Explanation

(d)
Given, $y = ae^{x} + bx\ e^{x} + cx^{2}e^{x}$ …(i)
On differentiating $w.r.t.x$, we get
$y^{'} = ae^{x} + b\left( xe^{x} + e^{x} \right) + c(x^{2}e^{x} + 2xe^{x})$
$\Rightarrow$ $y^{'} = ae^{x} + bxe^{x} + cx^{2}e^{x} + be^{x} + 2cxe^{x}$
$\Rightarrow$ $y^{'} = y + be^{x} + 2cxe^{x}$ …(ii)
Again, differentiating $w.r.t.x$, we get
$y^{''} = y^{'} + be^{x} + 2c(xe^{x} + e^{x})$
$\Rightarrow$ $y^{''} = y^{'} + be^{x} + 2cxe^{x} + 2ce^{x}$
$\Rightarrow$ $y^{''} = 2y^{'} - y + 2ce^{x}$ …(iii)
[from Eq. (ii)]
Again, differentiating $w.r.t.x$, we get
$y^{'''} = 2y^{''} - y^{'} + 2ce^{x}$
$\Rightarrow$ $y^{'''} = 2y^{''} - y^{'} + (y^{''} - 2y^{'} + y)$ [from eq. (iii)]
$\Rightarrow$ $y^{'''} - 3y^{''} + 3y^{'} - y = 0$