Question

解微分方程

Answer 1

xy'+y=y(lnx+lny)
(xy)'=yln(xy)
lnxy=u
xy=e^u
(xy)'=u'e^u=ue^u/x
u'=u/x
du/u=dx/x
ln|u|=ln|x|+C0
u=Cx
通解ln(xy)=Cx

还有一种分离变量法
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)