第1个回答 2017-03-26
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）.本回答被网友采纳
==>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）.本回答被网友采纳
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