设y'=p,则y''=pdp/dy
代入原方程得pdp/dy-p³-p=0 ==>p(dp/dy-p²-1)=0
∴p=0,dp/dy-p²-1=0
∵当p=0时,有dy/dx=0 ==>y=C (C是积分常数)
∴经检验y=C是原方程的解
∵当dp/dy-p²-1=0时,有dp/dy=p²+1
==>dp/(p²+1)=dy
==>arctanp=y+C1 (C1是积分常数)
==>p=tan(y+C1)
==>dy=tan(y+C1)dx
==>cos(y+C1)dy/sin(y+C1)=dx
==>d(sin(y+C1))/sin(y+C1)=dx
==>ln│sin(y+C1)│=x+ln│C2│ (C2是积分常数)
==>sin(y+C1)=C2e^x
∴sin(y+C1)=C2e^x也是原方程的解
故原微分方程的通解是y=C,或sin(y+C1)=C2e^x （C,C1,C2是积分常数）.