Verifying solutions to differential equations

Question

y"-2y'+5y=0 , y=e^xsin2x

Answer 1

y = e^xsin(2x)

 

y' = e^xsin(2x) + 2e^xcos(2x)

 

y" = [e^xsin(2x) + 2e^xcos(2x)] + 2[e^xcos(2x) - 2e^xsin(2x)]

 

    = 4e^xcos(2x) - 3e^xsin(2x)

 

So, y" -2y' + 5y

 

     = [4e^xcos(2x)-3e^xsin(2x)] - 2[e^xsin(2x) + 2e^xcos(2x)] +5[e^xsin(2x)]

 

     = 0