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