求方程的通解x∧2(y'+y∧2)+4xy+2=0感激不尽

Question

求方程的通解x∧2(y'+y∧2)+4xy+2=0 感激不尽

Answer 1

∵x^2(y'+y^2)+4xy+2=0
==>x^2y'+x^2y^2+4xy+2=0
==>x^2y'+(xy+2)^2-2=0
==>xt'=t(1-t) (令t=xy+2,再化简)
==>dt/t(1-t)=dx/x
==>[1/t-1/(t-1)]dt=dx/x
==>ln│t│-ln│t-1│=ln│x│+ln│C│ (C是积分常数)
==>t/(t-1)=Cx
==>t=Cx/(Cx-1)
==>xy+2=Cx/(Cx-1)
==>xy=(2-Cx)/(Cx-1)
==>y=(2-Cx)/(Cx^2-x)
∴原方程的通解是y=(2-Cx)/(Cx^2-x).