Question

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 $n \in N , y (2) \in[ n -1, n )$, then $n$ is equal to ______

Answer 1

Correct Answer: 3

The correct answer is 3
$\frac{dy}{dx}=\frac{4y^3+2y{x}^2}{3x{y}^2+x^3},y(1)=1$
$\frac{dy}{dx}=\frac{4(y/x{)}^3+2(y/x)}{3(y/x{)}^2+1}$
$y=xp$
$x\frac{dp}{dx}+p=\frac{4p^3+2p}{3p^2+1}$
$x\frac{dp}{dx}=\frac{p^3+p}{3p^2+1}$
$\int \frac{3p^2+1}{p^3+p}dp=\int \frac{dx}{x}$
$\ln \left(p^3+p\right)=\ln x+\ln C$
$p^3+p=xC$
${\left(\frac{y}{x}\right)}^3+\left(\frac{y}{x}\right)=xC$
$y^3+x^{2}y=x^{4}C$
$x=1,y=1$
$1+1=C\Rightarrow C=2$
$y^3+x^{2}y=2x^4$
Put $x=2$
$y^3+4y-32=0$
Having root between $2$ and $3$
$y(2)\in [2,3)$