Answer
$$y = 5{e^{3x - {x^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \left( {3 - 2x} \right)y;\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 5 \cr
& {\text{Separating variables leads to}} \cr
& \frac{{dy}}{y} = \left( {3 - 2x} \right)dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {\frac{1}{y}dy} = \int {\left( {3 - 2x} \right)dx} \cr
& {\text{integrating}} \cr
& \ln \left| y \right| = 3x - {x^2} + C\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = 5 \cr
& y\left( 0 \right) = 5{\text{ implies that }}y = 5{\text{ when }}x = 0 \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& \ln \left| 5 \right| = 3\left( 0 \right) - {\left( 0 \right)^2} + C \cr
& C = \ln 5 \cr
& {\text{substitute }}C = \ln 5{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& \ln \left| y \right| = 3x - {x^2} + \ln 5 \cr
& {\text{solve for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{3x - {x^2}}}{e^{\ln 5}} \cr
& y = 5{e^{3x - {x^2}}} \cr} $$
