a0âÎť(Îťâ1)(Îťâ2)...(Îťân+1)+...anâ2âÎť(Îťâ1)+anâ1âÎť+anâ=0
x=eu
(Îťâ1)Îť+2Îťâ12=0
Îť2+Îťâ12=0
Îť1â=3
Îť2â=â4
yâ˛â˛+yâ˛â12y=ue3u
y0â=Ce3u+e4uC1ââ
Solution for ue3u
a+bi=3, then s=1
y1â=u(Au+B)e3u
y1â˛â=(3Au2+(3B+2A)u+B)e3u
y1â˛â˛â=(9Au2+(9B+12A)u+6B+2A)e3u
14Aue3u+(7B+2A)e3u=ue3u
A=1/14
B=-1/49
y1â=(14uââ491â)ue3u
y=(14uââ491â)ue3u+Ce3u+e4uC1ââ
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