Question

求微分方程y"+y'+y＝3e^2x的通解

常微分方程求解 大学知识

Answer 1

y"+y'+y＝3e^(2x)
The aux. equation
r^2+r+1 =0
r=(-1+√3i)/2 or (-1-√3i)/2
let
yg= e^[(-1/2)x].[Acos(√3x/2) +Bsin(√3x/2)]
yp=Ce^(2x)
yp'=2Ce^(2x)
yp''=4Ce^(2x)
yp"+yp'+yp＝3e^(2x)
7C=3
C=3/7
yp=(3/7)e^(2x)
通解
y=yg+yp=e^[(-1/2)x].[Acos(√3x/2) +Bsin(√3x/2)] +(3/7)e^(2x)