求xy''-y'+xy'^2=0的通解

Question

求xy''-y'+xy'^2=0的通解

Answer 1

令x=e^t,则xy'=dy/dt,x²y''=d²y/dt²-dt/dt
于是,代入原方程得d²y/dt²-2dy/dt+(dy/dt)²=0.(1)
再令dy/dt=p,则d²y/dt²=dp/dt
于是,代入方程(1)得dp/dt-2p+p²=0
==>dp/(p(2-p))=dt
==>ln│p/(2-p)│=ln│2t│+ln│C1│ (C1是积分常数)
==>p/(2-p)=C1e^(2t)
==>dy/dt=p=2-2/(C1e^(2t)+1)
==>y=2t+ln│C1+e^(-2t)│+C2 (C2是积分常数)
==>y=2ln│x│+ln│C1+1/x²)│+C2
==>y=ln│C1x²+1│+C2
经验证y=C (C是积分常数)也是原方程的解
故 原方程的所有解是y=ln│C1x²+1│+C2,或y=C (C,C1,C2是积分常数)