Answer
$$\eqalign{
& y = \frac{2}{{1 - \left( {1/3} \right){e^{{x^2}}}}}\left( {{\text{Particular solution 1}}} \right) \cr
& y = \frac{2}{{1 + {e^{{x^2}}}}}\left( {{\text{Particular solution 2}}} \right) \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + 2xy = x{y^2} \cr
& {\text{The differential equation is written in the form}} \cr
& y' + P\left( x \right)y = Q\left( x \right){y^n} \cr
& \frac{{dy}}{{dx}} + 2xy = x{y^2} \to P\left( x \right) = 2x,{\text{ }}Q\left( x \right) = x,{\text{ }}n = 2 \cr
& {\text{From the book we have that the general solution of a Bernoulli }} \cr
& {\text{equation is:}} \cr
& {y^{1 - n}}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = \int {\left( {1 - n} \right)Q\left( x \right){e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }}dx} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Find }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} \cr
& {e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = {e^{\int {\left( {1 - 2} \right)\left( {2x} \right)dx} }} = {e^{ - \int {2x} dx}} = {e^{ - {x^2}}} \cr
& {\text{Substituting }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }},{\text{ }}n,{\text{ and }}Q\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {y^{1 - 2}}{e^{ - {x^2}}} = \int {\left( {1 - 2} \right)\left( x \right){e^{ - {x^2}}}dx} + C{\text{ }} \cr
& {y^{ - 1}}{e^{ - {x^2}}} = - \int {x{e^{ - {x^2}}}dx} + C{\text{ }} \cr
& {\text{Integrating}} \cr
& {y^{ - 1}}{e^{ - {x^2}}} = \frac{1}{2}{e^{ - {x^2}}} + C \cr
& {\text{Solve for }}y \cr
& \frac{1}{y} = \frac{1}{2} + C{e^{{x^2}}} \cr
& \frac{1}{y} = \frac{{1 + 2C{e^{{x^2}}}}}{2} \cr
& y = \frac{2}{{1 + 2C{e^{{x^2}}}}} \cr
& {\text{Use the initial condition }}\left( {0,3} \right) \cr
& 3 = \frac{2}{{1 + 2C}} \to C = - \frac{1}{6} \cr
& y = \frac{2}{{1 - \left( {1/3} \right){e^{{x^2}}}}}\left( {{\text{Particular solution 1}}} \right) \cr
& {\text{Use the initial condition }}\left( {0,1} \right) \cr
& 1 = \frac{2}{{1 + 2C}} \to C = \frac{1}{2} \cr
& y = \frac{2}{{1 + {e^{{x^2}}}}}\left( {{\text{Particular solution 2}}} \right) \cr} $$
