Explanation

(a)
Given, $y = Ae^{x} + Be^{2x} + Ce^{3x}$ …(i)
$\Rightarrow$ $y^{'}Ae^{x} + 2Be^{2x} + 3Ce^{3x}$
From Eq. (i),
$Ae^{x} = y - Be^{2x} - Ce^{3x}$
$\ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ y^{'} = y + Be^{2x} - Ce^{3x}$
$\therefore$ $y^{''} = y' + Be^{2x} + 6Ce^{3x}$ …(ii)
From Eq. (ii),
$Be^{2x} = y^{'} - y - 2Ce^{3x}$
$\therefore$ $y^{''} = y^{'} + 2y^{'} - 2y - 4Ce^{3x} + 6Ce^{3x}$
$\Rightarrow$ $y^{''} = 3y^{'} - 2y + 2Ce^{3x}$ …(iii)
Again, differentiating $w.r.t.x$, we get
$y''' = {3y}^{''} - 2y^{'} + 6Ce^{3x}$
From Eq. (iii),
$2Ce^{3x} = y^{''} - 3y^{'} + 2y$
$\therefore$ $y^{'''} = 3y^{''} - 2y^{'} + 3(y^{''} - 3y^{'} + 2y)$
$\Rightarrow$ $y^{'''} - 6y^{''} + 11y^{'} - 6y = 0$