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