$${{dy} \over {dx}} = {y \over x}{{(4{y^2} + 2{x^2})} \over {(3{y^2} + {x^2})}}$$

Put $$y = vx$$

$$ \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$

$$ \Rightarrow v + x{{dv} \over {dx}} = {{v(4{v^2} + 2)} \over {(3{v^2} + 1)}}$$

$$ \Rightarrow x{{dv} \over {dx}} = v\left( {{{(4{v^2} + 2 - 3{v^2} - 1)} \over {3{v^2} + 1}}} \right)$$

$$ \Rightarrow \int {(3{v^2} + 1){{dv} \over {{v^3} + v}} = \int {{{dx} \over x}} } $$

$$ \Rightarrow \ln |{v^3} + v| = \ln x + c$$

$$ \Rightarrow \ln \left| {{{\left( {{y \over x}} \right)}^3} + \left( {{y \over x}} \right)} \right| = \ln x + C$$

$$ \downarrow \,y(1) = 1$$

$$ \Rightarrow C = \ln 2$$

$$\therefore$$ for $$y(2)$$

$$\ln \left( {{{{y^3}} \over 8} + {y \over 2}} \right) = 2\ln 2 \Rightarrow {{{y^3}} \over 8} + {y \over 2} = 4$$

$$ \Rightarrow [y(2)] = 2$$

$$ \Rightarrow n = 3$$