-1/x积分得:-lnx+C1令f(x)=e^(lnx+C1),则f'(x)=e^(-lnx+C)/(-x)=f(x)/(-x)原式两边乘以f(x)f(x)y'-yf(x)/x=xe^x*f(x)f(x)y'+yf'(x)=x*e^x*e^(lnx+C1)=x*e^x*(x+e^C1)=x^2*e^x+x*e^x*e^C1两边积分:f(x)*y=x^2*e^x+(e^...