Find

y' and y''.

y = eαx sin βx

Given: y = e^ (αx) sin (βx)

To find the first derivative, simply use the Product and Chain Rules: y' = e^ (αx) cos (βx) (β) + sin (βx) (αe^ (αx)) y' = αe^ (αx) sin (βx) + βe^ (αx) cos (βx)

You could factor out the e^ (αx) if you wanted. It will make it easier when taking the second derivative. y' = e^ (αx) [αsin (βx) + βcos (βx) ]

To find the second derivative, simple take the derivative of y'. This one involves multiple Chain rules and multiple Product rules. y'' = e^ (αx) [[ (α) (cos (βx) (β) + sin (βx) (0) ] + [ (β) (-sin (βx) (β) + cos (βx) (0) ]] + [αsin (βx) + βcos (βx) ] (αe^ (αx))

That's a lot. Simplify: y'' = e^ (αx) [[αβcos (βx) ] + [-β²sin (βx) ]] + [αsin (βx) + βcos (βx) ] (αe^ (αx))

Distribute: y'' = αβe^ (αx) cos (βx) - β²e^ (αx) sin (βx) + α²e^ (αx) sin (βx) + αβe^ (αx) cos (βx)

And finally, add like-terms and order: y'' = α²e^ (αx) sin (βx) - β²e^ (αx) sin (βx) + 2αβe^ (αx) cos (βx)