Answer
$$y = - 2.5{e^{ - 0.2{x^2}}} + 1$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = x{e^{ - 0.2{x^2}}},{\text{ }}\left( {0, - \frac{3}{2}} \right) \cr
& {\text{Separate the variables}} \cr
& dy = x{e^{ - 0.2{x^2}}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {x{e^{ - 0.2{x^2}}}} dx \cr
& y = \frac{1}{{ - 0.4}}\int {\left( { - 0.4x} \right){e^{ - 0.2{x^2}}}} dx \cr
& y = - 2.5{e^{ - 0.2{x^2}}} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {0, - \frac{3}{2}} \right) \cr
& - \frac{3}{2} = - 2.5{e^{ - 0.2{{\left( 0 \right)}^2}}} + C \cr
& - \frac{3}{2} = - 2.5 + C \cr
& C = 1 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = - 2.5{e^{ - 0.2{x^2}}} + 1 \cr
& \cr
& {\text{Graph}} \cr} $$
