設y'=p,則y''=pdp/dy
代入原方程,得pdp/dy-p²=1
==>pdp/(1+p²)=dy
==>d(1+p²)/(1+p²)=2dy
==>ln(1+p²)=2y+ln(C1²) (C1是積分常數(shù))
==>1+p²=C1e^(2y)
==>p=±√[C1²e^(2y)-1]
==>dy/√[C1²e^(2y)-1]=±dx
==>e^(-y)dy/√[C1²-e^(-2y)]=±dx
==>d[e^(-y)]/√[C1²-e^(-2y)]=±dx
==>arcsin[e^(-y)/C1]=C2±x (C2是積分常數(shù))
==>e^(-y)=C1sin(C2±x)
故原方程的通解是e^(-y)=C1sin(C2±x) (C1,C2是積分常數(shù)).