Question

微分方程y''-y'-2y=e^x 的特解形式

RT
急急急，在线等

Answer 1

特征方程为r^2-r-2=0,r=2,-1
所以y1=C1e^(2x)+C2e^(-x)
所以设特解y2=Ae^x
则y2'=y2''=Ae^x
所以-2A=1,A=-1/2
所以y=y1+y2=C1e^(2x)+C2e^(-x)-(e^x)/2

Answer 2

The aux. equation
p^2-p-2=0
(p-2)(p+1) =0
p=2 or -1
yp = Ae^x
yp'' -yp' -2yp = e^x
-2Ae^x =e^x
A = -1/2
yp = (-1/2)e^x
y = Ce^(2x) +De^(-x) - (1/2)e^x