y'+(1/x)y=x^2
Замена y=uv
y'=u'v+uv'
u'v+uv'+1/x u'v = x^2
u(v'+1/x v)+u'v = x^2
Пусть v'+1/x v = 0
Тогда dv/dx+v/x=0 =>dv/v=-dx/x =>v=-x
xu'=x^2 => du/dx=x => u=1/2x^2+c
y=-x(1/2x^2+c)