Answer
$$y\left( t \right) = {C_1}\sin 4t + {C_2}\cos 4t{\text{ is a solution of the given differential equation}}$$
Work Step by Step
$$\eqalign{
& y\left( t \right) = {C_1}\sin 4t + {C_2}\cos 4t \cr
& {\text{differentiate}} \cr
& y'\left( t \right) = {C_1}\left( {4\cos 4t} \right) + {C_2}\left( { - 4\sin 4t} \right) \cr
& y'\left( t \right) = 4{C_1}\cos 4t - 4{C_2}\sin 4t \cr
& y''\left( t \right) = 4{C_1}\left( { - 4\sin 4t} \right) - 4{C_2}\left( {4\cos 4t} \right) \cr
& y''\left( t \right) = - 16{C_1}\sin 4t - 16{C_2}\cos 4t \cr
& {\text{replace }}y\left( t \right){\text{ and }}y''\left( t \right){\text{ in the differential equation }} \cr
& y''\left( t \right) + 16y = 0 \cr
& - 16{C_1}\sin 4t - 16{C_2}\cos 4t + 16\left( {{C_1}\sin 4t + {C_2}\cos 4t} \right) = 0 \cr
& {\text{simplify}} \cr
& - 16{C_1}\sin 4t - 16{C_2}\cos 4t + 16{C_1}\sin 4t + 16{C_{^2}}\cos 4t = 0 \cr
& 0 = 0 \cr
& {\text{then}} \cr
& y\left( t \right) = {C_1}\sin 4t + {C_2}\cos 4t{\text{ is a solution of the given differential equation}} \cr} $$
