(dy/dx)^3-4xy(dy/dx)+8y^2=0求通解

Question

(dy/dx)^3-4xy(dy/dx)+8y^2=0求通解

Answer 1

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
代入(1)式得 x=C1²/4+2√y/C1
==>y=(C1²/4)(x-C1²/4)²
==>y=C(x-C)² (令C=C1²/4)
当p³-4y²=0时,有p³=4y² ==>p=(4y²)^(1/3)
代入(1)式得 x=[(4y²)^(1/3)]²/(4y)+2y/(4y²)^(1/3)=3(y/4)^(1/3)
==>x³=27y/4
==>y=4x³/27
∴原方程的通解是y=4x³/27和y=C(x-C)² （C是积分常数）.