求yy"=(y')^2-（y')^3的通解

设y'=p,则y''=pdp/dy
代入原方程得ypdp/dy=p²-p³
==>p(ydp/dy-p+p²)=0
∴p=0.(1)
ydp/dy-p+p²=0.(2)
由(1)得,原方程的一个解是y=C (C是积分常数)
由(2)得,ydp/dy=p-p²
==>dp/(p-p²)=dy/y
==>ln│p/(1-p)│=ln│y│+ln│C1│ (C1是非零积分常数)
==>p/(1-p)=C1y
==>p=C1y/(C1y+1)
==>y'=C1y/(C1y+1)
==>(1+1/(C1y))dy=dx
==>y+ln│y│/C1=x+ln│C2│/C1 (C2是非零积分常数)
==>y=C2e^(C1(x-y))
即原方程另一解是y=C2e^(C1(x-y)) (C1,C2是非零积分常数)
故 综合上述所有解知,原方程的通解是y=C2e^(C1(x-y)) (C1,C2是积分常数).