第1个回答 2011-12-18
解:设t=xy,则y=t/x,y'=(xt'-t)/x²
代入原方程,得x((xt'-t)/x²)+t/x=(t/x)(lnx+ln(t/x))
==>t'-t/x+t/x=(t/x)lnt
==>t'=tlnt/x
==>dt/(tlnt)=dx/x
==>d(lnt)/lnt=dx/x
==>ln│lnt│=ln│x│+ln│C1│ (C1是积分常数)
==>lnt=C1x
==>t=e^(C1x)
==>xy=C^x (C=e^C1,也是积分常数)
故原方程的通解是xy=C^x (C是积分常数)。