（y^2-6x）y'+2y=0 求通解.
1个回答
∵(y^2-6x)y'+2y=0 ==>(y^2-6x)y'=-2y
==>(y^2-6x)dy/dx=-2y
==>dx/dy=(y^2-6x)/(-2y)
==>dx/dy=3x/y-y/2
==>dx/dy-3x/y=-y/2
∴先解齐次方程dx/dy-3x/y=0的通解
∵dx/dy-3x/y=0 ==>dx/dy=3x/y
==>dx/x=3dy/y
==>ln|x|=3ln|y|+ln|C| (C是积分常数)
==>x=Cy³
∴齐次方程dx/dy-3x/y=0的通解是x=Cy³ (C是积分常数)
于是,应用“常数变易法”,设原微分方程的通解为x=uy³ (u是关于y的函数)
∵dx/dy=y³du/dy+3uy²
∴把它代入dx/dy-3x/y=-y/2
得y³du/dy+3uy²-3uy³/y=-y/2
==>y³du/dy+3uy²-3uy²=-y/2
==>y³du/dy=-y/2
==>y²du/dy=-1/2
==>du=-dy/(2y²)
==>u=1/(2y)+C (C是积分常数)
把u=1/(2y)+C代入x=uy³,得x=[1/(2y)+C]y³=y²/2+Cy³
故原微分方程的通解是x=y²/2+Cy³ (C是积分常数).
