Explanation

(d)
Given that, $\frac{dy}{dx} + y = e^{- x}$
It is a linear differential equation, comparing with the standard equation
$\frac{dy}{dx} + Py = Q$
$\Rightarrow P = 1,\ Q = e^{- x}$
$\therefore IF = e^{\int_{}^{}{P\ dx}} = e^{x}$
$\therefore$ Required solution is
$ye^{x} = \int_{}^{}{e^{- x}\ e^{x}\ dx + c = \int_{}^{}{1\ dx + c}}$
$\Rightarrow ye^{x} = x + c$
At $x = 0,\ y = 0\ \therefore c = 0$
Hence, the required solution is
$ye^{x} = x\ \Rightarrow y = xe^{- x}$