Answer
See explanation.
Work Step by Step
$a)$ If $y = e^{2x}$ then $y' = -2e^{-2x}$ and $y'' = 4e^{-2x}$,
substitute into the equation $y'' + y' - 2y = 0 $
$4e^{-2x}+ (-2e^{-2x}) -2(e^{2x})= 0 $
$4e^{-2x}-2e^{-2x} -2e^{2x}= 0 $
$0 = 0$, Verified
If $y = e^x $ then $y' = e^x$ and $y'' = e^x$,
substitute into the equation $y'' + y' - 2y = 0 $
$e^x + e^x - 2e^x = 0 $
$2e^x - 2e^x = 0 $
$ 0 = 0$ Verified.
$b)$ If $y = c_1e^{-2x} +c_2e^x$ then $y' = -2c_1e^{-2x} +c_2e^x$ and $y'' = 4c_1e^{-2x} +c_2e^x,$
substitute into the equation $y'' + y' - 2y = 0 $
$(4c_1e^{-2x} +c_2e^x)+ (-2c_1e^{-2x} +c_2e^x) - 2(c_1e^{-2x} +c_2e^x) = 0$
$4c_1e^{-2x} -2c_1e^{-2x} - 2c_1e^{-2x} -2c_2e^x +c_2e^x +c_2e^x = 0$
$0 = 0$ verified.
