题目

求齐次方程的通解

提问时间：2021-03-18

答案

（1）令y=xt,则y'=xt'+t
代入原方程,得y'=(y/x)ln(y/x)
==>xt'+t=tlnt
==>xt'=t(lnt-1)
==>dt/[t(lnt-1)]=dx/x
==>d(lnt-1)/(lnt-1)=dx/x
==>ln│lnt-1│=ln│x│+ln│C│ (C是积分常数)
==>lnt-1=Cx
==>lnt=Cx+1
==>ln(y/x)=Cx+1
==>lny=lnx+Cx+1
故原方程的通解是lny=lnx+Cx+1 (C是积分常数).
（2）∵(x²+y²)dx-xydy=0
==>(2/x³)(x²+y²)dx=2ydy/x² (等式两端同乘2/x³)
==>2ydy/x²-2y²dx/x³=2dx/x
==>d(y²/x²)=2dx/x
==>y²/x²=ln(x²)+C (C是积分常数)
==>y²=x²[ln(x²)+C]
∴原方程的通解是y²=x²[ln(x²)+C] (C是积分常数).