求函数的通解y^3乘y''-1=0

Question

求函数的通解y^3乘y''-1=0

Answer 1

∵设y'=p,则y''=(dp/dy)(dy/dx)=pdp/dy
代入原方程y³pdp/dy-1=0 ==>pdp=dy/y³
==>p²/2=C1/2-1/(2y²) （C1是不等于零积分常数）
==>p²=C1-1/y²
==>p=±√(C1-1/y²)
∴y'=p=±√(C1-1/y²)==>dy/√(C1-1/y²)=±dx
==>ydy/√(C1y²-1)=±dx
==>(1/(2C1))d(C1y²-1)/√(C1y²-1)=±dx
==>(1/C1)√(C1y²-1)=C2/C1±x (C2是积分常数)
==>√(C1y²-1)=C2±C1x
==>C1y²-1=(C2±C1x)²
==>C1y²=(C2±C1x)²+1
故原微分方程的通解是C1y²-(C2±C1x)²=1 （C1(≠0)和C2是积分常数）