Explanation

(a)
Given, $\frac{dy}{dt} - \left( \frac{1}{1 + t} \right)y = \frac{1}{(1 + t)}$ and $y(0) = - 1$
$\therefore$ $IF = e^{\int_{}^{}{- \left( \frac{t}{1 + t} \right)dt}} = e^{- \int_{}^{}{\left( 1 - \frac{1}{1 + t} \right)dt}}$
$e^{- t + \log{(1 + t)}} = e^{- t}(1 + t)$
$\therefore$ Required solution is,
$ye^{- t}(1 + t) = \int_{}^{}\frac{1}{1 + t}\ e^{- t}\ (1 + t)dt + c$
$= \int_{}^{}e^{- t}\ dt + c$
$\Rightarrow ye^{- 1}(1 + t) = - e^{- 1} + c$
Since, $y(0) = - 1$
$\Rightarrow$ $c = 0$
$\therefore$ $y = - \frac{1}{(1 + t)}$
$\Rightarrow$ $y(1) = - \frac{1}{2}$