设y'=p
则代入原方程得 p³-4xyp+8y²=0
==>x=p²/4y+2y/p.(1)
令f(y,p)=p²/4y+2y/p
∵f'y=-p²/(4y²)+2/p
f'p=p/(2y)-2y/p²
∴代入公式 (1-f'y*p)dy-f'p*pdp=0
得 [1+(p²/(4y²)-2/p)p]dy-[p/(2y)-2y/p²]pdp=0
==>(p³-4y²)/(4y²)dy+(p³-4y²)/(2yp)dp=0
==>(p³-4y²)[dy/(4y²)-dp/(2yp)]=0
==>p³-4y²=0,或dy/(4y²)-dp/(2yp)=0
∵当dy/(4y²)-dp/(2yp)=0时,
有dp/p=dy/(2y) ==>ln│p│=ln│y│/2+ln│C1│ (C1是积分常数)
==>p=C1√y
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