Let y=y(x) be the solution of the differential equation dy dx= 4y^3+2yx^2 3xy^2+x^3,y(1)=1. If for some n∈ N,y(2)∈ [n-1,n), then n is equal to .

Given,

$$\frac{dy}{dx}=\frac{y}{x}\frac{(4{y}^{2}+2{x}^{2})}{(3{y}^{2}+{x}^{2})}$$dxdy​=xy​(3y2+x2)(4y2+2x2)​

Let $y=vx$y=vx

$$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$$⇒dxdy​=v+xdxdv​

So, $\frac{dy}{dx}=\frac{y}{x}\frac{(4{y}^{2}+2{x}^{2})}{(3{y}^{2}+{x}^{2})}$dxdy​=xy​(3y2+x2)(4y2+2x2)​

$$\Rightarrow v+x\frac{dv}{dx}=\frac{v(4{v}^{2}+2)}{(3{v}^{2}+1)}$$⇒v+xdxdv​=(3v2+1)v(4v2+2)​

$$\Rightarrow x\frac{dv}{dx}=v(\frac{(4{v}^{2}+2-3{v}^{2}-1)}{3{v}^{2}+1})$$⇒xdxdv​=v(3v2+1(4v2+2−3v2−1)​)

$$\Rightarrow \int (3{v}^{2}+1)\frac{dv}{{v}^{3}+v}=\int \frac{dx}{x}$$⇒∫(3v2+1)v3+vdv​=∫xdx​

$$\Rightarrow \mathrm{ln}|{v}^{3}+v|=\mathrm{ln}x+c$$⇒ln∣v3+v∣=lnx+c

$$\Rightarrow \mathrm{ln}|{(\frac{y}{x})}^{3}+(\frac{y}{x})|=\mathrm{ln}x+c$$⇒ln∣(xy​)3+(xy​)∣=lnx+c

Given, $y(1)=1$y(1)=1

$$\Rightarrow c=\mathrm{ln}2$$⇒c=ln2

So the equation becomes $\mathrm{ln}|{(\frac{y}{x})}^{3}+(\frac{y}{x})|=\mathrm{ln}x+\mathrm{ln}2$ln∣(xy​)3+(xy​)∣=lnx+ln2

Now for $y(2)$y(2)

So putting $x=2$x=2 in $\mathrm{ln}|{(\frac{y}{x})}^{3}+(\frac{y}{x})|=\mathrm{ln}x+\mathrm{ln}2$ln∣(xy​)3+(xy​)∣=lnx+ln2

We get, $\mathrm{ln}(\frac{{y}^{3}}{8}+\frac{y}{2})=2\mathrm{ln}2\Rightarrow \frac{{y}^{3}}{8}+\frac{y}{2}=4$ln(8y3​+2y​)=2ln2⇒8y3​+2y​=4

$$\Rightarrow [y(2)]=2$$⇒[y(2)]=2

So, $y(2)\in [2,3)$y(2)∈[2,3)

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

$$\Rightarrow n=3$$⇒n=3