积分因子法：两边同乘x^2,得x^2e^xdx+(3x^2y^2dx+2x^3ydy)=0x^2e^xdx=d[(x^2-2x+2)e^x]（分部积分计算）；3x^2y^2dx+2x^3ydy=d(x^3y^2)故有d[(x^2-2x+2)e^x+x^3y^2]=0,同解(x^2-2x+2)e^x+x^3y^2=C.其实它是贝努利方...