Question

求大神解答此题！

Answer 1

参考

追问

谢谢

Answer 2

y''+ay'+by = ce^x
y= (1/2)e^(2x) + (x-1/3) e^x 是解
The aux. equation
p^2+ap +b =0
解: p1=2 , p2 =1
p1+p2 = -a
a=-(p1+p2)= -(1+2)=-3
b= p1.p2 = 2
(a,b) = (-3,2)
y''-3y'+2y = ce^x
yp = (x-1/3) e^x
yp' = (x+2/3) e^x
yp''= (x+5/3) e^x
yp''-3yp'+2yp = ce^x
c= 5/3 - 3(2/3) -2/3
=5/3 - 6/3 -2/3
=-1
(a,b,c) =(-3,2, -1)
ans : A

追问

特征方程是怎么求出特征根的呢

还有求C时的yp是什么意思呢