急,求微分方程xy'+y=e^x在初始条件y(1)=e下的特解

Question

如题所述

Answer 1

xy'+y=e^x
(xy)'=e^x
d(xy)/dx=e^x
∫d(xy)=∫e^xdx
xy=e^x+c
y(1)=e ==> e=e+c ==>c=0
xy=e^x
y=e^x/x