∵xdy/dx-y=(x-1)e^x
==>xdy-ydx=(x-1)e^xdx
==>(xdy-ydx)/x^2=(x-1)e^xdx/x^2(等式两端同除x^2)
==>d(y/x)=d(e^x/x)
==>∫d(y/x)=∫d(e^x/x)
==>y/x=e^x/x+C(C是常数)
==>y=e^x+Cx
∴原方程的通解是y=e^x+Cx.