求y''+1/xy'+(y')²=0的通解

求y''+1/xy'+(y')²=0的通解

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)

在网上找到个和这题类似的，方法一样，你参考下 (x-2xy-y²)y' + y² = 0 (x-2xy-y²)dy + y²dx = 0 xdy - xd(y²) - y²dy + y²dx = 0 (x-y²)dy = xd(y²) - y²dx = x²d(y²/x) (1 - y²/x)dy = xd(y²/x) 令y²/x = u,即x = y²/u 代入得 (1-u)dy = y²du/u 即dy/y² = du/[u(1-u)] = du/u + du/(1-u) 积分得 -1/y = ln|u/(1-u)| + c |u/(1-u)| = ce^(-1/y) u = 1/[1±ce^(1/y)] 通解x = y²[1+ce^(1/y)]

如以上回答内容为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！

点此我要举报以上问答信息