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

Given,

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

Let $y = vx$

$⇒ \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

So, $\dfrac{dy}{dx} = \dfrac{y}{x}\dfrac{\left(4{y}^{2} + 2{x}^{2}\right)}{\left(3{y}^{2} + {x}^{2}\right)}$

$⇒ v + x\dfrac{dv}{dx} = \dfrac{v\left(4{v}^{2} + 2\right)}{\left(3{v}^{2} + 1\right)}$

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

$⇒ ∫ \left(3{v}^{2} + 1\right)\dfrac{dv}{{v}^{3} + v} = ∫ \dfrac{dx}{x}$

$⇒ \ln \left|{v}^{3} + v\right| = \ln x + c$

$⇒ \ln \left|{\left(\dfrac{y}{x}\right)}^{3} + \left(\dfrac{y}{x}\right)\right| = \ln x + c$

Given, $y\left(1\right) = 1$

$⇒ c = \ln 2$

So the equation becomes $\ln \left|{\left(\dfrac{y}{x}\right)}^{3} + \left(\dfrac{y}{x}\right)\right| = \ln x + \ln 2$

Now for $y\left(2\right)$

So putting $x = 2$ in $\ln \left|{\left(\dfrac{y}{x}\right)}^{3} + \left(\dfrac{y}{x}\right)\right| = \ln x + \ln 2$

We get, $\ln \left(\dfrac{{y}^{3}}{8} + \dfrac{y}{2}\right) = 2\ln 2 ⇒ \dfrac{{y}^{3}}{8} + \dfrac{y}{2} = 4$

$⇒ \left[y\left(2\right)\right] = 2$

So, $y\left(2\right) \in [ 2, 3 )$

So on comparing with $y\left(2\right) \in [ n - 1, n )$

$⇒ n = 3$