Answer
(i) $y'' - y = y''_{1} - y_{1} = e^{t} - e^{t} = 0$.
(ii) $y'' - y = y''_{2} - y_{2} = cosh(t) - cosh(t) = 0$.
Work Step by Step
We wish to verify that (i) $y_{1} = e^{t}$ and (ii) $y_{2} = cosh(t)$ are solutions to the differential equation y'' - y = 0.
For (i) we must find $y_{1}''$.
Since $D_{t} e^{t} = e^{t}$, It follows that $D_{t}''e^{t} = e^{t}$.
I.e., $y_{1}'' = e^{t}$. Thus,
$y'' - y = e^{t} - e^{t} = 0$.
For (ii), $y_{2} = cosh(t)$, we recall that by definition, $cosh(t) = \frac{e^{t} + e^{-t}}{2}$.
Then, $D_{t}(\frac{e^{t}+e^{-t}}{2}) = \frac{e^{t}+e^{t}}{2}$.
Then $D_{t}^{2}(\frac{e^{t}+e^{-t}}{2}) = D_{t}(\frac{e^{t}+e^{-t}}{2}) = \frac{e^{t}+e^{-t}}{2} = cosh(t)$.
So $y'' - y = cosh(t) - cosh(t) = 0$, as was to be shown.
