Answer
See explanation
Work Step by Step
$(a)$ If $y = e^{4x}$ then $y' = 4e^{4x}$ and $ y'' = 16e^{4x}$
Substitute into the equation $ y'' - 8y' + 16y = 0 $
$16e^{4x} - 8(4e^{4x}) + 16( e^{4x}) = 0 $
$16e^{4x} - 32e^{4x} + 16 e^{4x} = 0 $
$0 = 0$ verified
If $ y = xe^{4x}$ then $y' = (4x+1)e^{4x}$ and $y'' = (16x+8)e^{4x}$
Substitute into the equation $ y'' - 8y' + 16y = 0 $
$(16x+8)e^{4x} - 8(4x+1)e^{4x}+16(xe^{4x} ) = 0$
$16xe^{4x}+8e^{4x} - 32xe^{4x}- 8e^{4x}+16xe^{4x} ) = 0$
$ 0 = 0$ Verified
$(b)$ If $y = c_1e^{4x} +c_2xe^{4x} $
then $y' = 4c_1e^{4x} +c_2(4x+1)e^{4x} $
and $y'' = 16c_1e^{4x} +c_2(16x+8)e^{4x},$
Substitute into the equation $ y'' - 8y' + 16y = 0 $
$(16c_1e^{4x} +c_2(16x+8)e^{4x}) - 8(4c_1e^{4x} +c_2(4x+1)e^{4x}) + 16( c_1e^{4x} +c_2xe^{4x}) = 0$
$16c_1e^{4x} + 16c_2xe^{4x} + 8c_2 e^{4x } - 32c_1e^{4x} - 32c_2 x e^{4x} - 8c_2e^{4x} + 16c_1e^{4x} + 16c_2xe^{4x} = 0$
$0 = 0$ Verified
