Answer
$${\text{ }}y = {e^{ax}}\sin bx{\text{ is a solution}}$$
Work Step by Step
$$\eqalign{
& y'' - 2ay' + \left( {{a^2} + {b^2}} \right)y = 0 \cr
& {\text{Let }}y = {e^{ax}}\sin bx \cr
& {\text{Calculate }}y'{\text{ and }}y'' \cr
& y' = \frac{d}{{dx}}\left[ {{e^{ax}}\sin bx} \right] \cr
& y' = a{e^{ax}}\sin bx + b{e^{ax}}\cos bx \cr
& y' = {e^{ax}}\left( {a\sin bx + b\cos bx} \right) \cr
& y'' = \frac{d}{{dx}}\left[ {{e^{ax}}\left( {a\sin bx + b\cos bx} \right)} \right] \cr
& y'' = a{e^{ax}}\left( {a\sin bx + b\cos bx} \right) + {e^{ax}}\left( {ab\cos bx - {b^2}\sin bx} \right) \cr
& y'' = \left( {{a^2}\sin bx + 2ab\cos bx - {b^2}\sin bx} \right){e^{ax}} \cr
& y'' = 2ab{e^{ax}}\cos bx + \left( {{a^2} - {b^2}} \right){e^{ax}}\sin bx \cr
& {\text{Substitute }}y''{\text{ and }}y'{\text{ into }}y'' - 2ay' + \left( {{a^2} + {b^2}} \right)y = 0 \cr
& y'' = 2ab{e^{ax}}\cos bx + \left( {{a^2} - {b^2}} \right){e^{ax}}\sin bx \cr
& - 2ay' = - 2{a^2}{e^{ax}}\sin bx - 2ab{e^{ax}}\cos bx \cr
& \left( {{a^2} + {b^2}} \right)y = {e^{ax}}\sin bx\left( {{a^2} + {b^2}} \right) \cr
& {\text{Replacing and reducing}} \cr
& 2ab{e^{ax}}\cos bx + \left( {{a^2} - {b^2}} \right){e^{ax}}\sin bx - 2{a^2}{e^{ax}}\sin bx \cr
& - 2ab{e^{ax}}\cos bx + {e^{ax}}\sin bx\left( {{a^2} + {b^2}} \right) \cr
& {\text{Simplifying}} \cr
& 0 = 0 \cr} $$
