Explanation

(a)
Here, $\frac{dy}{dt} - \left( \ \frac{1}{1 + t}\ \right)y = \frac{1}{(1 + t)}$ and $y(0) = - 1$
Which represents linear differential equation of first order.
IF $= e^{\int_{}^{}{- \left( \ \frac{t}{1 + t}\ \right)dt}} = e^{- t + \log{|1 + t|}} = e^{- t}(1 + t)$
$\therefore$ Required solution is
$y(IF) = \lbrack\ \int_{}^{}{Q(IF)dt\ \rbrack + c}$
$\Rightarrow ye^{- t}(1 + t) = \int_{}^{}{\frac{1}{1 + t} \bullet e^{- t}(1 + t)dt + c = \int_{}^{}{e^{- t}dt + c}}$
$\Rightarrow ye^{- t}(1 + t) = - e^{- t} + c$
Since, $y(0) = - 1\ \Rightarrow \ - 1 \bullet e^{0}(1 + 0) = - e^{0} + c$
$\Rightarrow c = 0$
$\therefore y = - \frac{1}{(1 + t)}$ and $y(1) = - \frac{1}{2}$