解：令p=y'，则有：xp‘+p=0。即x dp/dx+p=0,xdp+pdx=0.得dpx=0,即px=c,即y'x=c.即y'=c/x 得y=lncx+c1,得y=lnx+C(C=lnc+c1)
xy''+y'=0y'=pxp'+p=0p'=-p/xdp/p=-dx/xlnp=ln(1/x)+Cp=C'/xdy/dx=C'/xy=C'lnx+C0
