求下列微分方程的同解

1.设y'/x=t,则y'=xt,y''= t+xdt/dx
∴x( t+xdt/dx)=xt*lnt ==>xdt/dx=t(lnt-1)
==>dt/[t(lnt-1)]=dx/x
==>d(lnt)/(lnt-1)=dx/x
==>ln│lnt-1│=ln│x│+ln│C1│ (C1是积分常数)
==>lnt-1=C1x
==>y'/x=e^(C1x+1)
==>y'=xe^(C1x+1)
故y=∫xe^(C1x+1)dx
=xe^(C1x+1)/C1-1/C1∫e^(C1x+1)dx (应用分部积分法)
=xe^(C1x+1)/C1-e^(C1x+1)/C1²+C2 (C2是积分常数)
=(x-1/C1)e^(C1x+1)/C1+C2 （C1,C2是积分常数）；
2.设y'=p,则y''=pdp/dy
∴ypdp/dy-p²=y²p ==>p(ydp/dy-p-y²)=0
==>p=0,或ydp/dy-p-y²=0
当p=0时,y'=0 ==>y=C (C是积分常数)
当ydp/dy-p-y²=0时,有ydp/dy-p=y².(1)
先求齐次方程ydp/dy-p=0的通解
∵ydp/dy-p=0 ==>dp/p=dy/y
==>ln│p│=ln│y│+ln│C│ (C是积分常数)
==>p=Cy
∴齐次方程ydp/dy-p=0的通解是p=Cy
于是,设方程(1)的通解为p=C(y)y (C(y)表示关于y的函数)
∵dp/dy=C'(y)y+C(y)
代入(1)得C'(y)y²+C(y)y-C(y)y=y²
==>C'(y)=1
==>C(y)=y+C1 (C1是积分常数)
∴方程(1)的通解是p=y(y+C1)
==>y'=y(y+C1)
==>dy/[y(y+C1)]=dx
==>[1/y-1/(y+C1)]dy=C1dx
==>ln│y│-ln│y+C1│=C1x+ln│C2│ (C2是积分常数)
==>y/(y+C1)=C2e^(C1x)
==>y=(y+C1)C2e^(C1x)
==>[1-C2e^(C1x)]y=C1C2e^(C1x)
==>y=C1C2e^(C1x)/[1-C2e^(C1x)]
故原微分方程的通解是y=C1C2e^(C1x)/[1-C2e^(C1x)],或y=C （C,C1,C2是积分常数）.