Answer
$$y = - \frac{1}{3}{\left( {4 - {x^2}} \right)^{3/2}} + 2$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = x\sqrt {4 - {x^2}} \cr
& {\text{Separate the variables}} \cr
& dy = x\sqrt {4 - {x^2}} dx \cr
& {\text{Integrate both sides with respect to }}x \cr
& \int {dy} = \int {x\sqrt {4 - {x^2}} } dx \cr
& \int {dy} = - \frac{1}{2}\int {\sqrt {4 - {x^2}} } \left( { - 2x} \right)dx \cr
& {\text{Let }}u = 4 - {x^2},{\text{ }}du = - 2xdx \cr
& \int {dy} = - \frac{1}{2}\int {\overbrace {\sqrt {4 - {x^2}} }^{{u^{1/2}}}} \overbrace {\left( { - 2x} \right)dx}^{du} \cr
& y = - \frac{1}{2}\left[ {\frac{{{{\left( {4 - {x^2}} \right)}^{3/2}}}}{{3/2}}} \right] + C \cr
& y = - \frac{1}{3}{\left( {4 - {x^2}} \right)^{3/2}} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {2,2} \right) \cr
& 2 = - \frac{1}{3}{\left( {4 - {2^2}} \right)^{3/2}} + C \cr
& 2 = - \frac{1}{3}{\left( 0 \right)^{3/2}} + C \cr
& C = 2 \cr
& {\text{Substitute 2 for }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = - \frac{1}{3}{\left( {4 - {x^2}} \right)^{3/2}} + 2{\text{ }}\left( {{\text{Particular solution}}} \right) \cr} $$
