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

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