Explanation

(d)
Given differential equation can be written as
$\frac{d^{2}y}{dx^{2}} - 3\ \frac{dy}{dx} + 2y = 0$
$\Rightarrow$ $\left( m^{2} - 3m + 2 \right)y = 0$
$\Rightarrow$ $(m - 1)(m - 2)y = 0$
$\Rightarrow$ $m = 1,2$
$\therefore$ Solution is $y = c_{1}e^{x} + c_{2}e^{2x}$
$y^{'} = c_{1}e^{x} + 2c_{2}e^{2x}$
From given condition
$y(0) = 1$
$\Rightarrow c_{1} + c_{2} = 1$…..(i)
And y’(0)=0
$\Rightarrow c_{1} + c_{2} = 1\ldots\ldots.(ii)$
On solving Eqs. (i) and (ii)we get
$- c_{2} = 1$
$\Rightarrow \ \ \ c_{2} = - 1$
And $c_{1} = 2$
$\therefore$y=$2e^{x} - e^{2x}$
$\therefore\ \ \ \ at\ x = \log_{e}2$
$y = 2e^{log2\ }–e^{2\log 2}$
$= 2 \times 2 - 2^{2} = 0$