这道伯努利方程怎么解

变形为dx/dy+yx=y³x²,两边同乘-1/x²,得
-(1/x²)dx/dy-y/x=-y³,令z=1/x,则得
dz/dy-yz=-y³ =>dz-yzdy=-y³dy,两边乘以积分因子e^(-y²/2),得
e^(-y²/2)dz-e^(-y²/2)yzdy=-y³e^(-y²/2)dy
=>d(ze^(-y²/2))=d(∫-y³e^(-y²/2)dy)
=>ze^(-y²/2)=(2+y²)e^(-y²/2)+C
=>x=[e^(-y²/2)]/[(2+y²)e^(-y²/2)+C]