Explanation

(c)
Given equation is $\frac{dy}{dx} - y = 1\ \Rightarrow \ \frac{dy}{1 + y} = dx$
On integrating both sides, we get
$\int_{}^{}{\frac{1}{1 + y}dy =}\int_{}^{}{dx}$
$\Rightarrow \log{(1 + y) = x + c}$
$\Rightarrow 1 + y = e^{x} \bullet e^{c}$ …(i)
At $x = 0,\ y = - 1$
Then $1 - 1 = e^{0} \bullet e^{c} \Rightarrow e^{c} = 0$
On putting the value of $e^{c}$ in Eq. (i).
Therefore, solution becomes
$1 + y = e^{x} \times 0\ \Rightarrow y(x) = - 1$