Answer
$$y = 6 - 6{e^{ - \frac{{{x^2}}}{2}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = x\left( {6 - y} \right),{\text{ }}\left( {0,0} \right) \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{6 - y}} = xdx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{1}{{6 - y}}dy} = \int x dx \cr
& - \ln \left| {6 - y} \right| = \frac{{{x^2}}}{2} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {0,0} \right) \cr
& - \ln \left| {6 - 0} \right| = \frac{{{{\left( 0 \right)}^2}}}{2} + C \cr
& - \ln 6 = C \cr
& C = \ln \left( {\frac{1}{6}} \right) \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& \ln \left| {6 - y} \right| = - \frac{{{x^2}}}{2} - \ln \left( {\frac{1}{6}} \right) \cr
& \ln \left| {6 - y} \right| = - \frac{{{x^2}}}{2} + \ln 6 \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| {6 - y} \right|}} = {e^{ - \frac{{{x^2}}}{2}}}{e^{\ln 6}} \cr
& 6 - y = 6{e^{ - \frac{{{x^2}}}{2}}} \cr
& y = 6 - 6{e^{ - \frac{{{x^2}}}{2}}} \cr
& \cr
& {\text{Graph}} \cr} $$
