y-x=xy2x2-1
2dydx∙y-xy2x2-1=x
z=y2=>z'=2y∙y'
z'-x∙zx2-1=x
z=uv, z'=u'v+uv'
u'v+uv'-x∙uvx2-1=x
uv'-x∙uvx2-1=0, u≠0
v'-x∙vx2-1=0
dvv=xdxx2-1
dvv=xdxx2-1
lnv=12lnx2-1+lnC
v=x2-1
u'x2-1=x
u'=xx2-1
du=xx2-1dx
u=x2-1+C
z=x2-1+C∙x2-1=x2-1+C∙x2-1
y=x2-1+C∙x2-1
Проверка:
y'=12x2-1+C∙x2-12x+Cxx2-1
2x+Cxx2-1x2-1+C∙x2-1-xx2-1+C∙x2-1-xx2-1+C∙x2-1x2-1=
=xx2-1+Cx2-1x2-1+C∙x2-1-xx2-1+C∙x2-1x2-1=
=xx2-1+Cx2-1-xx2-1+Cx2-1x2-1x2-1+C∙x2-1=0
Ответ:y=x2-1+C∙x2-1