Answer
$$\frac{{{y^2}}}{2} - 3y = {x^2} + x - 4$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{2x + 1}}{{y - 3}};\,\,\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 4 \cr
& {\text{Separating variables leads to}} \cr
& \left( {y - 3} \right)dy = \left( {2x + 1} \right)dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {\left( {y - 3} \right)dy} = \int {\left( {2x + 1} \right)} dx \cr
& {\text{integrate by using the power rule}} \cr
& \frac{{{y^2}}}{2} - 3y = {x^2} + x + C\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 0 \right) = 1 \cr
& \,y\left( 0 \right) = 4{\text{ implies that }}y = 4{\text{ when }}x = 0 \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& \frac{{{{\left( 4 \right)}^2}}}{2} - 3\left( 4 \right) = {\left( 0 \right)^2} + \left( 0 \right) + C \cr
& - 4 = 0 + C \cr
& C = - 4 \cr
& {\text{substitute }}C = - 4{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& \frac{{{y^2}}}{2} - 3y = {x^2} + x - 4 \cr} $$
