Let $y=y(x)$ be the solution of the differential equation

$\frac{d y}{d x}=\frac{4 y^{3}+2 y x^{2}}{3 x y^{2}+x^{3}}, y(1)=1$.

If for some $n \in \mathbb{N}, y(2) \in[n-1, n)$, then $n$ is equal to _____________.

<p>${{dy} \over {dx}} = {y \over x}{{(4{y^2} + 2{x^2})} \over {(3{y^2} + {x^2})}}$</p> <p>Put $y = vx$</p> <p>$\Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$</p> <p>$\Rightarrow v + x{{dv} \over {dx}} = {{v(4{v^2} + 2)} \over {(3{v^2} + 1)}}$</p> <p>$$ \Rightarrow x{{dv} \over {dx}} = v\left( {{{(4{v^2} + 2 - 3{v^2} - 1)} \over {3{v^2} + 1}}} \right)$$</p> <p>$\Rightarrow \int {(3{v^2} + 1){{dv} \over {{v^3} + v}} = \int {{{dx} \over x}} }$</p> <p>$\Rightarrow \ln |{v^3} + v| = \ln x + c$</p> <p>$$ \Rightarrow \ln \left| {{{\left( {{y \over x}} \right)}^3} + \left( {{y \over x}} \right)} \right| = \ln x + C$$</p> <p>$\downarrow \,y(1) = 1$</p> <p>$\Rightarrow C = \ln 2$</p> <p>$\therefore$ for $y(2)$</p> <p>$$\ln \left( {{{{y^3}} \over 8} + {y \over 2}} \right) = 2\ln 2 \Rightarrow {{{y^3}} \over 8} + {y \over 2} = 4$$</p> <p>$\Rightarrow [y(2)] = 2$</p> <p>$\Rightarrow n = 3$</p>