a0λ(λ−1)(λ−2)...(λ−n+1)+...an−2λ(λ−1)+an−1λ+an=0
x=eu
(λ−1)λ+2λ−12=0
λ2+λ−12=0
λ1=3
λ2=−4
y′′+y′−12y=ue3u
y0=Ce3u+e4uC1
Solution for ue3u
a+bi=3, then s=1
y1=u(Au+B)e3u
y1′=(3Au2+(3B+2A)u+B)e3u
y1′′=(9Au2+(9B+12A)u+6B+2A)e3u
14Aue3u+(7B+2A)e3u=ue3u
A=1/14
B=-1/49
y1=(14u−491)ue3u
y=(14u−491)ue3u+Ce3u+e4uC1
Comments