∵2y'-y=e^x
==>2dy-ydx=e^xdx
==>dy-ydx/2=e^xdx/2
==>e^(-x/2)dy-e^(-x/2)ydx/2=e^(x/2)dx/2 (等式两端同乘e^(-x/2))
==>e^(-x/2)dy+yd(e^(-x/2))=d(e^(x/2))
==>d(ye^(-x/2))=d(e^(x/2))
==>ye^(-x/2)=e^(x/2)+C (C是常数)
==>y=e^x+Ce^(x/2)
∴原方程的通解是y=e^x+Ce^(x/2).
2y'-y=e^x的通解
2dy-ydx=e^xdx==>dy-ydx/2=e^xdx/2==>e^(-x/2)dy-e^(-x/2)ydx/2=e^"}}}'>