老了不死;做代换y=e^z,则lny=z,dy=de^z=e^zdz
ylnydx+(x-lny)dy
=(e^z)zdx+(x-z)e^zdz
=(e^z)[zdx+(x-z)dz]=0
若e^z=0,即y=0
zdx+(x-z)dz
=zdx+xdz-zdz
=d(zx)-(1/2)dz^2
=d[zx-z^2/2]=0
得
zx-z^2/2=c(c为任意常数)
即原方程通解为
xlny-(lny)^2/2=c(c为任意常数)
1年前