If $$y\left( t \right)$$ is a solution $$\left( {1 + t} \right)\frac{{dy}}{{dt}} - ty = 1$$    and $$y\left( 0 \right) = - 1,$$   then $$y\left( 1 \right)$$ is equal to-

Solution :
The given differential equation is
$$\eqalign{ & \frac{{dy}}{{dt}} - \frac{t}{{1 + t}}y = \frac{1}{{1 + t}} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - \int {\frac{t}{{1\, + \,t}}\,dt} }} = {e^{ - \int {\left( {1\, - \,\frac{t}{{1\, + \,t}}} \right)\,dt} }} = {e^{ - \left( {t\, - \,\log \,\left( {1\, + \,t} \right)} \right)}} \cr & = {e^{ - t}}.{e^{\log \,\left( {1\, + \,t} \right)}} = \left( {1 + t} \right){e^{ - \,t}} \cr} $$
$$\therefore $$ Solution is
$$\eqalign{ & y.{e^{ - t}}\left( {1 + t} \right) = \int {\frac{1}{{\left( {1 + t} \right)}}{e^{ - t}}\left( {1 + t} \right)dt + C} \cr & \Rightarrow y.{e^{ - t}}\left( {1 + t} \right) = - {e^{ - t}} + C \cr & \Rightarrow y = - \frac{1}{{1 + t}} + \frac{{C{e^t}}}{{1 + t}} \cr & {\text{Given that }}y\left( 0 \right) = - 1 \cr & \Rightarrow - 1 = - 1 + C \cr & \Rightarrow C = 0 \cr & \therefore y = \frac{1}{{1 + t}} \cr & \therefore y\left( 1 \right) = - \frac{1}{{1 + 1}} = - \frac{1}{2} \cr} $$