Find the unique solution of the differential equation y''+8y'+16y=0 when y(0)=5 and y'(0)=-16?

  y'' + 8y' + 16y = 0  ...  y(0) = 5 , y'(0) = -16

assume: y = e^(λx)

     [e^(λx) ]''  +  8[e^(λx) ]'  +  16[e^(λx) ]  =  0

       λ²  e^(λx)  +  8 λ e^(λx)  +  16 e^(λx)  =  0

            e^(λx) • (λ² + 8 λ + 16)  =  0

    λ² + 8 λ + 16 = 0

          (λ + 4)² = 0

         λ = -  4 ... multiplicity = 2 , or it's a double zero

 yı = (Cı) • e^(-  4x) and yıı = (Cıı) • e^(-  4x) • x

    y = yı + yıı

    y = (Cı)  •  e^(-  4x)  +  (Cıı)  •  e^(-  4x)  •  x

  y' = -  4(Cı)  •  e^(-  4x)  +  (Cıı)  •  e^(-  4x)  +  -  4(Cıı)  •  x  •  e^(-  4x)

  y(0) = (Cı)  +  (Cıı)  •  x

  y'(0) = -  4(Cı)  +  (Cıı)  +  -  4(Cıı)  •  x

      5 = (Cı)  +  (Cıı)  •  x

       -16 = -  4(Cı)  +  (Cıı)  +  -  4(Cıı)  •  x

    ——————————————————<solve>

       Cı = 5 and Cıı = 4

  y = 5e^(-  4x)  +  4e^(-  4x)  •  x

  y = e^(-  4x) • [ 5  +  4x ]

Why does the y2 have and extra x multiplied to the general solution....that s the part my teacher didn t explain well