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' - 6y = 0 $
$4e^{-2x} - (-2e^{-2x}) - 6(e^{-2x}) = 0$
$6e^{-2x}- 6e^{-2x} = 0$
$ 0 = 0 $ Verified
If $y = e^{3x}$ then $ y' = 3e^{3x}$ and $y'' = 9e^{3x}$,
substitute into the equation $y'' - y' - 6y = 0 $
$9e^{3x} - 3e^{3x} - 6(e^{3x}) = 0$
$ 0 = 0$ verified
$b)$ If $y = c_1e^{-2x} + c_2e^{3x} $
then $y' = -2c_1e^{-2x} + 3c_2e^{3x}$
and $y'' = 4c_1e^{-2x} + 9c_2e^{3x} $
substitute into the equation $y'' - y' - 6y = 0 $
$(4c_1e^{-2x} + 9c_2e^{3x} ) - (-2c_1e^{-2x} + 3c_2e^{3x}) - 6(c_1e^{-2x} + c_2e^{3x} ) = 0$
$(4c_1e^{-2x} + 2c_1e^{-2x} - 6c_1e^{-2x}) + (9c_2e^{3x} - 3c_2e^{3x}-6c_2e^{3x} ) = 0$
$ 0 = 0 $ verified
