Answer
$${y^2} - 2y = {t^2} + t + 3$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = \frac{{2t + 1}}{{2y - 2}},\,\,\,\,\,\,\,\,y\left( 0 \right) = - 1 \cr
& {\text{separating the variables}} \cr
& \left( {2y - 2} \right)dy = \left( {2t + 1} \right)dt \cr
& \cr
& {\text{integrating both sides of the equation}} \cr
& \int {\left( {2y - 2} \right)} dy = \int {\left( {2t + 1} \right)} dt \cr
& {y^2} - 2y = {t^2} + t + C \cr
& \cr
& {\text{use the initial condition }}y\left( 0 \right) = - 1 \cr
& {\left( { - 1} \right)^2} - 2\left( { - 1} \right) = {\left( 0 \right)^2} + \left( 0 \right) + C \cr
& 3 = C \cr
& \cr
& Then,{\text{ the particular solution of the differential equation is}} \cr
& {y^2} - 2y = {t^2} + t + 3 \cr} $$
